Math4121 Lecture 34

Important:

$\mathfrak{M}=\{S\subset \mathbb{R}: S \text{ satisfies the caratheodory condition}\}$ , that is, for any $X$ of finite outer measure,

$m_e(X)=m_e(X\cap S)+m_e(X\cap S^c)$

In particular, the measure of sets can be infinite, not necessarily bounded. (We want to make the real line measurable.)

Lebesgue Integral

Simple Function

A function $\phi$ is called a simple function if

\phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x)

where $a_i\in \mathbb{R}$ and $\chi_{S_i}=\begin{cases}1, & x\in S_i \\ 0, & x\notin S_i\end{cases}$

where $\{S_i\}_{i=1}^{n}$ are pairwise disjoint each having finite measure.

constant function is not simple ( $\mathbb{R}$ is not finite measurable sets.)

Theorem 6.6

A function $f$ is measurable on $[a,b]$ if and only if there exists a sequence of simple functions $\{\phi_n\}$ such that $\lim_{n\to\infty} \phi_n(x)=f(x)$ almost everywhere on $[a,b]$ .

Proof:

Partition $[-n,n]$ into $n2^{n+1}$ pieces.

(These are just horizontal strips from $-n$ to $n$ with width $\frac{1}{2^n}$ .)

E_{n,k}=\{x\in[-n,n]:\frac{k}{2^n}\leq f(x)<\frac{k+1}{2^n}\}

for $-n2^n<k<n2^n$

E_{n,n2^n}=\{x\in[-n,n]:f(x)\geq n\}

E_{n,-n2^n}=\{x\in[-n,n]:f(x)<\frac{-n2^n+1}{2^n}\}

\phi_n(x)=\frac{k}{2^n}\chi_{E_{n,k}}(x)

is a simple function.

We need to justify that $\phi_n(x)\to f(x)$ for all $x\in\mathbb{R}$ .

Let $x\in\mathbb{R}$ . And choose $n_0$ large such that $x\in [-n_0,n_0]$ and $f(x)\in [-n_0,n_0]$ .

Then, for $n\geq n_0$ ,

|\phi_n(x)-f(x)|<\frac{1}{2^n}\to 0

as $n\to\infty$ .

QED

Integration

Given a measurable set $E$ and a simple function $\phi$ , we define

\int_E \phi dm=\sum_{i=1}^{n} a_i m(E\cap S_i)

Properties 6.10

Let $\phi$ and $\psi$ be simple functions, $c\in \mathbb{R}$ , and $E=E_1\cup E_2$ where $E_1\cap E_2=\emptyset$ and $E_1,E_2\in \mathfrak{M}$ . Then,

$\int_E c\phi dm=c\int_E \phi dm$ (linearity)
$\int_E (\phi+\psi)dm=\int_E \phi dm+\int_E \psi dm$ (additivity of simple functions)
if $\phi(x)\leq \psi(x)$ for all $x\in E$ , then $\int_E \phi dm\leq \int_E \psi dm$ (monotonicity)
$\int_E \phi(x)dm=\int_{E_1} \phi(x)dm+\int_{E_2} \phi(x)dm$ (additivity over disjoint measurable sets)

Proof:

Let $\phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x)$ and $\psi(x)=\sum_{j=1}^{m} b_j \chi_{T_j}(x)$ .

\phi+\psi=\sum_{i=1}^{n} a_i \chi_{S_i}+\sum_{j=1}^{m} b_j \chi_{T_j}

Without loss of generality, we may assume that $x\in E$ , $\bigcup_{i=1}^{n} S_i=\bigcup_{j=1}^{m} T_j=E$ .

\phi+\psi=\sum_{i,j=1}^{n,m}(a_i+b_j) \chi_{S_i\cup T_j}

is a simple function.

\begin{aligned} \int_E (\phi+\psi)dm&=\sum_{i,j=1}^{n,m}(a_i+b_j) m(E\cap S_i\cup T_j) \\ &=\sum_{i=1}^{n} a_i \sum_{j=1}^{m} m(E\cap S_i\cup T_j)+\sum_{j=1}^{m} b_j \sum_{i=1}^{n} m(E\cap S_i\cup T_j) \\ &=\sum_{i=1}^{n} a_i m(E\cap S_i)+\sum_{j=1}^{m} b_j m(E\cap T_j) \\ &=\int_E \phi dm+\int_E \psi dm \end{aligned}

\phi(x)=\sum_{i=1}^{n} a_i\sum_{j=1}^{m} \chi_{S_i\cap T_j}(x)

\psi(x)=\sum_{i=1}^{n} b_i\sum_{j=1}^{m} \chi_{S_i\cap T_j}(x)

If $x\in S_i\cap T_j$ , then $\phi(x)=a_i$ and $\psi(x)=b_j$ , therefore $a_i\leq b_j$ .

So,

\begin{aligned} \int_E \phi dm&=\sum_{i=1}^{n} \sum_{j=1}^{m} a_i m(E\cap S_i\cap T_j) \\ &\leq \sum_{i=1}^{n} \sum_{j=1}^{m} b_i m(E\cap S_i\cap T_j) \\ &=\int_E \psi dm \end{aligned}

QED

Back on Wednesday.