Posted by: Shuanglin Shao | November 5, 2020

## Measure and Integration theory, Lecture 15

Theorem 6.17. (Chebyshev’s inequality.) If $f\in L^p$ for $0, then for any $\alpha>0$,

$\mu(\{|f(x)|>\alpha\}) \le \frac {\|f\|_p^p}{\alpha^p}.$

Proof. Let $E_\alpha=\{|f(x)|>\alpha\}$. Then

$\mu(E_\alpha) = \int_{E_\alpha} 1 d\mu \le \int_X \frac {|f(x)|^p}{\alpha^p} =\frac {\|f\|_p^p}{\alpha^p}.$

Theorem 6.18. (Schur’s test.) Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be two $\sigma$-finite measure spaces. $K$ is $\mathcal{M}\times \mathcal{N}$ measurable function on $X\times Y$. Suppose that there exists $c>0$ such that

$\int_X |K(x,y)|d\mu(x)\le C<\infty, a.e., y;$

and

$\int_Y |K(x,y)|d\nu(y)\le C<\infty, a.e. x.$

Let $1\le p\le \infty$. If $f\in L^p(\nu)$, then the integral

$Tf(x) = \int K(x,y) f(y) d\nu(y)$

converges absolutely for $a.e. x$, and the function $Tf$ thus defined is in $L^p(\mu)$ and

$\|Tf\|_p\le C \|f\|_p.$

Proof. Suppose that $1. Let $q$ be the conjugate exponent to $p$, i.e., $\frac 1p+\frac 1q=1$. We write

$|K(x,y) f(y)| = |K(x,y)|^{1/q} (|K(x,y)|^{1/p} |f(y)|)$.

By applying the Holder inequality,

$\int |K(x,y) f(y)| d\nu(y) \le \left(\int |K(x,y)|d\nu(y) \right)^{1/q}\left( \int |K(x,y)||f(y)|^p d\nu(y)\right)^{1/p}$

$\le C^{1/q}\left( \int |K(x,y)||f(y)|^p d\nu(y)\right)^{1/p}$.

By Tonelli’s theorem,

$\int \left( \int |K(x,y) f(y)| d\nu(y) \right)^p d\mu(x) \le C^{p/q} \int |K(x,y)||f(y)|^p d\nu(y) d\mu(x)$

$\le C^{p/q+1} \int |f(y)|^p d\nu(y)<\infty.$

Thus by Fubini’s theorem for the mixed exponents (in this case, $L^p(d\mu)\bigl(L^1(d\nu)$)\bigr), $K(x, \cdot )f \in L^1(\nu)$ for $a.e. x$ and so that $Tf$ is well defined $a.e.$, and

$\|Tf\|_p \le C \|f\|_p.$

The proofs for $p=1, \infty$ are similar.

Theorem 6.19. (Minkowski’s inequality. ) Suppose that $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be two measure spaces, and $f$ be an $\mathcal{M}\times \mathcal{N}$ measurable function on $X\times Y$.

(a). If $f\ge 0$ and $1\le p<\infty$, then

$\left( \int \bigl( \int f(x,y) d\nu(y) \bigr)^pd\mu(x) \right)^{1/p}\le \int \left( \int f(x,y)^p d\mu(x)\right)^p d\nu(y).$

(b). If $1\le p\le \infty$, $f(\cdot, y) \in L^p(\mu)$ for $a..e. y$, and the function $y\mapsto \|f(\cdot, y)\|_p$ is in $L^1(\nu)$, then $f(x,\cdot) \in L^1(\nu)$ for $a.e. x$, the function $x\mapsto\int f(x,y) d\nu(y)$ is $L^(\mu)$, and

$\| \int f(\cdot, y) d\nu(y)\|_p \le \int \|f(\cdot, y)\|_p d\nu(y).$

Proof. Step 1. We first prove (a). If $p=1$, (a) is just the Tonelli’s theorem.

If $1, let $q$ be the conjugate exponent to $p$ and suppose that $g\in L^q(\mu)$. Then by Tonelli’s theorem and Holder’s inequality,

$\int \left( \int f(x,y) d\nu(y) \right) |g(x)| d\mu(x) = \int \int f(x,y) |g(x) |d\mu(x)d\nu(y)$

$\le \|g\|_{L^q } \int \left( \int f(x,y)^p d\mu(x)\right)^p d\nu(y).$

So by Theorem 6.14, (a) follows.

Step 2. When $p<\infty$, (b) follows from (a) and the Fubini’s theorem.

When $p=\infty$, it is obvious.

Definition. If $f$ is a measurable function on $(X,\mathcal{M}, \mu)$, we define its distribution function $\lambda_f: (0,\infty) \to [0,\infty)$ by

$\lambda_f(\alpha)= \mu(\{|f(x)|>\alpha\}).$

Proposition 6.22. (a). $\lambda_f$ is decreasing and right continuous.

(b). If $|f|\le |g|$, then $\lambda_f \le \lambda_g$.

(c). If $|f_n|$ increases to $|f|$, then $\lambda_{f_n}$ increases to $\lambda_f$.

(d). If $f=g+h$, then $\lambda_f(\alpha) \le \lambda_g(\frac 12 \alpha)+\lambda_h (\frac 12 \alpha).$

Proof. (a). Let $a_n\to \alpha$, then

$\{|f(x)|>\alpha\} =\cup_{n\ge 1} \{ |f(x)|> a_n\}.$

And the latter sets are increasing. So (a) follows.

(b) is obvious.

(c). For any $\alpha>0$,

$\{|f(x)|>\alpha\} = \cup_{n\ge 1} \{|f_n(x)|>\alpha\}.$

(d). If $|f|>\alpha$, then $|g|>\frac 12 \alpha$ or $|h|>\frac 12 \alpha$.

Proposition 6.24. If $0, then

$\int |f|^pd\mu= p\int_0^\infty \alpha^{p-1} \lambda_f (\alpha) d\alpha$.

Proof. For any $x$,

$|f(x)|^p = p\int_0^{|f(x)|} \alpha^{p-1}d\alpha.$

Then

$\int |f(x)|^p d\mu(x)= \int_X \int_0^{|f(x)|} p\alpha^{p-1} d\alpha d\mu(x) =\int_X\int_0^\infty 1_{\{0<\alpha<|f(x)|\}} p\alpha^{p-1}d\alpha d\mu$

$= \int_0^\infty \int_X 1_{\{0<\alpha<|f(x)|\}} p\alpha^{p-1} d\mu(x) d\alpha = \int_0^\infty p\alpha^{p-1} \lambda_f(\alpha) d\alpha.$

Note that $1_{\{0<\alpha<|f(x)|\}} p\alpha^{p-1}$ is an $\mathcal{L}\times \mathcal{M}$ measurable function.

Proposition 6.23. If $\lambda_f(\alpha)<\infty$ for all $\alpha>0$, and $\phi$ is a nonnegative Borel measurable function on $(0,\infty)$, then

$\int_X \phi\circ |f| d\mu = - \int_0^\infty \phi(\alpha) d\lambda_f(\alpha).$

Proof. We note that $-\lambda_f$ is increasing and right continuous. The construction in Theorem 1.16 gives a nonnegative and unique Borel measure $\nu$ on $(0,\infty)$ such that

$\nu((a,b]) = \lambda_f(a)-\lambda_f(b)$

for all $a. $\nu$ is the Lebesgue-Stieltjes measure associated with $\lambda_f$.

When $\phi=1_{(a,b]}$,

$LHS= \int_{a<|f(x)|\le b} d\mu = \mu((a<|f(x)|\le b)) = \lambda_f(a) -\lambda_f(b)$.

$RHS = -\int_a^b d\lambda_f = \nu((a,b]) = \lambda_f(a)-\lambda_f(b)$.

This proves the equation for characteristic function of a $h$ interval. By the uniqueness in Theorem 1.14 and $\lambda_f(\alpha)<\infty$, the equation is true for a Borel set; so follows the simple functions. Then by Theorem 2.10, the general case follows by the monotone convergence theorem.

Theorem 6.24. Let $\nu$ be a measure on the Borel sets of the positive real line $[0,\infty)$ such that $\phi(t) = \nu([0,t))$ is finite for every $t>0$. Note that $\phi(0)=0$ and that $\phi,$ being monotone, is Borel measurable. Let $(\Omega, \Sigma, \mu)$ be a measure space and $f$ an nonnegative measurable function on $\Omega$. Then

$\int_\Omega \phi(f(x)) \mu(dx) = \int_0^\infty \mu(\{|f(x)|>t\})\nu(dt)$.

In particular, by choosing $v(dt) = pt^{p-1}dt$ for $p>0$, we have

$\int_\Omega f(x)^p \mu(dx) = p\int_0^\infty t^{p-1} \mu(\{|f(x)|>t\})dt.$

Proof. We write

$\int_0^\infty \mu(\{|f(x)|>t\})\nu(dt) = \int \int_\Omega 1_{|f|>t}(x)\mu(dx) \nu(dt).$

Since $1_{|f(x)|>t}$ is a measurable function, Fubini’s theorem gives

$\int_\Omega \left(\int_0^\infty 1_{|f(x)|>t} \nu(dt)\right) \mu(dx)$.

However,

$\int_0^\infty 1_{|f(x)|>t} \nu(dt) =\int_0^{|f(x)|} \nu(dt) = \phi(f(x)).$

Definition. If $f$ is a measurable function on $X$ and $0, we define

$[f]_p = \left( \sup_{\alpha>0} \alpha^p \lambda_f(\alpha)\right)^{1/p}.$

The weak $L^p$ is defined to be $\{f:\, f \text{ is measurable and } [f]_p<\infty\}.$

Remark. (a). $L^p\subset \text{ weak }L^p$ but not equal. Example: $f(x)= x^{-1/p}$ on $(0,\infty)$.

(b). $[\cdot]_p$ is not a norm because the triangle inequality fails. Example: $p=2$. $f(x)= \frac 1x, 0 and $g(x)=1, 0.

We compute that

$\alpha^2 \lambda_f(\alpha) =\begin{cases}\alpha^2, 0<\alpha<1, \\ 1 \alpha\ge 1 \end{cases}$.

So $[f]_2= \sup_{\alpha}\alpha^2 \lambda_f(\alpha)=1.$

and similarly for $\lambda_g$.

However $\alpha^2\lambda_{f+g}(\alpha) = \begin{cases} \alpha^2, 0<\alpha\le 2, \\ \frac {\alpha^2}{\alpha-1}, 2<\alpha. \end{cases}$

So $\sup_\alpha \lambda^2 \lambda_{f+g} (\alpha) =4.$ So $[f+g]_2 =4.$

Posted by: Shuanglin Shao | October 31, 2020

## Measure and integration theory, Lecture 14

Let $(X,\mathcal{M}, \mu)$ be a measure space and $0.

Definition. $L^p(X)=\{f:\, f \text{ is measurable and } \int_X |f(x)|^pd\mu(x)<\infty.\}$ The $L^p$ norm is defined to be $\|f\|_{L^p}:=(\int_X |f(x)|^pd\mu(x))^{1/p}.$ When $p=\infty$, $\|f\|_\infty:=\inf\{a>0:\,\mu(\{|f(x)|>a\})=0\}.$ Define $L^\infty\{ f \text{ is measurable and } \|f\|_\infty<\infty. \}$

Remark. (1). $\|f\|_\infty \in \{a>0:\,\mu(\{|f(x)|>a\})=0\}$. Indeed, there exists $a_n\in \{a>0:\,\mu(\{|f(x)|>a\})=0\}$ such that $a_n\to \|f\|_\infty$. Then

$\{|f(x)|>\|f\|_\infty \}=\cup_{n\ge 1} \{|f(x)|>a_n\}$. Since the latter sets are increasing,

$\mu(\{a>0:\,\mu(\{|f(x)|>\|f\|_\infty \})=0\}) = \lim_{n\to\infty} \mu(\{a>0:\,\mu(\{|f(x)|>a_n\})=0\})$.

(b). For $0, $L^p(X)$ is a vector space. Indeed, $|f(x)+g(x)|^p\le 2^p\max \{|f(x)|, |g(x)|\}^p\le 2^p (|f(x)|^p+|g(x)|^p)$. Then

$\int |f(x)+g(x)|^p \le 2^p( \int |f(x)|^p+\int |g(x)|^p$.

Therefore $f+g\in L^p$.

Theorem. $(L^p(X), \|\cdot\|)$ is a normed vector space.

It is easy to verify that $\|f\|_{p}=0$ iff $f=0, a.e.$ and for any $\alpha\in \mathbb{C}, \|\alpha f\|_p=|\alpha| \|f\|_p$. It remains to prove the triangle inequality, i.e.,

$\|f+g\|_p\le \|f\|_p+\|g\|+\|g\|_p$.

It relies on the Holder inequality.

Theorem (Holder’s inequality. )

Suppose that $p \in [1,\infty]$ and $p'$ is the conjugate exponent such that $\frac {1}{p'}+\frac 1p=1.$ Then

$\|fg\|_1\le \|f\|_p\|g\|_{p'}$.

The equality holds iff $\alpha|f|^p=\beta |g|^{p'}, a.e.$ for some $\alpha, \beta$.

Assuming that Holder inequality is true.

$|f+g|^p\le|f+g|^{p-1} |f|+ |f+g|^{p-1} |g|$. For $p, p'=\frac {p}{p-1}$,

$\int |f+g|^p\le (\int |f+g|^p)^{\frac {1}{p'}}\|g\|_p+ (\int |f+g|^p)^{\frac {1}{p'}}\|g\|_{p}$.

Then if $\|f+g\|_p\neq 0$,

$\|f+g\|_p \le \|f\|_p+\|g\|_p.$ This proves the triangle inequality.

To prove Holder’s inequality,

Lemma. For $u,v\ge 0$ and $\frac 1p+\frac 1{p'}=1$, $uv\le \frac {u^p}{p}+\frac {v^{p'}}{p'}.$ Equality holds iff $u^p={v}^{p'}$.

Proof. Writing $\frac {u^n}{p}=\int_0^n x^{n-1}dx$ for $n=p,p'$. Then by the geometric observation the rectangle with side lengths $u,v$ has an area less than the sum of two areas under $y=x^{p-1}$ with respect to the $x, y$ axises.

Having this lemma, we normalize $\|f\|_p=1, \|g\|_{p'}=1$. Let $u(x)=|f(x)|, v(x)=|g(x)|$. Then

$|f(x)g(x)|\le \frac {|f(x)|^p}{p}+\frac {|g(x)|^{p'}}{p'}$. Integration on both sides gives

$\int |f(x)g(x)|\le \frac {1}{p} \int |f(x)|^p+\frac {1}{p'} \int |g(x)|^{p'} =\frac 1p+\frac 1{p'}=1. (*)$

The equality in $(*)$ forces

$\int \left( \frac {|f(x)|^p}{p} +\frac {|g(x)|^{p'}} {p'} - |f(x)g(x)|\right)=0$. On the other hand, we know that $\frac {|f(x)|^p}{p} +\frac {|g(x)|^{p'}} {p'} - |f(x)g(x)|\ge 0$. Therefore

$\frac {|f(x)|^p}{p} +\frac {|g(x)|^{p'}} {p'} =|f(x)g(x)|, a.e., x.$ That the equality holds iff $|f(x)|^p=|g(x)|^{p'}$.

Theorem 6.6. For $1\le p\le \infty$, $L^p$ is a Banach space.

Proof. We just prove the case for $1\le p<\infty$. Let $\{f_n\}$ be a Cauchy sequence in $L^p$. We choose a subsequence $\{f_{n_k}\}$ such that

$\|f_{n_{k=1}}-f_{n_k}\|_{p}\le 2^{-k}$.

