generated from smyalygames/quartz
65 lines
3.1 KiB
Markdown
65 lines
3.1 KiB
Markdown
---
|
|
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
|
|
$\mu(0) = 0$
|
|
$\mu(\cup_{n=1}^\infty A_{n}) = \Sigma_{n=1}^\infty \mu(A_{n})$
|
|
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 $s = \Sigma_{i=1}^{n} a_{i} \times X_{a_{i}}$ 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 $\int_{A} s \, d\mu \equiv \Sigma_{i=1}^{n} a_{i} \times \mu (A \cap A_{i})$ 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 $\mu(\cup A_{m}) = \mu(\cup B_{m}) = \Sigma_{m = 1}^{\infty} \mu (B_{m}) = \lim_{ N \to \infty } \Sigma_{m=1}^{N} \mu (B_{m})$.
|
|
# 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
|
|
$$
|
|
\int_{\cup B_{n}} s \, d\mu = \Sigma_{i} a_{i} \mu(\underbrace{A_{i} \cap (\cup B_{n})}_{\cup(A_{i} \cap B_{n})}) = \Sigma_{i} a_{i} \Sigma_{n}^{\infty} \mu (A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \Sigma_{i} a_{i} \mu(A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \int_{B_{n}} s \, d\mu
|
|
$$
|
|
QED |