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 $a_j\neq 0$.