Then $G(x):=\sum_{k=1}^\infty |f_{n_{k=1}}(x)-f_{n_k}(x)| \in L^p$ by the monotone convergence theorem. Then for $a.e. x$,

$\sum_{k=1}^\infty | f_{n_{k+1}}-f_{n_k}|<\infty$. Let $E$ be the measure zero set associated with $G$. Then we define

$latex g(x) =\begin{cases} \sum_{k=1}^\infty f_{n_{k+1}}-f_{n_k}, x\notin E \\ 0, x\in E. \end{cases}$

Then $g$ is measurable and $g\in L^p$ because $|g|\le G$.

We make two observations. $f_{n_k}$ converges to $g+f_{n_1}$ in $L^p$. This follows from the dominated convergence theorem. Indeed, let $S_k=\sum_{k=1}^K (f_{n_{k+1}}-f_{n_k}) =f_{n_{k+1}}-f_{n_1}.$ Then $S_K\le G.$ Then by the dominated convergence theorem,

$\|f_{n_k} -(f_{n_1}+g)\|_p\to 0, \text{ as K} \to \infty.$

Since $\{f_n\}$ is Cauchy, $\|f_n-(f_{n_1}+g)\|_p\to 0, \text{ as } n\to \infty.$

Proposition 6.7. For $1\le p<\infty$, the set of simple functions $f=\sum_{j=1}^\infty a_j1_{E_j}$, where $\mu(E_j)<\infty$, is dense in $L^p$.

This follows from that any measurable function can be approximated by a sequence of simple functions by Theorem 2.10.

Proposition 6.9. If $0, then $L^q\subset L^p+L^r$.

Proof. If $f\in L^q$, let $E= \{|f(x)|>1\}$. Then $f1_{E} \in L^p, f1_{E^c} \in L^r.$

Proposition 6.10. If $0, then $L^p\cap L^r\subset L^q$ and $\|f\|_{q}\le \|f\|_p^\lambda \|g\|_{p'}^{1-\lambda}$, where $\lambda \in (0,1)$ such that

$\frac 1q= \frac {\lambda }{p}+ \frac {1-\lambda}{r}$.

Proof. Write

$\int |f|^q= \int |f|^{q\lambda} |f|^{q(1-\lambda)}.$

By applying Holder’s inequality,

$\le ( \int |f|^{p})^{\lambda} (\int |f|^r )^{1-\lambda}.$

Let $\frac 1p+\frac 1q=1$ and $g\in L^q$. Define

$\phi_g(f)= \int fg.$

Then $\phi_g$ is a linear functional on $L^p$. If we define the operator norm

$\|\phi_g\|:=\sup\{\frac {|\int fg |}{\|f\|_p}:\, f\in L^p, f\neq 0. \}$

Then the Holder inequality shows that $\|\phi_g\|\le \|g\|_q$.

Proposition 6.13. Suppose that $p,q$ are conjugate exponent and $1\le q<\infty$. If $g\in L^q$, then

$\|g\|_q=\|\phi_g\|.$

Proof. If $g\neq 0$ and $0, define

$f= \frac {|g|^{q-1}\overline{sgn g}}{\|g\|_q^{q-1}}$.

Then $\|f\|_p^p =\frac {\int |g|^{(q-1)p}}{\|g\|_q^{(q-1)p}} =\frac {\int |g|^q}{\int |g|^q} =1.$

So $\|\phi_g\|\ge \int fg = \frac {\int |g|^q}{\|g\|_q^{q-1}} =\|g\|_q.$

If $q=\infty$, for $\epsilon>0$, let $A=\{x:\, |g(x)|>\|g\|_\infty-\epsilon\}.$ Then $\mu(A)>0$. So if $\mu$ is $\sigma$ finite, then there exists $B\subset A$ such that $0<\mu(B)<\mu(A)$. Let

$f= \frac {1_B\overline{sgn(g)} }{\mu(B)}.$

Then $\|f\|_1=1$. So

$\|\phi_g\| \ge \int fg =\frac {1}{\mu(B)} \int_B |g| \ge \|g\|_\infty -\epsilon.$

Theorem 6.14. Let $\frac 1p+\frac 1q=1$ and $\mu$ is $\sigma$ finite. Suppose that $g$ is a measurable function on $X$ such that $fg\in L^1$ for all $f$ in the space $\Sigma$ of simple functions that vanish outside a set of finite measure, and the quantity

$M_q(g)=\sup\{ |\int fg|:\, f\in \Sigma \text{ and } \|f\|_p=1\}$

is finite. Also suppose that $S_g=\{g(x)\neq 0\}$ is $\sigma$ finite. Then $g\in L^q$ and $M_q(g)=\|g\|_q$.

Proof. We first show that if $f$ is a bounded measurable function that vanishes outside of a set $E$ with finite measure, then for given $g$,

$\int |fg|\le M_q(g).$

Indeed, by Theorem 2.10, for $f$, there exists a sequence of simple functions $f_n$ such that

$|f_1|\le |f_2|\le \cdots |f_n|\le \cdots \le |f|$

and $f_n\to f$ pointwise. Since $|f_n|\le |f|1_E$ and $1_E g\in L^1$ by assumptions, then the dominated convergence theorem yields

$|\int fg| = \lim |\int f_ng|\le M_q(g).$

Secondly, suppose that $0. We may assume that $S_g$ is $\sigma$ finite. Let $\{E_n\}$ be an increasing sequence of sets of finite measure such that

$|\phi_1|\le |\phi_2|\le \cdots |\phi_n|\le \cdots \le |g|$

and $\phi_n\to g$ pointwise. Let $g_n=\phi_n1_{E_n}$. Then

$|g_1|\le |g_2|\le \cdots |g_n|\le \cdots \le |g|$

and $g_n\to g$ pointwise and $g_n$ vanishes outside $E_n$. Let

$f_n=\frac {|g_n|^{q-1}\bar{sgn g}}{\|g_n\|_q^{q-1}}.$

Then we see that $\|f_n\|_p=1$. By Fatou’s lemma,

$\|g\|_q\le \liminf \|g_n\|_q =\liminf \int |f_ng_n| \le \liminf \int |f_ng|$

$=\liminf \int f_ng \le M_q(g)$

by the remark at the beginning of the proof. On the other hand, $M_q(g) \le \|g\|_q$, so the proof is complete for the case $q<\infty$.

Now suppose that $q=\infty$. Given $\epsilon>0$, let $A=\{|g(x)|\ge M_\infty +\epsilon. \}$ If $\mu(A)>0$, we choose $B\subset A$ such that $0<\mu(B)<\mu(A)$. Let

$f=\frac {1}{\mu(B)} \int_B |g| \ge M_\infty +\epsilon.$

A contradiction to the remark at the beginning of the proof. Hence $\|g\|_\infty \le M_\infty (g)$.

Theorem 6.15. Let $p,q$ be conjugate exponent. If $1, for each $\phi\in (L^p)^*$ there exists $g\in L^q$ such that $\phi(f) =\int fg$ for all $f\in L^p$, and hence $L^q$ is isometrically isomorphic to $(L^p)^*$. The same conclusion holds for $p=1$ if $\mu$ is $\sigma$ finite.

Proof. First let us suppose that $\mu$ is finite, so that all simple functions are in $L^p$. If $\phi\in (L^p)^*$ and $E$ is a measurable set, let

$\nu(E) =\phi(1_E)$.

For any disjoint sequence $\{E_j\}$, if $E=\cup_{j=1}^\infty E_j$, we have that

$1_E =\sum_{j=1}^\infty 1_{E_j}$.

The series $\{\sum_{j=1}^n 1_{E_j}\}_{n\ge 1}$ converges in the $L^p$ norm. Indeed,

$\|1_E - \sum_{j=1}^n 1_{E_j}\|_p = \|\sum_{j=n+1}^\infty 1_{E_j}\|_{p} = \mu(\cup_{j=n+1}^\infty E_j)^{1/p}\to 0,$ as $n\to \infty$.

Hence if $\phi$ is a linear and continuous, then

$\nu(E) =\sum_{j=1}^\infty \phi (1_{E_j}) =\sum_{j=1}^\infty \nu(E_j),$

so that $\nu$ is a complex measure. The absolute convergence part in the definition of complex measures is easy to verify as $\mu$ is finite and one can decompose $\phi$ into real and complex parts, and then the positive and negative parts.

If $\mu(E)=0$, then $1_E=0 \in L^p$. So $\nu(E) =\phi (1_E)$. So $\nu\ll \mu$. By the Radon-Nikodym theorem, there exists $g\in L^1(\mu)$ such that $\phi(1_E)=\nu(E) =\int_E gd\mu$ for all $E$ and hence $\phi(f) =\int fg d\mu$ for all simple functions $f$. Moreover $|\int fg | \le \|\phi\|\|f\|_p$. So $g\in L^q$ by Theorem 6.14. By Proposition 6.7, the set of simple functions with finite measure sets is dense in $L^p$. Therefor $\phi(f) =\int fg$ for all $f\in L^p$.

Now suppose that $\mu$ is $\sigma$ finite. Write $X=\cup_{n\ge 1} E_n$ with increasing $\{E_n\}$ of finite measures. For each $n$, there exists $g_n\in L^q(E_n)$ such that for all $f\in L^p(E_n)$,

$\phi(f) = \int_{E_n} fg_n.$

By Proposition 6.13, $\|g_n\|_q\le \le \|\phi\|.$

From the preceding proof, $g_n=g_m, a.e.$ on $E_n$ and $m>n$. We define $g=g_n, a.e., \text{ on } E_n$. Then $|g|$ is the increasing limit of $g_n$ and so by monotone convergence theorem, $g\in L^p$ and $\|g\|_p\le \|\phi\|$. Moreover, $f1_{E_n} \to f$ in $L^p$, and hence

$\phi(f) = \lim \phi(f1_{E_n}) = \lim \int_{E_n} fg =\int fg.$

Posted by: Shuanglin Shao | October 18, 2020

## Measure and Integration theory, Lecture 13

Lemma 3.15. (Vitali type covering lemma. ) Let $\mathcal{C}$ be a collection of open balls in $\mathbb{R}^n$, and let $U = \cup_{B\in \mathcal{C}} B$. If $c, there exist disjoint $B_1, \cdots, B_k \in \mathcal{C}$ such that $\sum_{j=1}^k m(B_j)>3^{-n }c.$

Proof. The set $U$ is open, then for $c there exists a compact set $K$ such that

$K\subset U, c.

Since $K$ is compact, there exists finitely many balls $B_1, \cdots, B_n$ such that $K\subset \cup_{i=1}^n B_i$. Let $\mathcal{C}_1$ be the collection of the balls. Choose $B'_1 \in \mathcal{C}_1$ that has the largest radius. Let $\mathcal{C}_2$ be the collection of balls from $\mathcal{C}_1$ that is disjoint from $B'_1$. Choose the ball $B'_2 \in \mathcal{C}_2$ that has the largest radius. In general, suppose that $B'_1, B'_2, \cdots, B'_k, \mathcal{C}_1, \cdots, \mathcal{C}_k$ are choosen. Let $\mathcal{C}_{k+1}$ be the collection of balls in $\mathcal{C}_k$ that are disjoint from the previous $k$ balls $B'_1, \cdots, B'_k$. Choose $B'_{k+1}\in \mathcal{C}_{k+1}$ that has the largest radius. Since there are only $n$ balls, the process will stop. We assume that $B'_1, \cdots, B'_m$ are the finally chosen balls. Given any ball $B\in \mathcal{C}_1$, we claim that $B$ will intersect with some ball from $\{B'_j\}_{j=1}^m$. Otherwise, one more ball will be chosen. A contradiction. Hence

$\mathcal{C}_1 \subset \cup_{j=1}^m 3 B'_j.$

So $c.

Definition. A measurable function $f: \mathbb{R}^n\to \mathbb{C}$ is called locally integrable with respect to the Lebesgue measure if $\int_K |f(x)|dx <\infty$ for every bounded measurable set $K\subset \mathbb{R}^n.$ The space of such functions is denoted by $L^1_{loc}.$

Definition. If $f\in L^1_{loc}, x\in \mathbb{R}^n, r>0$, we define

$A_rf(x) = \frac {1}{m(B(x,r))} \int_{B(x,r)}f(y)dy$.

Lemma 3.16. If $f\in L^1_{loc}$, $A_rf(x)$ is jointly continuous in $r, x$.

Proof. Given $(x_0, r_0)$. Since $f\in L^1_{loc}$, $f\in L^1(B(x_0, r_0+1))$. By Corollary 3.6, given $\epsilon>0$, there exists $\delta>0$ such that if $m(E)<\delta$,

$\int_{E \cap B(x_0, r_0+1)} |f|<\epsilon.$

So given $\delta>0$, if $x\to x_0, r\to r_0$,

$m(B(x,r)\Delta B(x_0,r_0)) <\delta$.

So $| \int_{B(x,r)} f-\int_{B(x_0,r_0)}f |\le \int_{B(x,r)\Delta B(x_0,r_0)} |f|<\epsilon.$ So

$\int_{B(x,r)} f \to \int_{B(x_0,r_0)}f.$

Hence as $x\to x_0, r\to r_0$,

$\frac {1}{m(B(x,r))}\int_{B(x,r)} f \to \frac {1} {m(B(x_0,r_0)} \int_{B(x_0,r_0)}f.$

Definition. If $f\in L^1_{loc}$, the Hardy-Littlewood maximal function $H(f)$ by

$Hf(x) =\sup_{r>0} A_r|f|(x)= \sup_{r>0} \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)|dy.$

Remark. The function $H(f)$ is measurable for

$(Hf)^{-1} ((a,\infty))= \cup_{r>0} (A_r|f|)^{-1}(a,\infty)$

is open for any $a\in \mathbb{R}$.

Theorem 3.17. There is a constant $C>0$ such that for any $f\in L^1$ and $\alpha>0$,

$m(\{x:\, Hf(x)>\alpha \}) \le \frac {C}{\alpha} \int |f(x)|dx$.

Proof. Let $E_\alpha= \{x:\, Hf(x)>\alpha \}$. For $x\in E_\alpha$, there exists $B(x,r)$ such that

$\frac {1}{m(B(x,r))}\int_{B(x,r)} |f|>\alpha,\, m(B(x,r))\le \frac {1}{\alpha}\int_{B(x,r)} |f|$.

Let $U=\cup_{x\in E_\alpha} B(x,r)$. For any $c, by Vitali type lemma 3.15, there exists disjoint balls $B_1,\cdots, B_n$,

$c<3^{n}\sum_{j=1}^n m(B_j) \le 3^{n} \frac {1}{\alpha}\sum_{j=1}^n \int_{B_j} |f| \le \frac {3^{n}}{\alpha} \int_{\cup B_j} |f| \le \frac {3^{n}}{\alpha} \int |f|.$

Note that the right hand side does not depend on $c$. Let $c\to m(U)$, we prove that

$m(E_\alpha)\le m(U) \le \frac {3^{-n}}{\alpha} \int |f|.$

This proves Theorem 3.17.

We shall use the notion of limit superior for real valued functions of a real variable.

$\limsup_{r\to R} \phi(r) = \lim_{\epsilon\to 0} \sup_{0<|r-R|<\epsilon} \phi(r) = \inf_{\epsilon>0}\sup_{0<|r-R|<\epsilon} \phi(r).$

So $\lim_{r\to R} \phi(r) =c$ is equivalent to $\limsup_{r\to R} |\phi(r)-c|=0.$

Theorem 3.18. If $f\in L^1_{loc}$, then $\lim_{r\to 0} A_rf(x) =f(x), a.e. x$.

Proof. Since we are investigating continuity of a funtion at a point, it is a local property. So we may assume that $f\in L^1$. Secondly, it is easy to see that the claim is true for a continuous function.

