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<p<\infty, 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;


\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<p<\infty. 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<p<\infty, 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<p<\infty, 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.


\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<b, a,b \in (0,\infty). \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).


\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<p<\infty, 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<x<1 and g(x)=1, 0<x<1.

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<p<\infty.

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<p\le \infty, 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<p<q<r\le \infty, 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<p<q<r\le \infty, 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


Proof. If g\neq 0 and 0<q<\infty, 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<q<\infty. 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<p<\infty, 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<m(U), 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<m(u) there exists a compact set K such that

K\subset U, c<m(K).

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<m(K) \le 3^n \sum_{j=1}^m m(B'_j).

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<m(U), 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<m(V_\epsilon), 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<y,

(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|<N} H(x_j)=\sum_{|x_j|<N} (G(x_j)-F(x_j) ) = \sum_{|x_j|<N} (F(x_j+)-F(x_j)) \le F(N)-F(-N)<\infty.  (*)

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<x_0<x_1\cdots <x_n=x\}.

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<x_1\cdots <x_n=b\}. (*)

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<a<b<\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<b. 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<y, 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<x_1\cdots x_n=x 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<h<\delta,

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

For any such h, there exists x_0=x<x_1<\cdots<x_n=x+h 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<t_1<\cdots<t_m=x_1 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 <x_1<\cdots<x_n=b, 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<a<b<\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


\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).


\mu(A_j)\nu(B_j) = \sum_k \mu(C_k\cap A_j) \nu(D_k \cap B_j) .


\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,


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)<c_j\epsilon 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<a<b<\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<t_1<\cdots t_n=b\} 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 \int f = \lim_{n\to \infty } \int \phi_n.

For each \phi_n, for any \epsilon>0, define

A_n= \{f_k: f_k\ge (1-\epsilon) \phi_n\}.  Since \lim_{n\to \infty} f_n = f, \text{ and } 0\le \phi \le f and \phi_n is simple, there exists k_n such that

f_{k_n}\ge (1-\epsilon) \phi_n. Therefore \int f_{k_n} \ge (1-\epsilon) \int \phi_n. Since \{f_n\} is increasing,

\lim_{n\to \infty } \int f_{k_n} \ge (1-\epsilon) \int f.

This proves the claim.

Remark. (1). Positivity can not dropped. Example:

f= \begin{cases} 1, x\in [-1,0)\\ -\frac 1n, x\in [0,n] \end{cases}.

Then f_n\to f=\begin{cases} 1, x\in [-1,0), \\ 0, x\in [0,\infty). \end{cases}. It is clear that \int f_n =0 but \int f =1.

(2). Monotonicity can not be dropped. Example:

f_n(x) = 1_{[n,n+1]}. It converges to f=0. It is clear that

\int f_n =1 but \int f =0.

Theorem 2.15. If \{f_n\} is a finite or infinite sequence in L^+ and f=\sum f_n, then

\int f =\sum \int f_n.

The proof considers the partial sums of \sum_n f_n.

Theorem 2.16. If f\in L^+, then

\int f =0 \Leftrightarrow f=0, a.e.

The ``\Rightarrow" considers the set

A_n =\{f\ge \frac 1n\}, n\in \mathbb{N}.

If \mu(A_n)>0, because f\ge \frac 1n 1_{A_n}, then

\int f \ge \frac 1n \mu(A_n)>0. A contradiction.

It shows that \mu(A_n)=0.

Corollary 2.17. If \{f_n\}\in L^+, f\in L^+, and f_n increasingly converges to f a.e. x, then

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

Theorem 2.18. (Fatou’s lemma.) If f_n is any sequence in L^+, then

\int \liminf_{n\to \infty} f_n \le \liminf_{n\to \infty} \int f_n.

The proof relies on the definition that

\liminf_{n\to \infty} f_n(x) = \lim_{n\to \infty } \inf \{f_k(x):\, k\ge n\}.

Proof. By the monotone convergence theorem,

LHS = \lim_{n\to \infty} \int \inf\{ f_k(x):\, k\ge n\} \le \lim_{n\to\infty} \inf_{k\ge n} \{ \int f_k:\, k\ge n\}.

Proposition 2.20. If f\in L^+ and \int f<\infty, then \{x:\, f(x)=\infty\} is a null set and \{x:\, f(x)>0\} is \sigma finite.

The proof uses similar idea as Theorem 2.16.

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