---
lecture: 14
date: 2025-03-03
---
# 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:
$\mu(A) = \begin{cases}\#A & \text{When}\ A\ \text{is finite}\\ \infty & \text{When}\ n\ \text{is infinite}\end{cases}$

---
# 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 $a_{i}$

$X_{A}(x) = \begin{cases}1 & \text{If}\ x \in A\\ 0 & \text{If}\ x \notin A\end{cases}$

> [!note]
> If $X$ has $M$ then: $s$ is [[Measurable|measurable]] $\iff$ all $A_{i}$ are [[Measurable|measurable]]
> 
> $A_{i} = s^{-1}(\{ a_{i} \}) = s^{-1}(\mathbb{R} \setminus \{ a_{i} \})^{\complement}$

If we have a [[Measure|measure]] $\mu$ on $X$, define $\int_{A} s \, d\mu \equiv \sum_{i=1}^{n} a_{i} \times \mu (A \cap A_{i})$ for $A \in M$  (they are all $\in [0, \infty])$.
# Lebesgue Integral
Defines [[Lebesgue Integral]] in the lecture

$f : X \to [0, \infty]$ is $\int_{A} f \, d\mu \equiv \sup_{0 \leq s \leq f} \int_{A} s \, d\mu$

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

---
# Lemma
Given [[Measure|measure]] $\mu$, and
$A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]] sets.
> [!info]- What this subset of a subset of a subset... looks like
> ![[Drawing 2025-03-03 11.49.17.excalidraw.dark.svg]]
%%[[Drawing 2025-03-03 11.49.17.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-03 11.49.17.excalidraw.light.svg|light exported image]]%%

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 $A_{0} \equiv \emptyset$.
So $A_{n} = B_{1} \cup \dots \cup B_{n}$, and $\cup B_{n} = \cup A_{n}$, and $B_{n} \cap B_{m} = \emptyset$ when $n \neq m$.
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|measurable]] [[Simple Function|simple function]].
Then $A \mapsto \int_{A} s \, d\mu$ defines a [[Measure|measure]] on $X$.
## Proof
$A = \emptyset \implies$ measure is $0$. Given [[Continuous|continuous]] disjoint union $\cup B_{n}$ with $B_{n}$ [[Measure|measure]], then
$$
\int_{\cup B_{n}} s \, d\mu = \sum_{i} a_{i} \mu(\underbrace{A_{i} \cap (\cup B_{n})}_{\cup(A_{i} \cap B_{n})}) = \sum_{i} a_{i} \sum_{n}^{\infty} \mu (A_{i} \cap B_{n}) = \sum_{n}^{\infty} \sum_{i} a_{i} \mu(A_{i} \cap B_{n}) = \sum_{n}^{\infty} \int_{B_{n}} s \, d\mu 
$$
QED