Now given $\epsilon>0$, there exists a continuous function $g$ such that $\|f-g\|_{L^1}<\epsilon.$ Consider

$\limsup_{r\to 0} |A_rf(x)-f(x)| =\limsup_{r\to 0} |A_r(f-g)(x)+A_r(g)(x)-g(x)+g(x)-f(x)|$

$\le \limsup_{r\to 0} |A_r(f-g)(x) | + |g(x)-f(x)|.$

Hence $\limsup_{r\to 0} |A_rf(x)-f(x)| \le H(f-g)(x)+ |f(x)-g(x)|$

Given any $\alpha>0$,

$E_\alpha=\{\limsup_{r\to 0} |A_rf(x)-f(x)| >\alpha\} \subset \{ H(f-g)>\frac \alpha2\}\cup \{|f-g|>\frac \alpha 2\}.$

By the maximal theorem,

$m(E_\alpha) \le \frac {2C\|f-g\|_{L^1}}{\alpha} + \frac {2\|f-g\|_{L^1}}{\alpha} \le \frac {2(C+1)\epsilon}{\alpha}.$

Let $\epsilon\to 0$, we see that $m(E_\alpha)=0$. We rationalize $\alpha$, we see that

$m( \{\limsup_{r\to 0} |A_rf(x)-f(x)| > 0 \}) =0.$

Definition. For any $f\in L^1_{loc}$, define the Lebesgue set $L_f$ of $f$ to be

$L_f=\{ x:\, \lim_{r\to 0} \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-f(x)|dy =0\}$.

Theorem. If $f\in L^1_{loc}$, then $m((L_f)^c) =0.$

Proof. For any $a\in \mathbb{Q}$, $|f-a|\in L^1_{loc}$. Then for $a.e. x$,

$\lim_{r\to 0} \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-a|dy =|f(x)-a|.$

Let $E_a=\{x:\,\lim_{r\to 0} \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-a|dy =|f(x)-a| \}$. Then $m(E_a^c)=0.$

For any $x\in E:=\cap_{\mathbb{Q}} E_a$, there exists $a_n\in \mathbb{Q}$ such that $a_n\to f(x)$,

$\frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-f(x)|dy \le \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-a_n|dy+ |f(x)-a_n|.$

So $\limsup_{r\to 0} \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-f(x)|dy\le |f(x)-a_n|.$

Let $n\to \infty$,

$\lim_{r\to 0} \frac {1}{m(B(x,r))} \int_{B(x,r)} |f(y)-f(x)|dy=0.$

Thus $E\subset L_f$. Since $m(E^c)=0$, $m(L_f^c)=0.$

Definition. A Borel measure $\nu$ is called “regular" if (i). $\nu(K)<\infty$ for every compact $K$. (ii). $\nu(E) =\inf\{\nu(U):\, U \text{ open }, E\subset U \}$, for any $E\in \mathcal{B}_{\mathbb{R}^n}$. A signed measure or complex measure $\nu$ is called “regular” if $|\nu|$ is regular.

Theorem 3.22. Let $\nu$ be a regular signed or complex Borel measure on $\mathbb{R}^n$, and $\nu= \lambda +fdm$ be its Lebesgue-Radon-Nikodym representation. Then for $m a.e. x\in \mathbb{R}^n$,

$\lim_{r\to 0} \frac {\nu(B(x,r))}{m(B(x,r))}=f(x)$.

Proof. We make two observations. (i). If $\nu$ is regular, so are $\lambda$. (ii). We may assume that $\lambda>0$, because $|\lambda(B(x,r))| \le |\lambda|(B(x,r))$.

We know that $\lambda, m$ are mutually singular. Let $A$ be a Borel set such that $\lambda(A) = m(A^c) =0.$ Let

$F_k =\{ x\in A:\, \limsup_{r\to 0} \frac {\lambda(B(x,r))}{m(B(x,r))} >\frac 1k\}$. We will show that $m(F_k)=0$ for all $k$, which will complete the proof with the aid of the Lebesgue-Radon-Nikodym theorem.

By the regularity of $\lambda$, for any $\epsilon>0$, there exists open set $U_\epsilon$ such that $A\subset U_\epsilon$ and $\lambda(U_\epsilon)<\epsilon.$ For any $x\in F_k$, there exists $B(x,r)\subset U_\epsilon$ such that $\lambda(B(x,r)) >k^{-1} m(B(x,r))$. Let $V_\epsilon= \cup_{x\in F_k} B(x,r)$ and $c, there exists $x_1, \cdots, x_J$ and disjoint balls $B(x_1, r_1), \cdots, B(x_J, r_J)$ such that

$c<3^n \sum_{j=1}^J m(B(x_j, r_j))\le 3^n k \sum_{j=1}^J \lambda(B(x_j, r_j))$

$\le 3^n k\lambda(V_\epsilon) \le 3^n k\lambda (U_\epsilon) \le 3^n k\epsilon.$

Let $c\to m(V_\epsilon)$. We see that $m(V_\epsilon)\le 3^n k\epsilon$. Since $F_k \subset V_\epsilon$, $m(F_k) \le 3^n k\epsilon$. So $m(F_k)=0.$

Theorem 3.23. Let $F: \mathbb{R}\to \mathbb{R}$ be increasing, and let $G(x) = F(x+)$.

(a). The set of points at which $F$ is discontinuous is countable.

(b). $F$ and $G$ are differentiable $a.e.$, and $F'=G', a.e.$

Proof. (a). Let $A:=\{x:\, F(x+)-F(x-)>0\}$. It is clear that $A$ is the set of points at which $F$ is discontinuous. Since $F$ is increasing, for $x,

$(F(x-), F(x+))\cap (F(y-), F(y+)) = \emptyset.$

Also for $x\in A$, the interval $(F(x-), F(x+))$ contains a rational number. So $A$ is at most countable.

(b). We observe that $G$ is increasing and right continous. $G=F$ except perhaps where $F$ is discontinuous. Moreover

$G(x+h)-G(x) = \begin{cases} \mu_G((x,x+h]), \text{ if }h>0, \\ -\mu_G((x+h,x]), \text{ if }h<0\end{cases}.$

We know that $\mu_G$ is regular. So by Theorem 3.22, $G'$ exists $a.e.$.

Let $H=G-F$. We need to show that $H'$ exists and equals zero $a.e.$ Let $\{x_j\}$ be an enumeration of points at which $H\neq 0$, i.e., $H(x_j)>0, \forall j$. Given any $N\in \mathbb{N}$,

$\sum_{j:\, |x_j|

Let $\mu=\sum_j H(x_j) \delta_{x_j}$. Then $\mu$ is a Borel measure that is finite on compact sets by $(*)$. Then $\mu$ is a Lebesgue-Stieltjes measure.

We know that

$|\frac {H(x+h)-H(x)}{h}|\le \frac {H(x+h)+H(x)}{|h|} \le 4 \frac {\mu(x-2|h|, x+2|h|)}{4|h|}$. This goes to zero as $h\to 0$ for $a.e. x$ by Theorem 3.22. Indeed, let $E= \{x_j\}$. Then $m(E)=0$ since $E$ is at most countable; $\mu(E^c)=0$. So $\mu\perp m$. Thus $H'$ exists $a.e.$, and $H'=0, a.e.$

Definition. If $F: \mathbb{R}\to \mathbb{C}$ and $x\in \mathbb{R}$, we define

$T_F(x)=\sup\{\sum_{j=1}^n |F(x_j)-F(x_{j-1})|:\, n\in \mathbb{N}, -\infty

$T_F$ is called the total variation of $F$.

Remark. It is easy to observe that for $a<b$,

$T_F(b) = T_F(a) +\sup\{\sum_{j=1}^n |F(x_j)-F(x_{j-1})|:\, n\in \mathbb{N}, a=x_0

Thus $T_F$ is an increasing function with values in $[0,\infty]$.

Definition. (1). If $T_F(\infty)<\infty$, $F$ is said to be of bounded variation on $\mathbb{R}$. We denote the space of all such functions by $BV$.

(2). The supremum in $(*)$ is called the total variation of $F$ on $[a,b]$. If it is finite, then we say that $F\in BV([a,b])$.

Examples. (1). If $F:\mathbb{R}\to \mathbb{R}$ is bounded and increasing, then $F\in BV.$

(2). If $F, G\in BV$ and $a, b\in \mathbb{C}$, then $aF+bG\in BV.$

(3). If $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded, then $F\in BV([a,b])$. Hint: using the mean value theorem.

(4). If $F=\sin x$, then $F\in BV([a,b])$ for $-\infty, but $F\notin BV.$ One can see it by taking $x_{n} = n\pi+\frac \pi 2, n \in \mathbb{N}$.

(5). If $F(x)= x\sin x^{-1}$ for $x\neq 0$ and $F(0)=0$, then $F\notin BV([a,b])$ for $a<0. One can see it by taking $x_{n}= \frac {1}{n\pi+\pi/2}$.

Lemma 3.26. If $F\in BV$ and is bounded, then $T_F\pm F$ is increasing.

Proof. If $x, then by $(*)$,

$T_F(y)-T_F(x)\ge |F(x)-F(y)|$.

Theorem 3.27. (1). $F\in BV$ iff $Re F, Im F \in BV$.

(2). If $F: \mathbb{R}\to \mathbb{R}$, then $F\in BV$ iff $F$ is the difference of two bounded increasing functions. For the $\Rightarrow$ direction, we take $\frac 12 (T_F\pm F)$.

(3). If $F\in BV$, then $F(x+) =\lim_{y\to x+} F(y), F(x-)=\lim_{y\to x-} F(y)$ exist for all $x\in \mathbb{R}$. This follows from (1) and (2).

(4). If $F\in BV$, the set of points at which $F$ is discontinuous is countable. This follows from (1) and (2).

(5). If $F\in BV$ and $G(x)=F(x+)$, then $F', G'$ exist and are equal a.e. This follows from (1), (2) and Theorem 3.23.

Definition. NBV=\{F\in BV:\, F \text{ is right continuous and } F(-\infty) =0\}.

Remark. If $F\in BV$, then the function $G$ defined by $G(x) = F(x+)-F(-\infty)$ is in NBV and $G'=F', a.e.$. Indeed, $F\in BV$; so $F= F_1-F_2$ the difference of two increasing functions. $G(x) = F_1(x+)-( F_2(x+) +F(-\infty))$ is again the difference of two increasing functions. So $G\in BV$. $G(-\infty) =0$ is obvious.

Theorem 3.28. If $F\in BV$, then $T_F(-\infty) =0$. If $F$ is also right continuous, so is $T_F$.

Proof. Given $\epsilon>0$ and $x\in \mathbb{R}$, then by the definition of $T_F$, there exists $x_0 such that

$\sum_{j=1}^n |F(x_j)-F(x_{j-1})| \ge T_F(x) -\epsilon.$

So by $(*)$,

$T_F(x) -T_F(x_0) \ge T_F(x) -\epsilon$. This proves that $T_F(x_0)\le \epsilon.$ Since $T_F$ is increasing, $T_F(y)\le \epsilon$ for all $y\le x_0$. So $T_F(-\infty) =0$.

For the second claim, we define $\alpha= T_F(x+)-T_F(x)$ and assume that $\alpha>0$. For any $\epsilon>0$, since $F$ is right continuous and $T_F(x+)$ is defined, there exists $\delta>0$ such that for $0,

$|F(x+h)-F(x)|<\epsilon, T_F(x+h)-T_F(x+)<\epsilon.$

For any such $h$, there exists $x_0=x such that

$\sum_{j=1}^n |F(x_j)-F(x_{j-1})|\ge \frac 34 |T_F(x+h)-T_F(x)|\ge \frac 34 \alpha$, and so

$\sum_{j=2}^n |F(x_j)-F(x_{j-1})| \ge \frac 34 \alpha -|F(x_1)-F(x_0)|\ge \frac 34 \alpha -\epsilon.$

Similarly on the interval $[x,x_1]$, we apply the same reasoning to conclude that there exists $t_0=x so that

$\sum_{j=1}^m |F(t_j)-F(t_{j-1})| \ge \frac 34 \alpha$.

So on $[x, x+h]$,

$T_F(x+h)-T_F(x) \ge \sum_{j=2}^n |F(x_j)-F(x_{j-1})|+ \sum_{j=1}^m |F(t_j)-F(t_{j-1})|\ge \frac 32 \alpha-\epsilon.$

On the other hand, $T_F(x+h)-T_F(x) <\alpha+\epsilon.$

So that $\alpha<4\epsilon$. Since $\epsilon$ is arbitrary, a contradiction. Therefore $\alpha=0$.

Theorem 3.29. If $\mu$ is a complex measure on $\mathbb{R}$ and $F(x) = \mu((-\infty, x])$, then $F\in NBV$. Conversely if $F\in NBV$, there is a unique complex Borel measure $\mu_F$ such that $F(x) = \mu_F((-\infty, x])$; moreover $|\mu_F| = \mu_{T_F}$.

Proof. Suppose that $\mu$ is a complex measure. By decomposing into real and complex parts, and then considering the positive and negative parts of measures,

$\mu= \mu_1^+-\mu_1^1+i(\mu_2^+-\mu_2^-)$. Since $\mu$ is a complex measure, $\mu^\pm_j, j=1,2$ are finite positive measures. Let $F_j^\pm = \mu^\pm_j ((-\infty,x])$. Then $F^\pm_j$ are right continuous, increasing functions. Also $F_j^\pm(-\infty) =0$ and $F_j^\pm (\infty) <\infty$. By Theorem 3.27 (1) and (2), $F=F_1^+-F_1^-+i(F_2^+-F_2^-) \in NBV.$

Conversely, for $F$, we write $F=F_1^+-F_1^-+i(F_2^+-F_2^-)$ by Theorem 3.27 (1) and (2). Then by Theorem 3.27, each $F_j^\pm, j=1,2$ is bounded and increasing. Then $F_j^\pm\in BV$. By Theorem 3.28, $F$ is right continuous and so $T_F$ is right continuous. So each $F_j^\pm$ is right continuous by Theorem 3.28. Also it is easy to see that each $F_j^\pm(-\infty) =0$ by Theorem 3.28 again. So by Theorem 1.16, $F_j^\pm(x)= \mu_{F_j^\pm}((-\infty, x])$ for some finite Borel measure $\mu_{F_j^\pm}$. Let $\mu= \mu_1^+-\mu_1^-+i(\mu_2^+-\mu_2^-)$.

The last step is to show that $|\mu_F| = \mu_{T_F}$. It is contained in Exercise 28 and 21 in Folland’s book.

Theorem 3.30. If $F\in NBV$, then $F'\in L^1(m)$. Morevoer $\mu_F\perp m$ iff $F'=0, a.e.$, and $\mu_F\ll m$ iff $F(x)=\int_{-\infty}^x F'(t)dt.$

Proof. Since $F\in NBV$, then $\mu_F$ in Theorem 3.29 is a complex Borel measure that is also regular. Then by the Radon-Nikodym theorem,

$\mu_F= \lambda+ F'dm$, where $F'=\lim_{r\to 0} \frac {\mu_F(E_r)}{m(E_r)} \in L^1$, where $E_r=(x,x+r], \text{ or } (x-r, x]$ by Theorem 3.22. The rest conclusions follows easily from the Radon-Nikodym representation above.

Proposition 3.32. If $F\in NBV$, then $F$ is absolutely continuous iff $\mu_F\ll m$.

Proof. $\Leftarrow.$ If $\mu_F\ll m$, then by Theorem 3.5, we see that the claim holds.

