# Math4121 Lecture 35

## Continue on Lebesgue Integration

### Lebesgue Integration

#### Definition of Lebesgue Integral

For simple functions $\phi = \sum_{i=1}^{n} a_i \chi_{S_i}$, given a measure $E$, the Lebesgue integral is defined as:

$$
\int_{\mathbb{R}^n} \phi \, dm = \sum_{i=1}^{n} a_i m(S_i\cap E)
$$


Given a non-negative measurable function $f$ and a measurable set $E$.

Define $\int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is a simple function and } \phi \leq f \right\}$

(**We do allows $\int_E f \, dm = \infty$**)

For general measurable function $f$, we can define $f^-(x)=\max\{0,-f(x)\}$, $f^+(x)=\max\{0,f(x)\}$. (The positive part of the function and the negative part of the function, both non-negative)

Then $f=f^+-f^-$.

We say $f$ is integrable if $\int_E f^+ \, dm < \infty$ and $\int_E f^- \, dm < \infty$. (both finite) If at least one is finite, define

$$
\int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm
$$

We allow for $A-\infty = -\infty$ and $A+\infty = \infty$ for any $A\in \mathbb{R}$. But not $\infty-\infty$.

#### Immediate Properties of Lebesgue Integral

If $f$ is measurable and $m(E)=0$, then $\int_E f \, dm = 0$.

If $E=E_1\cup E_2$ and $E_1\cap E_2=\emptyset$, then $\int_E f \, dm = \int_{E_1} f \, dm + \int_{E_2} f \, dm$.

#### Corollary

If $f\leq g$ almost everywhere, ($f\leq g$ except for a set of measure 0), then $\int_E f \, dm \leq \int_E g \, dm$.

Proof:

Let $F=\{x\in E: f(x)>g(x)\}$. Then $m(F)=0$.

$$
\begin{aligned}
\int_E f \, dm &= \int_{E\setminus F} f \, dm + \int_F f \, dm\\
&\leq \int_{E\setminus F} g \, dm+\int_E g \, dm\\
&= \int_E g \, dm
\end{aligned}
$$

QED

#### Proposition 6.13

If $f$ is non-negative and $\int_E f \, dm =0$, then $f=0$ almost everywhere on $E$, $f(x)=0$ $\forall x\in E\setminus F$, where $m(F)=0$.

Proof:

Let $E_n=\{x\in E: f(x)\geq \frac{1}{n}\}$. Then $\frac{1}{n}\chi_{E_n}(x)\leq f(x)$ for all $x\in E$.

By definition $\frac{1}{n}m(E_n)=\int_E \frac{1}{n}\chi_{E_n} \, dm \leq \int_E f \, dm =0$.

Therefore, $m(E_n)=0$ for all $n$.

Now $U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n$, and $E_n\subseteq E_{n+1}$ for all $n$.

Therefore, $m(U)=m(\bigcup_{n=1}^{\infty} E_n)=\lim_{n\to\infty} m(E_n)=0$.

QED

### Convergence Theorems

When does $\lim_{n\to\infty} \int_E f_n \, dm = \int_E \lim_{n\to\infty} f_n \, dm$?

#### Theorem 6.14 Monotone Convergence Theorem

Let $\{f_n\}$ be a monotone increasing sequence of measurable functions on $E$ and $f_n\to f$ almost everywhere on $E$. ($f_n(x)\leq f_{n+1}(x)$ for all $x\in E$ and $n$)

If there exists $A>0$ such that $\left|\int_E f_n \, dm\right|\leq A$ for all $n\in \mathbb{N}$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists for almost every $x\in E$ and it is integrable on $E$ and

$$
\int_E f \, dm = \lim_{n\to\infty} \int_E f_n \, dm
$$

Proof:

First to show the limit exists almost everywhere. It suffices to show

$$
\mathcal{U}=\{x\in E: f_n(x) \text{ is unbounded}\}
$$

has measure 0.

Let $\epsilon>0$ and write

$$
U=\bigcup_{n=1}^{\infty} E_n
$$

where $E_n=\{x\in E: |f_n(x)|\geq \epsilon\}$.

Then $U\subseteq \mathcal{U}$ and $m(U)<\epsilon$.


CONTINUE NEXT TIME.

QED

