Lecture 14

Recap

Going over Lecture 13 - Measure Theory

• Definition of Sigma-Algebra
  • Concept of Measurable function
  • If $F\subset \wp (X)$ the $\sigma$-algebra generated by $F$ is ${\cap }_{F\subset M}M$ $\sigma$-algebra
  • New?: Borel Sigma-Algebra
• Definition of Pointwise

Measure $\mu$ on $X$ with $M$:
$\mu :M\to [0,\infty ]$ such that
$\mu (0)=0$
$\mu ({\cup }_{n=1}^{\infty }{A}_n)={\sum }_{n=1}^{\infty }\mu (A_n)$
For $A_n\in M$ pairwise disjoint

Example $M=\wp (X)$ define measure:
$\mu (A)=\{\begin{array}{ll}\#A& \text{When}A\text{is finite}\\ \infty & \text{When}n\text{is infinite}\end{array}$

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Simple Function

A simple function on $X$ is a function $s:X\to \mathbb{R}$ of the form $s={\sum }_{i=1}^n{a}_i\times X_{a_i}$ for pairwise disjoint $A_i\subset X$ and distinct real numbers $a_i$

$X_A(x)=\{\begin{array}{ll}1& \text{If}x\in A\\ 0& \text{If}x\notin A\end{array}$

Note

If $X$ has $M$ then: $s$ is measurable $⟺$ all $A_i$ are measurable

$A_i=s^{-1}(\{a_i\})=s^{-1}(\mathbb{R}\setminus \{a_i\}{)}^{\complement }$

If we have a measure $\mu$ on $X$, define ${\int }_{A}sd\mu \equiv {\sum }_{i=1}^n{a}_i\times \mu (A\cap A_i)$ for $A\in M$ (they are all $\in [0,\infty ])$.

Lebesgue Integral

Defines Lebesgue Integral in the lecture

$f:X\to [0,\infty ]$ is ${\int }_{A}fd\mu \equiv {\sup }_{0\le s\le f}{\int }_{A}sd\mu$

Cut out $A$ when $A=X$.
$\int fd\mu \equiv \int f(x)d\mu (x)$
Behaves nice under limits
$\int \lim f_{n}d\mu =\lim \int f_{n}d\mu$

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Lemma

Given measure $\mu$, and
$A_1\subset A_2\subset A_3\subset \dots$ measurable sets.

What this subset of a subset of a subset... looks like

Then $\mu (A_n)\to \mu ({\cup }_{m=1}^{\infty }{A}_m)$ as $n\to \infty$.

Proof

Consider $B_n=A_n\setminus A_{n-1}$ with $A_0\equiv \varnothing$.
So $A_n=B_1\cup \cdots \cup B_n$, and $\cup B_n=\cup A_n$, and $B_n\cap B_m=\varnothing$ when $n\ne m$.
Hence $\mu (\cup A_m)=\mu (\cup B_m)={\sum }_{m=1}^{\infty }\mu (B_m)={\lim }_{N\to \infty }{\sum }_{m=1}^N\mu (B_m)$.

Lemma

$X$, $\mu$, $s:X\to [0,\infty ]$ measurable simple function.
Then $A\mapsto {\int }_{A}sd\mu$ defines a measure on $X$.

Proof

$A=\varnothing ⟹$ measure is $0$. Given continuous disjoint union $\cup B_n$ with $B_n$ measure, then

$${\int }_{\cup B_n}sd\mu =\sum _i{a}_i\mu (\underset{\cup (A_i\cap B_n)}{\underbrace{A_i\cap (\cup B_n)}})=\sum _i{a}_i\sum _n^{\infty }\mu (A_i\cap B_n)=\sum _n^{\infty }\sum _i{a}_i\mu (A_i\cap B_n)=\sum _n^{\infty }{\int }_{B_n}sd\mu$$

QED