$\Rightarrow.$ We need to prove that for $E\in \mathcal{B}_\mathbb{R}$, if $m(E)=0$, then $\mu(E)=0$. Since $F$ is absolutely continuous, then for any $\epsilon>0$, there exists $\delta>0$ such that for any finite disjoint intervals $(a_j,b_j), 1\le j\le N$, then if $\sum_{j=1}^N (b_j-a_j)<\delta$,

$\sum_{j=1}^N |F(b_j)-F(a_j)|<\epsilon.$

Since $m(E)=0$, there exists a sequence of open sets $U_j$ such that $U_1 \supset U_2 \supset \cdots \supset U_n \cdots$, $m(U_j)<\delta. On the other hand, since$latex \mu_F$is regular, we can find a sequence of open sets $U'_j$ such that $U'_1\supset U'_2 \supset \cdots\supset U'_n \cdots$ such that $|\mu_F|(U'_j) \to |\mu_F|(E)$. That is to say, $|\mu_F(U_j\setminus E)| \le |\mu_F|(U'_j\setminus E) \to 0$ as $j\to \infty$. Therefore $U_j\cap U'_j$ is a sequence of open sets that are decreasing and contains $E$, and moreover $\mu_F(U_j\cap U'_j) \to \mu_F(E)$. We abbreviate it by $U_j$. Each $U_j$ is an at most countable union of disjoint open intervals $(a_j^k, b_j^k)$. For any $N$, $\sum_{j=1}^N |\mu_F(a_j^k,b_j^k)| \le \sum_{j=1}^N |F(b_j^k)-F(a_j^k)| <\epsilon.$ So $| \mu_F(U_j)|\le \epsilon.$ Since $\mu_F(U_j)\to \mu(E)$, we see that $\mu_F(E) \le \epsilon.$ Since $\epsilon$ is arbitrary, $\mu_F(E) =0$. Corollary 3.33. If $f\in L^1(m)$, then the function $F(x) = \int_{-\infty}^x f(t)dt$ is in NBV and is absolutely continuous, and $f=F', a.e.$ Conversely if $F\in NBV$ is absolutely continuous, then $F'\in L^1(m)$ and $F(x) = \int_{-\infty}^x F'(t)dt$. Proof. Suppose that $f\in L^1(m)$. Then $F \in BV$, right continuous and $F(-\infty) =0$ by dominated convergence theorem. So $F\in NBV.$ That $F$ is absolutely continuous follows from Corollary 3.6. By Theorem 3.32, the deduced measure $\mu_F\ll m$ and also $\mu_F ((-\infty, x) =F(x)$. $\mu_F = F'dm$ by Theorem 3.30. On the other hand, $fdm, f\in L^1$ gives rise to the same function $F$. By the uniqueness in Theorem 3.29,$fdm$and$F’dm$are the same complex Borel measures. Hence $f=F' a.e.$ Conversely, if $F\in NBV$ and is absolutely continuous, then $\mu_F\ll m$ by Theorem 3.32. Hence by Theorem 3.30, $F(x)=\int_{-\infty}^x F'(x)dx.$ Considering the real and imaginary parts of complex valued functions, and positive and negative parts of real-valued functions, we see that $F'\in L^1$ because $\mu_F$ is a complex Borel measure. Lemma 2.34 If $F$ is absolutely continuous on $[a,b]$, then $F\in BV([a,b])$. Proof. We know that $F$ is absolutely continuous. Let $\epsilon>0$ be given. There exists $\delta>0$ such that for disjoint intervals $(a_i,b_i), 1\le i\le n$ with $\sum_{i=1}^n (b_i-a_i)<\delta$, then $\sum_{i=1}^n |F(b_i)-F(a_i)|<\epsilon.$ Let$N$be the smallest integer that is larger than $\frac {b-a}{\delta}$. We divide $[a,b]$ into $N$ consecutive segments with length that is at most $\delta$. For any points $x_0=a , if necessary adding more endpoints of the previous consecutive segments , so that these points can be grouped into $N$ subgroups. On each subgroup, if the points are denoted by $x_1, \cdots, x_k$, $\sum_{j=}^k |F(x_j)-F(x_{j-1})|<\epsilon.$ So the total sum is majorized by $N\epsilon.$ This holds true for any partition of $[a,b]$. So $F\in BV([a,b])$. Theorem 3.35 (The fundamental theorem of calculus for Lebesgue integrals.) If $-\infty and $F:[a,b]\to \mathbb{C}$, the following are equivalent: (a) $F$ is absolutely continuous on$[a,b]$(b) $F(x)-F(a) = \int_a^x f(t)dt$ for some $f\in L^1([a,b], m)$. (c) $F$ is differentiable $a.e.$ on $[a,b]$, $F'\in L^1([a,b], m)$, and $F(x)-F(a) =\int_a^x F'(t)dt$. Proof. $(a)\Rightarrow (c).$ By subtracting $F(a)$ and extending to $\mathbb{R}$ trivially, we may assume that $F\in NBV$ by Lemma 3.34. So by Corollary 3.33, $(c)$ follows. $(c)\Rightarrow (b)$ is trival. $(b)\Rightarrow (a).$ Extending $f$ trivially to $\mathbb{R}$, i.e., $f(t)=0, t\notin [a,b]$. Then we invoke Corollary 3.33. Posted by: Shuanglin Shao | October 17, 2020 ## Measure and Integration theory, Lecture 12 Definition. A complex measure on a measurable space $(X, \mathcal{M})$ is a map $\nu:\, \mathcal{M}\to \mathbb{C}$ such that (a). $\nu(\emptyset)=0$. (b). If $\{E_j\}$ is a sequence of disjoint sets in $\mathcal{M}$, then $\nu(\cup_{j=1}^\infty E_j) =\sum_{j=1}^\infty \nu(E_j)$ where the series converges absolutely. Remark. (1). A positive measure is a complex measure only if it is finite. Indeed, $E_1=X, E_j=\emptyset, j\ge 2$. $\nu(X)$ converges absolutely. That is to say, $\nu(X)<\infty.$ (2). If $\nu$ is a complex measure, the range of $\nu$ is a bounded set. Indeed, We write $\nu=\nu_r+i\nu_i$. One can prove that $\nu_r, \nu_i$ are signed measures. The total variations $|\nu_r|(X), |\nu_i|(X)$ are bounds for $\nu_r, \nu_i$. So the range of $\nu$ is bounded. Definition. Let $\nu$ be a complex measure. Define $L^1(\nu) = L^1(\nu_r)\cap L^1(\nu_i)$. For $f\in L^1(\nu)$, $\int fd\nu = \int fd\nu_r+ i\int f d\nu_i$. Definition. If $\mu, \nu$ are complex measures, we say that $\mu\perp \nu$ if $\mu_a\perp \nu_b$ for$a, b= r, i.$If $\lambda$ is a positive measure, we say that $\nu\ll \lambda$ if $\nu_r\ll \lambda$ and $\nu_i \ll \lambda$. Theorem 3.12. (The Lebesgue-Radon-Nikodym theorem for complex measures. ) If $\nu$ is a complex measure and $\mu$ is a $\sigma$-finite positive measures on $(X, \mathcal{M})$, there is a complex measure $\lambda$ and an $f\in L^1(\mu)$ such that $\lambda \perp \mu$ and $d\nu= d\lambda + fd\mu.$ If also $\lambda'\perp \mu,$ and $d\nu= \lambda'+f'd\mu$, then $\lambda =\lambda'$ and $f=f', \mu, a.e.$. The idea of the proof is to apply Theorem 3.8 to real and imaginary parts separately. Proof. We write $\nu=\nu_r+i\nu_i$. For $\nu_r$ and $\mu$, by the real Lebesgue-Radon-Nikodym theorem, we see that $\nu_r = \lambda_r+\mu_r, \lambda_r\perp \mu, \mu_r\ll \mu$, and there exists $f_r$, an extended $\mu$ integrable function such that $\mu_r=f_rd\mu$. We know that at most one of $\int f_r^+d\mu$ and $\int f_r^-d\mu$ is infinite. Suppose that $\int f_r^+d\mu= \infty$, we will obtain a contradiction to that $\nu_r$ is finite. Let $E = \{x\in X:\, f_r(x)\ge 0\}$ measurable. Since $\lambda_r, \mu$ are mutually singular, there exist a disjoint decomposition of $X$, $A, B$, such that $A$ is null for $\lambda_r$ and $B$ is null for $\mu$. $\infty = \int f_r^+d\mu = \int_{E\cap A} f_r^+ d\mu + \int_{E\cap B} f_r^+d\mu=\int_{E\cap A} f_r^+d\mu.$ and so $\nu_r (E\cap A) = \lambda_r(E\cap A) +\mu_r(E\cap A) = \mu_r(E\cap A) =\int_{E\cap A} f_r^+d\mu =\infty$. A contradiction to that $\nu_r$ is finite. Thus $f_r\in L^1(\mu)$. The same applies to $\nu_i$, $\nu_i =\lambda_i+\mu_i, \mu_i = f_i d\mu$ and $f_i\in L^1(\mu)$. Thus $f= f_r+if_i \in L^1(\mu)$. Now $\nu=\nu_r+i\nu_i = (\lambda_r+i\lambda_i)+(\mu_r+i\mu_i)$. Let $\lambda= (\lambda_r+i\lambda_i), f=f_r+if_i$. It is easy to verify that that is the desired decomposition. Definition. The total variation of a complex measure $\nu$ is the positive measure $|\nu|$ determined by the property that if $d\nu=fd\mu$ where $\mu$ is a positive measure, then $d|\nu| = |f|d\mu$. To see that this definition is well defined. (1). Every complex measure $\nu$ is of the form $fd\mu$ for a finite measure $\mu$ and some $f\in L^1$. Indeed, one can take $\mu= |\nu_r|+|\nu_i|$. Then Theorem 3.12 can be applied to obtain the existence of $f$. (2). If $d\nu =f_1 d\mu_1 = f_2d\mu_2,$ let $\rho= \mu_1+\mu_2$, then by Proposition 3.9, $f_1\frac {d\mu_1}{d\rho} d\rho=d\nu = f_2\frac {d\mu_2}{d\rho} d\rho,$ so that $f_1\frac {d\mu_1}{d\rho} = f_2\frac {d\mu_2}{d\rho}, \rho, a.e.$ Since $\frac {d\mu_i}{d\rho}$ is nonnegative, we see that $|f_1|\frac {d\mu_1}{d\rho} = | f_1\frac {d\mu_1}{d\rho}| = | f_2\frac {d\mu_2}{d\rho} | = |f_2|\frac {d\mu_2}{d\rho}, \rho a.e.$, and thus $|f_1|d\mu_1 =|f_1|\frac {d\mu_1}{d\rho} d\rho = |f_2|\frac {d\mu_2}{d\rho} d\rho = |f_2|d\mu_2.$ Hence the definition of $|\nu|$ is independent of $\mu$ and $f$. Remark. When $\nu$ is a signed measure, $|\nu| =\nu_++\nu_{-}$. Take $\mu = |\nu|$ and $f= 1_P-1_N$, where $P,N$ are the Hahn decomposition for $\nu$. It is clear that $|f| = 1_P+1_N=1_X.$ So the new definition of the total variation agrees with the old definition. Proposition 3.13. Let $\nu$ be a complex measure on $(X, \mathcal{M})$. (a). $|\nu(E)|\le |\nu|(E)$ for all $E\in \mathcal{M}$. (b). $\nu\ll |\nu|$ and $\frac {d\nu}{d|\nu|}$ has absolute value $1, |\nu| a.e.$ (c). $L^1(\nu) = L^1(|\nu|)$ and if $f\in L^1(\nu)$, then $|\int fd\nu| \le \int |f|d|\nu|$. Proof. (a). Since $\nu$ is a complex measure, the tatol variation of $\nu$ is defined to be $d|\nu| = |f|d\mu$ if $f= fd\mu$, where $\mu$ is some $\sigma$-finite measure. Then $\nu(E) =\int_E fd\mu, f\in L^1(\mu)$ and so $|\nu(E)| \le \int_E |f|d\mu = |\nu|(E)$. (b). From part (a), $\nu\ll |\nu|$ is obvious. Let $g= \frac {d\nu}{d|\nu|}$. Then $g\in L^1$ by Theorem 3.12 and $d\nu= gd|\nu|$. If $d\nu= fd\mu$, $d|\nu| = |f|d\mu$ for some $f\in L^1(\mu)$ and $\mu$ is $\sigma$ finite positive measure, then $d\nu = gd|\nu| = g|f|d\mu.$ From the uniqueness in Theorem 3.12, $f=g|f|, a.e.$ So if $|f|\neq 0$, $|g| =1.$ Let $E= \{x: |f|= 0\}$. Then $|\nu|(E) = \int_E |f|d\mu =0.$ Thus $|g|=1, |\nu| a.e.$ Proposition 3.14. If $\nu_1, \nu_2$ are complex measures on $(X, \mathcal{M})$, then $|\nu_1+\nu_2| \le |\nu_1|+ |\nu_2|$. Posted by: Shuanglin Shao | October 8, 2020 ## Measure and Integration theory, Lecture 11 Definition. Let $(X, \mathcal{M})$ be a measure space. A signed measure on $(X, \mathcal{M})$ is a function $\nu: \mathcal{M}\to [-\infty, \infty]$ such that (a). $\nu(\emptyset) =0.$ (b). $\nu$ assumes at most one of the values $\pm \infty$. (c). If $\{E_j\}$ is a sequence of disjoint sets in $\mathcal{M}$, then $\nu(\cup_{j=1}^\infty E_j) = \sum_{j=1}^\infty \nu(E_j),$ where the latter sum converges absolutely if it is finite. Example 1. Every measure is a signed measure. We shall sometimes refer to measures as positive measures. Example 2. If $\mu_1, \mu_2$ are measures on $\mathcal{M}$ and at least one of them is finite, then $\nu= \mu_1-\mu_2$ is a signed measure. Indeed, Let $E_j$ be a sequence of disjoint sets in $\mathcal{M}$, then if both of them are finite, $\mu_1(\cup_{j=1}^\infty E_j) = \sum_{j=1}^\infty \mu_1(E_j)$ and $\mu_2(\cup_{j=1}^\infty E_j) = \sum_{j=1}^\infty \mu_2(E_j)$, $\mu_1(\cup_{j=1}^\infty E_j)-\mu_2(\cup_{j=1}^\infty E_j) = \sum_{j=1}^\infty (\mu_1-\mu_2)(E_j).$ Also one can easily see that the sum converges absolutely. This proves that $\nu=\mu_1-\mu_2$ is a signed measure. The case where one of them is infinite is proved similarly. Example 3. Let $(X,\mathcal{M},\mu)$ be a measure space; let $f:X\to [-\infty, \infty]$ is a measurable function such that at least one of $\int f^+, \int f^-$ is finite, then $\nu \mapsto \nu(E) = \int_E fd\mu$ is a signed measure. Indeed, for $f\in L^+$, $E\to \int_E fd\mu$ is a measure on $\mathcal{M}$. This follows from the monotone convergence theorem. The general case follows from the Example 2. Proposition 3.1. Let $\nu$ be a signed measure on $(X,\mathcal{M})$. If $\{E_j\}$ is an increasing sequence in $\mathcal{M}$, then $\nu(\cup_{j=1}^\infty E_j) = \lim_{j\to \infty } \nu(E_j)$. Proof. The proof is to observe that for $E_j, 1\le j\le n$ disjoint, then $\nu(\cup_{j=1}^n E_j) =\sum_{j=1}^n \nu(E_j).$ Then we write $\cup_{j=1}^\infty E_j = \cup_{j=1}^\infty (E_j\setminus E_{j-1})$, where $E_0=\emptyset.$ It is easy to see that $\{ E_j\setminus E_{j-1}\}$ is disjoint. Then $\nu(\cup_{j=1}^\infty E_j) = \sum_{j=1}^\infty \nu(E_j\setminus E_{j=1})$ $=\lim_{n\to \infty} \sum_{j=1}^n \nu(E_j\setminus E_{j-1}) =\lim_{n\to \infty } \nu(E_n)$. For decreasing sets $\{E_j\}$, consider $E_1\setminus E_j.$ If $\nu(E_1)$ is finite, then we have continuity from above. Definition. Let $\nu$ be a signed measure on $(X, \mathcal{M}, \nu)$. A set$E$is called positive for $\nu$ if for all subsets $F\subset E, F\in \mathcal{M}$, then $\nu(F) \ge 0$. Similarly for negative sets and null sets. Example. Let $f:\, X\to [-\infty, \infty]$ be a measurable function. Let $\nu(E) :=\int_E fd\nu$, for $E\in \mathcal{M}$. Then if $E$ is positive, then $f\ge 0, a.e.$ on $E$. Indeed, for any $F\subset E, F\in \mathcal{M}$, then $\int_F fd\nu \ge 0.$ If there exists $F\subset E$ with $\nu(F)>0$, but $f<0$. We derive a contradiction. Writing $F=\cup_{n\in \mathbb{N} } \{ x\in F: -2^{n+1}\le f(x)<-2^{-n} \}$, since $\nu(F)>0$, there exists $n_0$ such that $\nu(\{ x\in F:\, -2^{n_0+1}\le f(x)<-2^{n_0}\}) \neq 0$. Call this set $F_0$. Then $F_0\in \mathcal{M}$. Then $\int_{F_0} fd\mu \le -2^{-n_0} \mu(F_0)<0.$ A contradiction. Thus $f\ge 0, a.e.$ on $E$. Lemma 3.2. The union of any countable family of positive sets is positive. Let $\{F_j\}$ be a sequence of positive sets. The proof follows by writing $\cup_{j=1}^\infty F_j$ as a disjoint union. Theorem 3.3 (Hahn decomposition theorem. ) If $\nu$ is a signed measure on $(X, \mathcal{M})$, there exists a positive set $P$ and a negative set $N$ such that $P\cup N=X$ and $P\cap N=\emptyset$. If$P’, N’$is another such pair, then $P\Delta P'= N\Delta N'$ is null for $\nu$. Proof. (a). Uniqueness is easy. We write $P\Delta P'= (P\setminus P')\cup (P'\setminus P)$. Two observations. $P\setminus P'\subset P$ and $P\setminus P'= P\cap N'\subset N'$. Thus $\nu(P\setminus P') =0$. This applies to any measurable subset of $P\setminus P'$. (b). Assume that $\nu$ does not take $+\infty$. Let $m=\sup\{ \nu(E):\, E\in \mathcal{M}, E \text{ is postive for } \nu\}.$ Then there exists $\{E_j\}$ such that $\mu(E_j) \to m$. If setting $P=\cup_i E_i$. We may assume that $E_i$ is increasing. Then $\nu(P) = \lim_{n\to \infty} \nu(E_i) =m.$ We show that $N=P^c$ is negative. (i)It is obvious that $N$ does not contain any positive set with nonzero measure. (ii). Let $A\subset N$ with $0<\nu(A)<\infty$. We claim that there exists $B\subset A$ such that $\nu(B)>\nu(A)$. Otherwise one can show that $A$ is a positive set with nonzero measure. A contradiction to (i). (iii). Suppose that $N$ is not negative: There exists $A_1\subset N$ such that $\mu(A_1)>0$. By (ii), there exists $B\subset A_1$ satisfying $\nu(B)>\nu(A_1)$. Let $(n, B)$ be the pair such that $B\subset A_1, \nu(B)>\nu(A_1)$ and $n$ is the smallest integer such that $\nu(B)\ge \nu(A_1)+\frac 1n.$ Let $n_1=\inf\{n:\, (n, B) \text{is the pair for} A_1. \}$ Note that such $n_1$ is attained. Let $A_2$ be the corresponding set. Suppose that $A_1\supset A_2\cdots\supset A_k, n_1, n_2, \cdots, n_{k-1}$ are chosen. Let $n_k =\inf\{n:\, B\subset A, \nu(B)-\nu(A_{k-1})\ge \frac 1n, n \text{ is the smallest such integer}\}.$ We write $A_1=A_n\cup(A_{n-1}\setminus A_n) \cup \cdots\cup (A_1\setminus A_2)$. Then $\nu(A_1) = \nu(A_k)+\sum_{k=1}^{n-1} \nu(A_k\setminus A_{k+1}) =\nu(\cap_{j=1}^n A_j)+\sum_{k=1}^{n-1} \nu(A_k\setminus A_{k+1}) .$ Let $n\to \infty$. Then $0<\nu(A_1) \le \nu(\cap_{j=1}^\infty A_j) -\sum_{j=1}^\infty \frac {1}{n_j}$. Obviously we have $\cap_{j=1}^\infty A_j \neq \emptyset.$ Then $\sum_{j=1}^\infty \frac {1}{n_j} \le \nu(\cap_{j=1}^\infty A_j)<\infty.$ This implies that $\lim_{j\to \infty }n_j=\infty.$ For $\cap_{j=1}^\infty A_j$, there exists $B$ and the smallest integer$N$such that $\nu(B)- \nu(\cap_{k=1}^\infty A_k) \ge \frac 1N$. Thus there exists $n_K>N$ such that $\frac {1}{n_K}<\frac 1N<\nu(B)-\nu(\cap_{k=1}^\infty A_k)\le \nu(B)-\nu(A_{K-1}) .$ This contradicts to the choice of $n_K$. Definition. $X= P\cup N$ of $X$ is the disjoint union of a positive set and a negative set is called a Hahn decomposition for $\nu$. Definiton. Two signed measures $\mu, \nu$ on $(X, \mathcal{M})$ are mutually singular, or that $\nu$ is singular with respect to $\mu$, or vice versa, if there exist $E, F\in \mathcal{M}$ such that $E\cup F =X, E\cap F =\emptyset$, $E$ is null for $\mu$, and $F$ is null for $\nu$. This is denoted by $\mu \perp \nu$. Informally speaking, ”$\mu, \nu$ live on disjoint sets. “ Theorem 3.4. (The Jordan Decomposition Theorem. ) If $\nu$ is a signed measure, there exist unique positive measures $\nu^+, \nu^-$ such that $\nu=\nu^ - \nu^-$ and $\nu^+\perp \nu^-.$ Proof. Firstly there is a Hahn decomposition for $\nu$: $N\cup P =X, N\cap P=\emptyset.$ For any $E\in \mathcal{M}$, we define $\nu^+=\nu(E\cap P), \nu^-= \nu(E\cap N).$ (a). The measures $\nu^+, \nu^-$ are positive measures on $\mathcal{M}$. (b). The set $N$ is null for $\nu^+$; $P$ is null for $\nu^-$. Thus $\nu^+, \nu^-$ are singular, i.e., $\nu^+\perp \nu^-$. (c). The uniqueness: Suppose that $\nu=\nu^+-\nu^-= \mu^+-\mu^-$. The new measures $\mu^+, \mu^-$ live on different sets $P_1, N_1$ respectively. It is not hard to show that $P_1$ is positive for $\nu$, and $N_1$ is negative for $\nu$. By the uniqueness of the Hahn decomposition, $P\Delta P_1 , N\Delta N_1$ is null for $\nu$. By using this information, for any $\nu\in \mathcal{M}$, $\nu^- (E) = \nu(E\cap N) = \nu(E\cap N\cap n_1)+\nu(E\cap N\cap N_1^c) = \. On the other hand,$latex \mu^ \nu(E\cap N_1)$ also equals the previous sum. Also $\nu(E\cap N_1) = \mu^+( E \cap N_1)+\mu^-(E\cap N_1) =\mu^-(E)$. Thus $\nu^- = \mu^-$. Likewise, $\mu^+=\nu^+$.

Definition. The measures $\mu^+, \nu^+$ are called the positive and negative variations of $\nu$. The expression $\nu=\nu^+-\nu^-$ is called the Jordan decomposition of $\nu$. The total variation of $\nu$ is the measure, $|\nu|=\nu^++\nu^-$.

Remark. (1). If $\nu$ is a signed measure, then $E\in \mathcal{M}$ is $\nu$ null for $\nu$ is equivalent to $|\nu|(E) =0$.

(2). If $\mu, \nu$ are two signed measures, $\nu\perp \mu$ is equivalent to $|\nu|\perp \mu$ is equivalent to $\nu^+\perp \mu, \nu^-\perp \mu.$

(3). If the range of $\nu$ is contained in $\mathbb{R}$, then $\nu$ is bounded.

Definition. Integration with respect to a signed measure $\nu$:

$\nu^1(\nu):= L^1(\nu^+) \cap L^1(\nu^-)$

and $\int f \nu = \int f d\nu^+- \int fd\nu^-$.

Definition. A signed measure is called finite (resp. $\sigma$ finite) if $|\nu|$ is finite (resp. $\sigma$ finite).

Definition. Suppose that $\nu$ is a signed measure and $\mu$ is a positive measure on $(X, \mathcal{M})$. We say that $\nu$ is absolutely continuous with respect to $\mu$ and write $\nu \ll \mu$ if $\nu(E)=0$ for every $E\in \mathcal{M}$ for which $\mu(E)=0$.

Remark. (1). $\nu\ll \mu$ is equivalent to $|\nu|\ll \mu$ is equivalent to $\nu^+\ll \mu, \nu^-\ll \mu.$

(2). $\nu\perp \mu$ and $\nu\ll \mu$ implies that $\nu=0$.

Theorem 3.5. Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X, \mathcal{M})$. Then $\nu\ll \mu$ is equivalent to, for any $\epsilon>0$, there exists $\delta>0$ such that for $\mu(E)<\delta$, $|\nu(E)|<\epsilon$.

Proof. $\Rightarrow.$ Suppose the contrary. For $\epsilon=1$, for any $2^{-k}$, there exists $\mu(E_k)<2^{-k}$, $|\nu|(E_k)\ge 1$. Let $F_n=\cup_{k\ge n} E_k$. Then $F_n$ is decreasing and $\mu(F_n)\le \sum_{k\ge n} 2^{-k} =2^{-n+1}$. Then

$\mu(\cap_{n\ge 1} F_n) = \lim_{n\to \infty} \mu(F_n)=0$ because $F_n$ is decreasing and $\mu(F_1)\le 1$. Then because $\nu\ll \mu$ is equivalent to $|\nu|\ll \mu$, $|\nu|(\cap_{n\ge 1}F_n)=0$. However since $|\nu|$ is bounded, we have

$\lim_{n\to \infty} |\nu| (\cap_{n\ge 1}F_n) = \lim_{n\to\infty} |\nu|(F_n)\ge 1$. A contradiction.

$\Leftarrow.$ Let $\epsilon, \delta$ be given as above. For $E\in \mathcal{M}$, $\mu(E)=0$. Then $\mu(E)<\delta$ for any $\delta>0$. Thus $|\nu|(E)<\epsilon.$ This is true for any $\epsilon>0$. Therefore $|\nu|(E)=0$. Hence $\nu(E)=0$.

Remark. (1). Let $\mu$ be a measure on $\mathcal{M}$ and $f\to [-\infty, \infty]$ is a measurable function such that at least one of $\int f^+d\mu, \int f^-d\mu$ is finite. In this case, we shall call $f$ and extended $\mu$ integrable function. For $E\in \mathcal{M}$, define $\nu(E) = \int_E fd\mu = \int_E f^+d\mu-\int_E f^-d\mu$. Then $\nu$ is a signed measure. Also $\nu\ll \mu.$

For any complex valued function $f\in L^1(\mu)$, the preceding theorem can be applied to the $Re(f)$ and $Im(f)$.

Corollary 3.6. If $f\in L^1(\mu)$. For any $\epsilon>0$, there exists $\delta>0$ such that for $\mu(E)<\delta$,

$|\int_E fd\mu|<\epsilon.$

Definition. For any $E\in \mathcal{M}$, $\nu(E) = \int fd\mu$ is equivalent to $d\nu= fd\mu.$

Lemma 3.7. (Dichotomy.) Suppose that $\nu$ and $\mu$ are finite measures on $(X, \mathcal{M})$. Then either $\nu\perp \mu$ or there exists $\epsilon>0$ and $E\in \mathcal{M}$ such that $\mu(E)>0$ and $\nu\ge \epsilon \mu$ on $E$.

Proof. Let $X= P_n\cup N_n$ be a Hahn decomposition for $\nu-n^{-1}\mu$, and let $P= \cup_{n=1}^\infty P_n$ and $N= \cap_{n=1}^\infty N_n=P^c$. Then for all $n$, $N$ is the negative set for $\nu-n^{-1}\mu$. Thus $0\le \nu(N)\le n^{-1}\mu(N)$. Since $\mu$ is a finite measure, $\nu(N)=0$.

If $\mu(P)=0$, then $\nu\perp \mu$.

If $\mu(P)>0$, there for some $n$, $\mu(P_n)>0$; therefore $P_n$ is a positive set for $\nu-n^{-1}\mu$.

Theorem 3.8. (Lebesgue-Radon-Nikodym theorem.) Let $\nu$ be a $\sigma$-finite signed measure and $\mu$ a $\sigma$-finite measure on $(X, \mathcal{M})$. There exist unique $\sigma$ measures $\lambda, \rho$ on $(X, \mathcal{M})$ such that

$\lambda \perp \mu, \rho \ll \mu,$ and $\nu=\lambda + \rho$.

Moreover there exists an extended $\mu$ integrable function $f:\, X\to \mathbb{R}$ such that $d\rho = f d\mu$, and any two such functions are equal a.e.

Proof. (1). We first assume that $\mu$ a $\sigma$ finite measure and $\nu$ a finite positive measure. Let

$\mathcal{F} =\{f:\, X\to [0,\infty] \text{ is measurable } \int_E fd\mu \le \nu(E) \text{ for all } E\in \mathcal{M}\}$.

(i). $\mathcal{F} \neq \emptyset$ because $0\in \mathcal{F}$.

(ii). Let $a:= \sup\{ \int f d\mu:\, f\in \mathcal{F}\}$. Then $a \le \nu(X)<\infty$. There exists a sequence of $\{f_n\}\in \mathcal{F}$, such that $\int f_n \to a$. Let $g_n = \max \{f_1,\cdots, f_n\}$. Then $g_n\in \mathcal{F}$. Aso $g_n$ is increasing. Let $g=\lim_{n\to \infty} g_n$. By the monotone convergence theorem, $a =\int g.$

Define $\lambda (E) =\nu(E) -\int_E gd\mu, E\in \mathcal{M}$. Then $\lambda \perp \mu$. Otherwise there exists $E\in \mathcal{M}$ and $\epsilon>0$ with $\mu(E)>0$ such that

$\lambda (F) -\epsilon \mu(F) >0$

for $F\subset E$ and $F\in \mathcal{M}$. That is to say,

$\nu(F) - \int_F gd\mu - \epsilon \mu(F)>0$. Let $g_1 = g+\epsilon 1_E$. Then $g_1 \in \mathcal{F}$. However $\int g_1 =\int g+\epsilon\mu(E) >\int g =a$. A contradiction.

Define $\rho (E) =\int_E gd\mu$. Then $\rho \ll \mu.$

Thus $\nu= \lambda +\rho, \rho \perp \mu, \rho \ll \mu$.

(2). Uniqueness. Let $\nu = \lambda +\rho = \lambda_1+\rho_1$, $\rho\perp \mu, \rho\ll \mu, \lambda_1\perp \mu, \rho_1\ll \mu$. Thus $\lambda -\lambda_1 =\rho_1-\rho$ and therefore $\lambda-\lambda_1 \perp \mu, \lambda-\lambda_1 \ll \mu$. This implies $\lambda= \lambda_1$. Similarly $\rho = \rho_1$. Next we need to prve the uniqueness of $g$. We make two observations. (i). $g\ge 0$.

(ii). Both $\rho$ and $\lambda$ are positive measures. So $\rho(X) = \int g d\mu<\infty$. So $g\in L^1$. Suppose there are two functions $g, g_1\in L^1(\mu)$ such that

$\int_E gd\mu = \int_E g_1 d\mu, \text{ for all } E\in \mathcal{M}$. Thus $g=g_1, a.e.$.

(3). Suppose that $\mu, \nu$ are $\sigma$ finite measures. Thus $X=\cup_i X_i$, $\nu(X_i) <\infty$ and $X_i$ are disjoint. Let $\nu_i(E) = \nu(E\cap X_i)$. Then $\nu_i$ is a measure. By (1) and (2) above, there exist $\lambda_i, \rho_i$ such that $\lambda_i \perp \mu$ and $\rho_i \ll \mu$ such that $\nu_i = \lambda_i + \rho_i$. Let $\lambda =\sum \lambda_i, \rho= \sum_j \rho_i$, where $\rho_i =f_i d\mu$. Let $f=\sum f_i$. One can verify that $\lambda\perp \mu,$ and $\rho\ll \mu$. Indeed, by definition, $\rho(E) = \sum_i \rho_i(E) =\sum \int f_i d\mu= \int_E \sum f_i d\mu=\int_E fd\mu$. For $\lambda =\sum \lambda_i$ and for each $i$, there exists a Hahn decomposition $E_i, F_i$ satisfying that $E_i\cup F_i=X, E_i\cap F_i=\emptyset$, and $\lambda_i, \mu$ live on $E_i, F_i$ respectively. Let $E = \cap E_i$. Then $E^c= \cup F_i$. We verify that $\lambda, \mu$ live on $E$ and $E^c$ respectively: for each $F\subset E, \lambda(E) = \sum \lambda_i (E)=0$ and for each $F\subset E^c= F_1\cup (F_2\setminus F_1) \cup (F_3\setminus (F_1\cup F_2))\cdots$, then $\mu(F)=0.$ This proves that $\lambda \perp \mu$. For uniqueness of $g$, let $E\subset X_i$, $f=g a.e.$ on $X_i$, which implies $f=g, a.e.$ on $X$.

(4). In general, $\mu$ a $\sigma$-finite measure and $\nu$ is a signed measure. $|\nu|= \nu^++\nu^-$ is $\sigma$ finite. $\nu=\nu^+-\nu^-$ and $\nu^+, \nu^-$ are $\sigma$ finite and $\nu^+\perp \nu^-$. We have

$\nu^+=\lambda^++\rho^+, \lambda^+\perp \mu, \rho^+\ll \mu, \rho^+=f^+d\mu$

and

$\nu^-=\lambda^-+\rho^-, \lambda^-\perp \mu, \rho^-\ll \mu, \rho^-=f^-d\mu$.

Let $\lambda = \lambda^+-\lambda^-, \rho= \rho^+-\rho^-, f= f^+-f^-$. Then one can verify that $\lambda\perp \mu, \rho \ll \mu$ and $\rho = fd\mu.$ Indeed, if $\int_X f^+ = \int f^- d\mu=\infty$. Then

$\nu^+(X)= \nu^+(P)=\infty, \nu^-(X)=\nu^-(N) =\infty$

where $P, N$ is the Hahn decomposition for $\nu$.

Since $\nu^+\perp \nu^-$, there exist $E_1, F_1$ such that $\nu^+, \nu^-$ live on $F_1, E_1$ respectively. So

$\infty = \nu^+(P)= \nu^+(P\cap E_1)+ \nu^+(P\cap F_1)=\nu^+(P\cap F_1)$

and $\infty = \nu^- (N)= \nu^-(N\cap F_1)$. Thus $\nu$ assumes both $\infty$ and $-\infty$. A contradiction to the definition of $\nu$. This proves that $\mu$ is an extended $\mu$-integrable function.

The uniqueness is proved similarly. For any $E\in \mathcal{M}$, $\int_E fd\mu = \int_E gd\mu$. Take $E=F_1$. Then

$\int_{F_1} f^+d\mu=\int_{F_1} gd\mu$. Therefore $f^+= g1_{F_1}, a.e.$ because part (1) and part (2) apply to $\nu^-$. Similarly $f^-=g1_{E_1}, a.e.$ Therefore $f=g, a.e.$ on $X$.

Definition. (1). Given $\mu$ a $\sigma$ finite positive measure and $\nu$ a signed measure. Then $\nu= \lambda+\rho$ with $\lambda \perp \mu$ and $\rho \ll \mu$ is called the Lebesgue decomposition of $\nu$ with respect to $\mu$.

(2). Let $\nu, \mu$ be as above and $\nu\ll \mu$, then Theorem 3.8 says that there exists $f$ such that $d\nu = fd\mu$. In this case, $f$ is called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$.

Remark. Let $\nu_1, \nu_2$ be two $$\sigma$ finite signed measures and $\mu$ be a $\sigma$ finite positive measure on $(X, \mathcal{M})$. Then $\frac {d\nu_1+d\nu_2}{d\mu} = \frac {d\nu_1}{d\mu}+ \frac {d\nu_2}{d\mu}$. Proposition 3.9. Suppose that $\nu$ is a $\sigma$ finite signed measure and $\mu, \lambda$ are $\sigma$ finite measures on $(X, \mathcal{M})$ such that $\nu\ll \mu, \mu\ll \lambda$. (a). If $g\in L^1(\nu)$, then $g\frac {d\nu}{d\mu} \in L^1(\mu)$ and $\int gd\nu = \int g\frac {d\nu}{d\mu} d\mu$. (b). We have $\nu\ll \lambda$ and $\frac {d\nu}{d\lambda } = \frac {d\nu}{d\mu } \frac {d\mu}{d\lambda}, \lambda a.e.$ Proof. Without loss of generality we assume that $\nu\ge 0$. For $E\in \mathcal{M}$, $\nu(E) = \frac {d\nu}{d\mu} \mu(E)$. So $\int g d\nu = \int g \frac {d\nu}{d\mu} d\mu$ holds true for $g=1_E$. Then by usual procedure, it is true for $g\in L^1(\nu)$. That $g\frac {d\nu}{ d\mu} \in L^1(\mu)$ follows from the equation by considering $g^+$ and $g^-$. This proves (a). (b). By applying (a), for $E\in \mathcal{M}$ twice, $\nu(E) = \int_E \frac {d\nu}{d\mu} d\mu= \int_E \frac {d\nu}{d\mu}\frac {d\mu}{d\lambda} d\lambda$. Note that the equation in (a) does not require $g\in L^1$. Thus $\frac {d\nu}{d\mu} =\frac {d\nu}{d\mu}\frac {d\mu}{d\lambda}. \lambda. a.e.$ Corollary 3.10. If $\mu\ll \lambda$ and $\lambda \ll \mu$, then $\frac {d\lambda } {d\mu} \frac {d\mu}{d\lambda} =1, a.e.$ (with respect to either $\lambda$ or $\mu$. ) Posted by: Shuanglin Shao | October 7, 2020 ## Measure and Integration theory, Lecture 10 Lemma 2.43. If $U$ is open in $\mathbb{R}^n$, $U$ is countable union of cubes with disjoint interiors. Proof. (a). On $\mathbb{R}$, any open set$U$can be written as an at most countable union of open intervals. (b). Since $U$ is open, the projection of $U$ to each axis is open. Thus $U$ can be written as a countable union of cubes with disjoint interiors. Note the cubes are general rectangles. To write them as a union of cubes, one has to work a little bit more. For an interval $(a,b)$, there exists a sequence $a_n,b_n$, both are dyadic numbers, so that $a_n$ decreasing to $a$ and $b_n$ increasing to $b$. Thus $(a,b)$ can be written as a countable union of dyadic intervals with disjoint interiors. Thus $U$ can be written as a countable union of cubes. Theorem 2.44. Suppose $T\in GL(n, \mathbb{R})$, an invertible linear transform from $\mathbb{R}^n$ to $\mathbb{R}^n$. (a). If $f$ is a Lebesgue measurable function on $\mathbb{R}^n$, so is $f\circ T$. If $f\ge 0$ or $f\in L^1(m)$, then $\int f (x)dx = |{det}(T)| \int f\circ T(x) dx, (*)$. (b). If $E\in \mathcal{L}^n$, then $T(E) \in \mathcal{L}^n$ and $m(T(E)) = |det{T}| m(E)$. Proof. (i). The transform $T$ is an invertible linear transform and so it is continuous. Therefore if $f$ is Borel measurable, then $f\circ T$ is Borel measurable. (ii). Let $T_1, T_2, T_3$ denote the three elementary linear transforms. $T_1(x_1, \cdots, x_j, \cdots, x_n) =(x_1, \cdots, cx_j, \cdots, x_n) , c\neq 0$ and $T_2(x_1, \cdots, x_j, \cdots, x_n) =(x_1, \cdots, x_j+cx_k, \cdots, x_n)$ and $T_3(x_1, \cdots, x_j,\cdots, x_k, \cdots, x_n) =(x_1, \cdots, x_k, \cdots, x_j, \cdots, x_n).$ It is easy to see that the integration formula (*) is true for the three linear transforms because any invertible linear transforms can be written as a product of the three linear transforms. Now we check the integration formula (*) is true for $T_1, T_2, T_3.$ These are the consequences of Fubini’s theorem. Indeed, for $T_3$, we interchange the order of integration of $x_j$ and $x_k$. For $T_1,$ we integrate first with the $x_j$ axis and use the one dimensional formula, $\int f(cx) dx = |c|^{-1} \int f(x)dx$. Note that it is true for simple functions, and then the result for measurable functions follows from approximation by simple functions. We can also verify the formula for $T_2$. (iii). The claim in (b) follows if $E$ is Borel, and taking $f=1_E$ in (a). (iv). For general Lebesgue measurable functions, we first observe that Borel null sets are invariant under $T$ and $T^{-1}$ by using (b). Then the general result follows by approximations. Corollary 2.46. Lebesgue measure is invariant under rotations. Now we introduce the polar coordinates. Let $\Phi: \mathbb{R}^n\setminus \{0\}\to (0,\infty)\times S^{n-1}$ defined by $x\to (r, x')$ where $r=|x|$ and $x'=\frac {x}{|x|}$. It is easy to see that $\Phi$ is a continuous bijection. Definition. For any $E\in \mathcal{B}_{(0,\infty)\times S^{n-1}}$, define $m_*(E) = m(\Phi^{-1}(E))$, the Borel measure on $(0,\infty)\times S^{n-1}$. Definition. Define $\rho=\rho_n$ on $(0,\infty)$ by $\rho(E) = \int_E r^{n-1} dr$, for any Borel set $E$ in $(0,\infty)$. Theorem. There is a unique Borel measure $\sigma = \sigma_{n-1}$ on $S^{n-1}$ such that $m_*= \rho\times \sigma$. If $f$ is Borel measurable on $\mathbb{R}^n$ and $f\ge 0$ or $f\in L^1(m)$, then $\int_{\mathbb{R}^n} f(x)dx = \int_0^\infty \int_{S^{n-1}} f(rx') r^{n-1} d\sigma(x')dr. (**)$ Proof. (1). For any Borel measurable set $B$ on $S^{n-1}$, define $\sigma(B):= \frac {m(\Phi^{-1}((0,1)\times B))}{\rho((0,1))}= n m(\Phi^{-1}((0,1)\times B)).$ It is guessed from the equation (**) by taking $f=1_{(0,1)\times B}$. We prove that $\sigma$ is a measure on $S^{n-1}$. Indeed, (i). That $\sigma (\emptyset) =0$ is clear. (ii).It is clear that $\sigma$ is closed under countable additions. (2). Next we need to prove that $m_*= \rho \times \sigma$. For $(a,b]\times B$, where $B$ is Borel on $S^{n-1}$, on one hand, $\rho \times \sigma ((a, b]\times B) = \rho((a,b]) \sigma(B) = \frac {b^n-a^n}{n} \sigma(B)$. On the other hand, for any$b>a>0$, $(a,b]\times B= (0,b]\times B\setminus (0,a]\times B$ and $(0,a]$. Let $E_a= \Phi^{-1} ( (0,a]\times B)$ and $E_b= \Phi^{-1}((0,b]\times B)$ So $m_* ((a,b]\times B)= m(E_b\setminus E_a) =m(E_b)-m(E_a).$ By the dilation property of Lebesgue measure, $m(E_a) = a^n m(\Phi^{-1} ((0,1]\times B))$. Thus $m_* ((a,b]\times B) = \frac {b^n-a^n}{n} \sigma(B)$. This proves that $m_*= \rho\times \sigma$ on $(a,b]\times B$. Next we fix $E\in \mathcal{B}_{S^{n-1}}$ and let $\mathcal{A}_E$ be the collection of finite disjoint unions of sets of the form $(a,b]\times E$. We know that $m_*, \rho\times\sigma$ both agree on $\mathcal{A}_E$. On the other hand, the $\sigma$ algebra $\mathcal{M}_E=\{A\times E:\, A \in \mathcal{B}_{(0,\infty)}\}$ on $(0,\infty) \times E$ is generated by $\mathcal{A}_E$. By the uniqueness theorem in Theorem 1.14, $m_*, \rho\times \sigma$ both also agree on it. Next we know that $\cup\{M_E:\, E\in \mathcal{B}_{S^{n-1}}\}$ is the set of all the Borel measurable rectangles on $(0,\infty)\times S^{n-1}$ and $m_*, \rho\times \sigma$ all agree on this union, so another application of the uniqueness theorem shows that $m_*, \rho \times \sigma$ agree on the Borel sets of $(0,\infty)\times S^{n-1}$. Posted by: Shuanglin Shao | September 25, 2020 ## Measure and integration theory, Lecture 9 Let $(X,\mathcal{M}, \mu)$ and $(Y,\mathcal{N}, \nu)$ be measure spaces. We have discussed the product $\sigma$ algebra $\mathcal{M}\otimes \mathcal{N}$ on $X\times Y$, i.e., the $\sigma$ algebra generated by $\{\pi_\alpha^{-1} (E_\alpha):\, E_\alpha \in \mathcal{M} \text{ or } \mathcal{N}\}$. Definition. We define a (measurable) rectangle to be a set of the form $A\times B$ with $A\in \mathcal{M}$ and $B\in \mathcal{N}$. Lemma. The collection $\mathcal{A}$ of finite disjoint unions of rectangles is an algebra. Proof. We just need to verify the collection of rectangles forms an elementary family. (a). $(A_1\times B_1) \cap (A_2\times B_2) = (A_1\cap A_2) \times (B_1\cap B_2)$. (b). $(A\times B)^c = (A^c \times B )\cup (X\times B^c)$. One can easily verify that the $\sigma$ algebra generated by $\mathcal{A}$ is $\mathcal{M} \times \mathcal{N}$. Suppose that $A\times B$ is a rectangle that is (finite or countable) disjoint union of rectangles $A_j\times B_j$. Then $\mu(A) \nu(B) = \sum_j \mu(A_j) \nu(B_j)$. Indeed, for $x\in X$ and $y\in Y$, $1_A(x) 1_B(y) = 1_{A\times B} = \sum_j 1_{A_j\times B_j} = \sum_j 1_{A_j} 1_{B_j}$. If we integrate with respect to $x$, $\mu(A) 1_B(y) =\int 1_A(x)1_B(y) d\mu(x)$ $= \sum_j \int 1_{A_j}(x) 1_{B_j}(y) d\mu(x) = \sum_j \mu(A_j) 1_{B_j}(y)$. The integration in $y$ yields $\mu(A)\nu(B) = \sum_j \mu(A_j) \nu(B_j)$. So if $E\in \mathcal{A}$ is the disjoint union of rectangles $A_1\times B_1, \cdots, A_n\times B_n$, then $\pi(E)=\sum_j \mu(A_j)\nu(B_j)$ with the convention that $0\cdot \infty =0.$ Lemma. $\pi$ is well defined on $\mathcal{A}$, and $\pi$ is a premeasure on $\mathcal{A}$. Proof. (a). $E\in \mathcal{A}$. We write $E=\cup_j A_j\times B_j= \cup_k C_k\times D_k$, disjoint. Then for each $j$, $A_j\times B_j = \cup_k (C_k\times D_k) \cap (A_j\times B_j) = \cup_k (C_k\cap A_j) \times (D_k\cap B_j)$. Thus $\mu(A_j)\nu(B_j) = \sum_k \mu(C_k\cap A_j) \nu(D_k \cap B_j)$. Thus $\pi(E) = \sum_k\sum_j \mu(C_k\cap A_j) \nu(D_k \cap B_j).$ This proves that $\sum_j \mu(A_j)\nu(B_j) = \sum_k \mu(C_k) \nu(D_k)$. The way to show that $\pi$ is a premeasure on $\mathcal{A}$ is similar as above. Thus Theorem 1.14 applies to obtain a measure on $\mathcal{M}\times \mathcal{N}$ that extends $\pi$. Definition. The measure above is called the product measure that is denoted by $\mu\times \nu$. Remark. If $\mu,\nu$ are $\sigma$ finite, then $\mu\times \nu$ is $\sigma$ finite. Indeed, $X= \cup_j X_j, Y=\cup_k Y_k$ with $\mu(X_j), \nu(Y_k) <\infty$, then $X\times Y = \cup_{j,k} X_j\times Y_k,$ $\mu\times \nu (X_j\times Y_k)= \mu(X_j)\nu(Y_k)<\infty$. So $\mu\times \nu$ is $\sigma$ finite. Definition. Let $(X\times Y, \mathcal{M}\otimes \mathcal{N}, \mu\times \nu)$ be a product space. If $E\subset X\times Y$, for $x\in X, y\in Y$, we define the $x$-section $E_x$ and the $y-$section $E^y$ of $E$ by $E_x=\{y\in Y:\, (x,y)\in E\}, E^y=\{x\in X:\, (x,y)\in E\}$. If $f$ is a function on $X\times Y$, we write the $x-$section and the $y-$section of $E$ by $f_x(y)= f^y(x)= f(x,y)$. Remark. It is easy to see that $(1_E)_x= 1_{E_x}, (1_E)^y= 1_{E^y}$. Proposition 2.34. (a). If $E\in \mathcal{M}\otimes \mathcal{N}$, then $E_x\in \mathcal{N}$ for all $x\in X$ and $E^y\in \mathcal{M}$ for all $y\in Y$. (b). If $f$ is $\mathcal{M}\otimes \mathcal{N}$ measurable, then $f_x$ is $\mathcal{N}$ measurable for all $x\in X$ and $f^y$ is $\mathcal{M}$ measurable for all $y\in Y$. Proof. The proof is to work on measurable rectangles and then consider the general case. If $E=A\times B$ with $A\in \mathcal{M}$ and $B\in \mathcal{N}$, for $x\in X$, $E_x=\begin{cases} B, x\in A, \\ \emptyset, x\notin A \end{cases}$ So $E_x\in \mathcal{N}$. If $E$ is finite disjoint unions of rectangles $A_j\times B_j$, i.e., $E= \cup_{J=1}^n A_j\times B_j$, if$x\in \cap_{j} A_j$for $j$ in a subset of $\{1,2,\cdots, n\}$, then $E_x \in \cup_j B_j$ for $j$ in the same subset. If $x\notin \cup_{j=1}^n A_j$, then $E_x= \emptyset.$ So $E_x \in \mathcal{N}$. We recall that $\mathcal{A}$ is the algebra of finite disjoint unions of rectangles $A\times B$, with $A\in \mathcal{M}$ and $B\in \mathcal{N}$. We have proved that if $E\in \mathcal{A}$, then $E_x\in \mathcal{N}$. We define $\sigma = \{E\subset X\times Y:\, E_x\in \mathcal{N}\}.$ (i). $\mathcal{A} \subset \sigma$. (ii). The collection $\sigma$ is closed under complements. If $E\in \sigma$, then $E_x\in \mathcal{N}$. Since $(E^c)_x= (E_x)^c$, we see that $E^c \in \sigma$. (iii). Since $(\cup_{n=1}^\infty E_n)_x = \cup (E_n)_x$, if $E_n\in \sigma$, then $\cup_n E_n\in \sigma$. This proves $\sigma$ is closed under countable unions. Thus $\sigma$ is itself a $\sigma$ algebra. Thus $\sigma$ contains the $\sigma$ algebra generated by $\mathcal{A}$ that is $\mathcal{M}\otimes \mathcal{N}$. Therefore the claim in (a) is proved. To prove (b), we first consider a real valued function $f$. Note that for $x\in X, y\in Y$, $\{y: f_x(y)>a\} = \{y: f(x,y)>a\} = \{(x,y): f(x,y)>a\}_x$ and $\{x: f^y(x)>a\} = \{x: f(x,y)>a\} = \{(x,y): f(x,y)>a\}^y$. This proves that for a measurable function $f$, $f_x, f^y$ are both measurable. The proof is similar for complex valued functions. Definition. A monotone class on $X$: a subset $\mathcal{C}$ of $\mathcal{P}(X)$ that is closed under countable increasing unions and countable decreasing intersections. That is to say, if $E_1\subset E_2\subset \cdots \subset E_n\subset \cdots$ and $E_j\in \mathcal{C}$, then $\cup_j E_j\in \mathcal{C}$. Likewise for intersections. Example. (1). Every $\sigma$ algebra is a monotone class. (2). The intersection of any family of monotone classes is a monotone class. (3). For $\mathcal{E} \subset \mathcal{P}(X)$, there is a unique smallest monotone class containing $\mathcal{E}$, namely, the intersection of all monotone classes that contain $\mathcal{E}$. This is called the monotone class generated by $\mathcal{E}$. Theorem 2.35. (The monotone class lemma. ) If $\mathcal{A}$ is an algebra of subsets of $X$, then the monotone class $\mathcal{C}$ generated by $\mathcal{A}$ coincides with the $\sigma$ algebra $\mathcal{M}$ generated by $\mathcal{A}$. Proof. Firstly $\mathcal{C} \subset \mathcal{M}$ because $\mathcal{M}$ is itself an algebra. To prove that $\mathcal{M} \subset \mathcal{C}$, we show that $\mathcal{C}$ is a $\sigma$ algebra. To this end, for $E\in \mathcal{C}$, $\mathcal{C}(E) =\{ F\in \mathcal{C}:\, E\setminus F, F\setminus E, E\cap F\in\mathcal{C}\}$. (i). It is easy to see that $\mathcal{C}(E)$ is a monotone class. For instance, if $E_n\in \mathcal{C}(E)$ and $E_n$ is increasing, then $E\setminus ( \cup_n E_n) = \cap_n(E\setminus E_n)$ and $E\setminus E_n$ is decreasing. We know that $\mathcal{C}$ is closed under decreasing intersections, we see that $E\setminus ( \cup_n E_n) \in \mathcal{C}$. Similarly for $( \cup_n E_n)\setminus E, (\cup_n E_n) \cap E$. So $\cup_n E_n \in \mathcal{C}(E)$. (ii). The collection $\mathcal{C}(E)$ contains $\mathcal{A}$. Thus $\mathcal{C}(E)$ contains $\mathcal{C}$. Indeed, we first observe that if $E, F\in \mathcal{C}$, then $E\in \mathcal{C}(F)$ is equivalent to $F\in \mathcal{C}(E)$. If $E\in \mathcal{A}$, then $\mathcal{A}\subset \mathcal{C}(E)$ because $\mathcal{A}$ is an algebra. Thus $\mathcal{C} \subset \mathcal{C}(E)$. That is to say, for $F\in \mathcal{C}$, $F\in \mathcal{C}(E)$. So $E\in \mathcal{C}(F)$. Thus $\mathcal{C} \subset \mathcal{C}(F)$. This proves the claim. (iii). To prove $\mathcal{C}$ is closed under complements, we observe that $X\in \mathcal{C}$. And for $E\in \mathcal{C}$, $X\in \mathcal{C} \subset \mathcal{C}(E)$. Thus $E^c= X\setminus E \in \mathcal{C}$. That $\mathcal{M}$ is closed under countable unions is clear. Thus $\mathcal{C}$ is an algebra. We now come to the main results of this section, the Tonelli-Fubini theorem. Theorem 2.26. Suppose that $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ are $\sigma$ finite measure spaces. If $E\in \mathcal{M}\otimes \mathcal{N}$, then the functions $x\mapsto \nu(E_x), y\mapsto \mu(E^y)$ are measurable on $X, Y$, respectively, and $\mu\times \nu (E) = \int \nu(E_x) d\mu(x)= \int \mu(E^y)d\nu(y).$ Proof. Step 1. Let $\mathcal{A}$ be the algebra consisting of finite disjoint unions of rectangles in $\mathcal{M}\otimes \mathcal{N}$. For $E= A\times B \in \mathcal{A}$, $E_x = \begin{cases}B, x\in A, \\ \emptyset, x\notin A \end{cases}$ so $\nu( E_x) = 1_A(x)|B|$ is measurable. This reasoning can be generalized to any element in $\mathcal{A}$. Step 2. Let $\mu(X), \nu(Y)<\infty$. By Theorem 2.35, it suffices to prove the claim for sets in the monotone class $\mathcal{C}$ generated by $\mathcal{A}$ since $\mathcal{M}\otimes \mathcal{N}$ agrees with this monotone class. Let $\{E_j\}_{j\ge 1} \in \mathcal{C}$ and $E_j$ increasing. Since $(\cup E_j)_x = \cup (E_j)_x$ and $(E_j)_x$ is increasing, we see that $\nu((\cap E_j) _x) = \lim_{j\to \infty} \nu((E_j)_x)$. Since each $\nu((E_j)_x)$ is measurable, then $\nu((\cap E_j)_x)$ is measurable. Likewise for decreasing intersection since the measures of $X, Y$ are finite. Of course, the same claims hold true for $\mu(E^y)$. Step 3. In general, $X\times Y= \cup_{i\ge 1} Z_i$ and $\{Z_i\}$ is disjoint and $\mu\times \nu(Z_i)<\infty$. For $E\in \mathcal{M}\otimes \mathcal{N}$, $E= \cup_i E_i, \text{ where } E_i = E\cap Z_i.$ By Step 1 and 2, $\nu((E_i)_x)$ is measurable. So $\nu(E_x) = \sum_i \nu ((E_i)_x)$ is measurable. Likewise for $\mu(E^y)$. Lastly the same procedure can be used to establish the integral claim with an additional input of monotone convergence theorem. Theorem 2.37. (The Fubini-Tonelli Theorem. ) Suppose that $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ are $\sigma$ finite measure spaces. (a). (Tonelli.) If $f\in L^+(X\times Y)$, then the functions $g(x) = \int f_x(y) d\mu(y), h(y) = \int f^y(x) d\mu(y)$ are in $L^+(X), L^+(y)$ respectively, and $\int f d(\mu\times \nu) = \int [\int f(x,y) d\nu(y)]d\mu(x) = \int [\int f(x,y)d\mu(x)]d\nu(y). (*)$ (b). (Fubini.) If $f\in L^1(\mu\times \nu)$, then $f_x \in L^1(\nu)$ for $a.e. x\in X$, and $f^y\in L^1(\mu)$ for $a.e. y\in Y$, then $a.e.$ defined functions $g(x) = \int f_x d\nu, h(y) = \int f^yd\mu$ are in $L^1(\mu), L^1(\nu)$ respectively, and $(*)$ holds. Proof. (a). Step 1. For $f\in L^+(X\times Y)$, there exists $\{\phi_n\}$ simple functions on $X\times Y$ such that $\phi_1\le \phi_2 \le \cdots \le \phi_n\le \cdots \le f$ and $\phi_n\to f$ pointwise. Then for any $x$, $(\phi_n)_x \to f_x$ pointwise in $Y$. So by monotone convergence theorem, $\int (\phi_n)_xd\nu(y) \to \int f_x d\nu(y)$. If $\phi_n = \sum_k a_n^k 1_{E_n^k}$, then $\int (\phi_n)_xd\nu(y) = \sum_k a_n^k \nu((E_n^k)_x)$, which is measurable. So $\int f_x d\nu$ is measurable. Likewise for $h(y) = \int f^y d\mu(x)$. Step 2. Let $\phi_n$ and $f$ be as above. Then $\int \phi_n d(\mu\times \nu) \to \int fd(\mu\times \nu)$. On the other hand, by Theorem 2.36, $\int \phi_n d(\mu\times \nu) =\sum a_n^k \mu\times \nu(E_n^k) =\sum_k \int \nu((E_n^k)_x)d\mu(x)$, which converges to $\int [\int f_x d\nu(y)]d\mu(x)$. Hence we have $\int fd(\mu\times \nu=\int [\int f_x d\nu(y)]d\mu(x)$. Likewise, $\int fd(\mu\times \nu=\int [\int f^y d\mu(x)]d\nu(y)$. Step 3. Let $f$ be real valued and $f\in L^1(\mu\times \nu)$.. We have $f=f^+-f^-$. Then by Step 1 and 2, $\int f^+ d(\mu\times \nu) = \int_X [\int_Y f_x^+d\nu(y)]d\mu(x)<\infty$. Thus the function $\int_Y f^+_x(y)d\nu(y)\in L^1(d\mu)$. Thus for $a.e. x$, $f_x^+ \in L^1(d\nu)$. Similarly for $f_x^-$. Thus for $a.e. x$, $f_x\in L^1(d\nu)$; and $g(x):=\int_Y f_x(y)d\nu(y)\in L^1(d\mu)$ is an $a.e.$ defined function and is in $L^1(d\mu)$. The similar claim holds for $f^y$ and $h$. Step 4. Let $f$ be a complex valued function and $f\in L^1(\mu\times \nu)$. We write $f= Re(f)+ i Im(f)$ with $Re(f), Im(f) \in L^1(\mu\times \nu)$. By Step 3, for $a.e. x$, $f_x \in L^1(d\nu)$; moreover $\int f_x d\nu(y)$ is an $a.e.$ defined function and is in $L^1(d\mu)$. This proves Fubini’s theorem. Posted by: Shuanglin Shao | September 22, 2020 ## Measure theory and integration, Lecture 8 Let $\{f_n\} \to f$ as $n\to \infty$. There are pointwise convergence, uniform convergence, convergence in $L^1$, and convergence in measure (to be defined). Examples. (a). $f_n= n^{-1} 1_{[0,n]}$. (b). $f_n= 1_{[n,n+1]}$. (c). $f_n = n1_{[0,\frac 1n]}$. (d). The typewriter example. 1st level: $f_1=1_{[0,1]}$. 2nd level, $f_2= 1_{[0,\frac 12]}, f_3 = 1_{[\frac 12, 1]}$. 3rd level: $f_4=1_{[0,\frac 14]}, f_5 = 1_{[\frac 14, \frac 12]}, f_6= 1_{[\frac 12, \frac 34]}, f_4 = 1_{[\frac 34, 1]}$. kth step: $f_n= 1_{[\frac {j}{2^k}, \frac {j+1}{2^k}]},$ where $n=(1+2+\cdot+2^{k-1}-1)+(j+1)= 2^k+j.$ Definition. A sequence of measurable functions complex valued functions on $(X,\mathcal{M}, \mu)$ is Cauchy in measure: for any $\epsilon>0$, $\mu(\{x:\, |f_n(x) - f_m(x)|\ge \epsilon\})\to 0$ as $n,m\to \infty$. Definition. $\{f_n\}$ converges in measure to $f$ as $n\to \infty$: for any $\epsilon>0$, $\mu(\{x:\, |f_n(x)-f(x)|\ge \epsilon\}) \to 0$. Example. $f_n= \frac 1n 1_{(0,n)}, f=0.$ Proposition 2.29. If $f_n\to f$ in $L^1$, then $f_n \to f$ in measure. This follows from the Chebyshev’s inequality. Let $\epsilon>0$ and $E_\epsilon:=\{|f_n(x)-f(x)|\ge \epsilon\}$. Then $\mu(E_\epsilon) =\int_{E_\epsilon} 1 dx \le \int_X \frac {|f_n(x)-f(x)|}{\epsilon} = \frac {\|f_n-f\|_{L^1}}{\epsilon}$ as $n\to \infty.$ Theorem 2.30. Suppose that $\{f_n\}$ is Cauchy in measure. Then there exists a measurable function $f$ such that $f_n\to f$ in measure, and there is a subsequence $\{f_{n_j}\}_{j\ge 1}$ that converges to $f a.e..$ Moreover if there exists $f_n\to g$ in measure, then $g=f, a.e.$. Proof. Step 1. Since $f_n\to f$ in measure, there exists a subsequence $\{f_{n_k}\}$ such that $\mu(|f_{n_{k+1}}-f_{n_k}|\ge 2^{-k})<2^{-k}.$ We denote the set above by $E_k.$ Step 2. The set $E= \cap_{n=1}^\infty \cup_{k\ge n} E_k$ has measure$0$as $\mu(\cup_{k\ge n} E_k )\le 2^{-n+1}$ and the sets are decreasing. On $E^c$, one can prove that $\{f_{n_k}\}$ is Cauchy pointwise. Indeed, for $x\in E^c$, there exists$n$such that for $k\ge n$, $|f_{n_{k+1}}(x)-f_{n_k}|<2^{-k}$. This proves $\{f_{n_k}\}$ is Cauchy. Therefore $\lim_{k\to \infty} f_{n_k}(x) =f(x).$ Step 3. To prove $f_n\to f$ in measure, we just need to prove that $f_{n_k} \to f$ in measure. Indeed, given $n\ge 1$, on $\left( \cup_{k\ge n} E_k\right)^c$, for$l\ge k\ge n$$|f_{n_{l}}(x)-f_{n_k}(x)|\le 2^{-k+1}.$ Let $l\to\infty$. $|f(x)-f_{n_k}(x)|<2^{-k+1}$. Therefore $\mu(|f(x)-f_{n_k}(x)|\ge 2^{-k+1} ) <2^{-n+1}.$ This proves $\{f_{n_k}\}$ converges to $f$ in measure. Step 4. The uniqueness is easy. Corollary 2.32. If $f_n\to f$ in $L^1$, there is subsequence $f_{n_j}$ such that $f_{n_j} \to f, a.e.$. Egoroff’s Theorem. Suppose that $\mu(X)<\infty$, and $f_n\to f a.e..$ Then for any $\delta>0$, there exists $E\subset X$ such that $\mu(E)<\delta$, and $f_n\to f$ uniformly on $E^c$. Proof. Given $\epsilon>0$. Since $f_n\to f a.e.$, $\cap_{N=1}^\infty \cup_{n\ge N} \{|f_n(x)-f(x)|>\epsilon\} \subset A.$ Here $\mu(A)=0$. We denote $E_N = \cup_{n\ge N} \{|f_n(x)-f(x)|>\epsilon\}$. For any $\delta>0$, there exists $N$ such that $\mu(E_N)<\delta$, $|f_n(x)-f(x)|<\epsilon$ on $E_N^c$. Take $\epsilon= \frac 1k$, there exists $E_{N_k}$ such that $\mu(E_{N_k})<2^{-k} \delta$. So $\mu(\cup_{k=1}^\infty E_{N_k})<\delta$. On $\cap_{k=1}^\infty E_{N_k}^c = \cap_{k}\cap_{n\ge N_k} \{|f_n-f|\le \frac 1k\},$ for any $\epsilon>0$, there exists $k_0$ such that for $n\ge N_{k_0}$ $|f_n-f| \le \frac 1{k_0}<\epsilon.$ This proves uniform convergence. Posted by: Shuanglin Shao | September 18, 2020 ## Measure and Integration theory, Lecture 7 Let $f$ be a real valued measurable function. Let $f^+=\max \{f, 0\}\ge 0; f^-= -\min\{f,0\}\ge 0$. These are the positive and negative parts of $f$. An easy observation is $f= f^+-f^-, |f| = f^++f^-.$ Definition. Let $f$ be a real valued measurable function. If $\int f^+$ or $\int f^-$ is finite, we denote $\int f = \int f^+ - \int f^-.$ We say $f$ is integrable if both terms are finite. Proposition 2.21. The set of integrable real valued functions on $X$ is a real vector space, and the integral is a linear functional on it. Proof. We need to prove two claims. (1). $\int \alpha f =\alpha \int f$ for any $\alpha \in \mathbb{R}.$ This is easy: we distinguish three cases, $\alpha>0, =0,<0.$ (2). $\int f+g = \int f + \int g.$ It is to observe that, if$h = f+g,$then $h^+-h^- = f^+-f^- +g^+-g^-$. We rearrange it to obtain $h^++f^-+g^- = h^- +f^++g^+$. Taking integrals on both sides yields (2). Definition. The complex valued function $f$ is integrable if $Re f$ and $Im f$ are both integrable. Remark. It is easy to see that the space of complex valued integrable functions is a complex vector space and the integral is a linear functional over it. It is denoted by $L^1$. Remark. We will regard $L^1$ as a set of equivalence classes of a.e. defined integrable functions on $X$, where $f$ and $f$ are considered to be equivalent if and only if $f=g, a.e.$. Proposition 2.22. If $f\in L^1$, then $|\int f | \le \int |f|$. Proof. When $f$ is real valued, the proof is easy. For complex valued $f$, there exists $\theta$, $|\int f| = e^{i\theta} \int f = \int e^{i\theta} f$ by linearity of integrals. It further equals $= \int Re( e^{i\theta} f ) \le \int | Re( e^{i\theta} f ) | \le \int |f|.$ Proposition 2.24. (The dominated convergence theorem. ) Let $\{f_n\}$ be a sequence in $L^1$ such that (a). $f_n\to f, a.e.,$ (b). There exists $g\in L^1$ s.t. $|f_n|\le g$ for a.e. for all $n$. Then $f\in L^1$ and $\int f = \lim_{n\to \infty} \int f_n.$ Proof. Without loss of generality, we assume that $f$ is real valued. Step 1. The claim that $f\in L^1$ is easy. (2). By Fatou’s theorem, $\int g-f = \int \liminf_{n\to \infty} (g-f_n) \le \liminf ( \int g -\int f_n) = \int g -\limsup \int f_n.$ This yields, $\limsup \int f_n \le \int f.$ The same process can be applied to$\int g+f $to show $\int f \le\liminf \int f_n.$ Remark. Examples. (1). $f_n = n1_{[0,\frac 1n]}$ and $f=0.$ (2). $f_n= 1_{[n,n+1]}$ and $f=0$. (3). $f_n = \frac {1}{n} 1_{[0,n]}$ and $f=0.$ Theorem 2.26. If $f\in L^1$ and $\epsilon>0$, there is an integrable simple function $\phi=\sum a_j 1_{E_j}$ such that $\int |f-\phi|<\epsilon.$ If $\mu$ is the Lebesgue-Stieljes measure on $\mathbb{R}$, the sets $E_j$ in the definition of $\phi$ can be taken to finite unions of open intervals; moreover there is a continuous function $g$ that vanishes outside a bounded interval such that $\int |f-g|<\epsilon.$ Proof. Step 1. For an integrable function $f$, there exists a sequence of simple functions $\phi_n$ such that $0\le |\phi_1| \le |\phi_2| \le \cdots \le |f|$ and $\phi_n\to f$ pointwise. By DCT, $|f-\phi_n|\le 2|f|$, $\int |f-\phi_n| \to 0.$ Step 2. From Step 1, $\int |\phi|\le \int |f|+\epsilon.$ That is to say $\sum |a_j| m (E_j)<\infty$ for $a_j\neq 0$. Thus $m(E_j)<\infty$ for each $j$.$ For each $j$, by Theorem 1.20, there exists a set $A_j$ that is a finite unions of open intervals such that $m(A_j\Delta E_j) for some small constant $c_j$. By taking $c_j$ small, we can take the sets in the definition of $\phi$ to be finite unions of open intervals.

Step 3. Since we can approximate each $1_{E_j}$, where $E_j$ is an open interval of finite length, by continuous functions $f_j$ with the following property

$\int | 1_{E_j} - f_j| \le d_j \epsilon$

for some constant $d_j$. By taking $d_j$ small, we can replace $\phi$ by a continuous function $g$ that vanishes outside a bounded interval such that $\int |f-g| <\epsilon.$

Theorem 2.27. Suppose that $f: X\times [a,b]\to \mathbb{C}, -\infty and that $f(\cdot, t):\, X\to \mathbb{C}$ is integrable for each $t\in [a,b]$. Let $F(t) =\int_X f(x,t) d\mu(x)$.

(a). Suppose that there exists $g\in L^1(\mu)$ such that $|f(x,t) |\le g(x)$ for all $x,t$. If $\lim_{t\to t_0} f(x,t) =f(x,t_0)$ for every $x$, then

$\lim_{t \to t_0} F(t) =F(t_0)$. In particular, if $f(x, \cdot)$ is continuous for each $latrex x$, then $F$ is continuous.

(b). Suppose that $\frac {\partial f}{\partial t}$ exists and there is a $g\in L^1(\mu)$ such that $|\frac {\partial f}{\partial t} (x,t)|\le g(x)$ for all $x,t$. Then $F$ is differentiable and

$F'(t) = \int \frac {\partial f}{\partial t} (x,t) d\mu(x).$

Proof. (a). We are proving the claim in (a) by using sequential characterization of continuity. Let $t_n\to t_0$, the sequence of measurable functions $\{f(x,t_n)\}$ is uniformly by a $L^1$ function and it converges to $f(x,t_0)$ pointwise. Therefore by the dominated convergence theorem,

$\lim_{t\to t_0} \int f(x,t_n)d\mu(x) = \int f(x,t_0) d\mu, i.e., \lim_{n\to \infty} F(t_n)= F(t_0).$

(b). The proof is similar. We need to prove that, for each $t\in [a,b], h_n\to 0$,

$\lim_{n\to \infty} \frac {F(t+h_n)-F(t)}{h_n} = \int \frac {\partial f}{\partial t}(x,t) d\mu(x).$

By applying the mean function for $f$ in $t$, we see that the sequence $\{ \frac {f(x,t+h_n) -f(x,t)}{h_n}\}$ is uniformly bounded by a $L^1$ function $g$; moreover it converges to $\frac {\partial f} { \partial t}$. Again by the dominated convergence theorem,

$\lim_{n\to \infty} \frac {F(t+h_n)-F(t)}{h_n} = \int \frac {\partial f} { \partial t}(x,t ) d\mu=\int \frac {\partial f}{\partial t}(x,t) d\mu(x).$

We next discuss the relation between Riemann integrable functions and Lebesgue measurable function.

Theorem. Let $f$ be a bounded real-valued function on $[a,b]$. If $f$ is Riemann integrable, then it is Lebesgue integrable. Moreover the two integrals are equal.

Remark. Without loss of generality, we assume that $f$ is a nonnegative function. As in Theorem 2.10, there is a sequence of simple functions $\{\phi_n\}$ that satisfies,

$\phi_1 (x) \le \phi_2(x) \le \cdots \le f$

and $\phi_n\to f$ uniformly on $[a,b]$. Although $f$ is a pointwise limit of $\{\phi_n\}$, the Lebesgue measurability of $\phi_n$ is not immediately clear. We should seek to prove the Lebesgue measurability $f$ in other ways.

Proof. Let $P= \{a= t_0 by a partition of $[a,b]$. Let

$G_P= \sum M_j 1_{[t_{j-1}, t_j]}, g_P = \sum m_j1_{[t_{j-1}, t_j]},$

where $M_j, m_j$ are supremum and infimum of $f$ on $[a,b]$. Since $f$ is Riemann integrable, we can choose a sequence of partitions $\{P_n\}$ whose mesh size $\max_j (t_j-t_{j-1}) \to 0$ and

$\lim_{n\to \infty} \int_{[a,b]} G_{P_n} = \lim_{n\to \infty} \int_{[a,b]} g_{P_n} = \int_{[a,b]} f (x)dx$.

These are understood that the two sequences $\sum M_j (t_j-t_{j-1})$ and $\sum m_j (t_j-t_{j-1})$ has the same limit.

This implies

$\int_{[a,b]} (G_{P_n}-g_{P_n})dx \to 0$ as $n\to \infty$.

On the other hand, by the monotone convergence theorem for functions, $g:= \lim g_{P_n} \le f \le G:= \lim G_{P_n}.$

Therefore by the dominated convergence theorem (the following integrals are understood as in the Lebesgue sense. )

$\int_{[a,b]} (G-g) dx= \lim_{n\to \infty} \int_{[a,b]} (G_{P_n} - g_{P_n}) dx$

$=\lim_{n\to \infty} \left( \sum M_j (t_j-t_{j-1})-\sum m_j (t_j-t_{j-1})\right) =0.$

Since $G\ge g$ pointwise, $G=g, a.e.$, which yields that $G=g=f, a.e.$. Since $G$ is Lebesgue measurable as simple functions and $G=f a.e.,$ $f$ is Lebesgue measurable.

That the two integrals are equal is immediate.

Posted by: Shuanglin Shao | September 13, 2020

## Measure and Integration Theory, Lecture 6

Definition. Let $(X, \mathcal{M}, \mu)$ be a measure space. Let

$L^+=$ the space of all measurable functions from $X$ to $[0,\infty]$. Let $\phi\in L^+$ and $\phi$ is simple, $\phi= \sum_{j=1}^n a_j 1_{E_j}$. We define the integral of $\phi$ with respect to $\mu$ by

$\int \phi d\mu = \sum_{j=1}^n a_j \mu(E_j)$.

With the convention, $0\cdot \infty =0$.

Definition. If $A \in \mathcal{M}$, we define

$\int_A \phi d\mu = \int \phi 1_A d\mu.$

Remark. The definition of $\int_A$ makes sense because

$\phi1_A = \sum a_j 1_{E_j \cap A}$.

Proposition. Let $\phi$ and $\psi$ be simple functions in $L^+$.

(a). If $c\ge 0$, $\int c\phi = c\int \phi$.

(b). $\int \phi + \psi = \int \phi + \int \psi.$

(c). If $\phi \le \psi$, then $\int \phi \le \int \psi.$

(d). The map $A \mapsto \int_A \phi d\mu$ is a measure on $\mathcal{M}$.

Proof. (a) is easy.

(b). We write $\phi = \sum a_j 1_{E_j}, \, \psi= \sum b_k 1_{F_k}$, where $\{E_j\}$ are disjoint sets in $X$ and $\cup E_j =X$; so are $\{F_k\}$. The observation is that $\phi+\psi$ takes values $a_j+b_k$ on $E_j \cap F_k$. Then

$\int \phi+\psi = \sum_{j,k} (a_j+b_k) \mu (E_j\cap F_k) =\sum_j a_j \mu(E_j)+\sum_k b_k \mu(F_k).$

The latter sum is equal to $\int \phi+\int \psi.$ Here we have used that

$\sum_k \mu(E_j \cap F_k) = \mu(E_k)$.

(c) follows from (b) by observing that $\psi = \phi+ (\psi-\phi)$.

The proof of (d) is standard.

Definition. The integral $f\in L^+$,

$\int fd\mu = \sup\{\int \phi d\mu:\, 0\le \phi \le f, \phi \text{ simple functions} \}.$

Remark. (1). This definition coincides with the old definitions when $f$ is simple.

(2). If $f\le g, f,g\in L^+$, $\int f \le \int g.$

(3). For $c\in [0,\infty)$, $\int cf = c\int f.$ Indeed, for $c=0$, it is true. For $c>0$, $0\le \phi \le cf,$ we have $0\le \frac 1c \phi \le f$. Then

$\int f \ge \int \frac 1c \phi =\frac 1c \int \phi$. The latter equality is because $\phi$ is simple. Then we prove

$c\int f \ge \int cf.$

By the same analysis, $\int cf \ge c\int f.$

Thus $\int cf = c\int f$.

Theorem 2.14. (The monotone convergence theorem. ) If $\{f_n\}$ is a sequence in $L^+$ such that $f_j \le f_{j+1}$ for all $j$, and $f= \lim_{n\to \infty} f_n$, then

$\int f = \lim_{n\to \infty} \int f_n.$

Proof. The idea is to work with an approximate function to $f$. To begin it, we choose a sequence of simple functions $\phi_n$, $0\le \phi_n \le f$ and