generated from smyalygames/quartz
64 lines
3.1 KiB
Markdown
64 lines
3.1 KiB
Markdown
---
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lecture: 14
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date: 2025-03-03
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---
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# Recap
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Going over [[Lecture 13 - Measure Theory]]
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- Definition of [[Sigma-Algebra]]
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- Concept of [[Measurable]] function
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- If $F \subset \wp(X)$ the $\sigma$-algebra generated by $F$ is $\cap_{F \subset M} M$ $\sigma$-algebra
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- New?: [[Borel Sigma-Algebra]]
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- Definition of [[Pointwise]]
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Measure $\mu$ on $X$ with $M$:
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$\mu : M \to [0, \infty]$ such that
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$\mu(0) = 0$
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$\mu(\cup_{n=1}^\infty A_{n}) = \Sigma_{n=1}^\infty \mu(A_{n})$
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For $A_{n} \in M$ **pairwise disjoint**
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Example $M = \wp(X)$ define measure:
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$\mu(A) = \begin{cases}\#A & \text{When}\ A\ \text{is finite}\\ \infty & \text{When}\ n\ \text{is infinite}\end{cases}$
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---
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# Simple Function
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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 $a_{i}$
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$X_{A}(x) = \begin{cases}1 & \text{If}\ x \in A\\ 0 & \text{If}\ x \notin A\end{cases}$
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> [!note]
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> If $X$ has $M$ then: $s$ is [[Measurable|measurable]] $\iff$ all $A_{i}$ are [[Measurable|measurable]]
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>
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> $A_{i} = s^{-1}(\{ a_{i} \}) = s^{-1}(\mathbb{R} \setminus \{ a_{i} \})^{\complement}$
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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 $\in [0, \infty])$.
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# Lebesgue Integral
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Defines [[Lebesgue Integral]] in the lecture
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$f : X \to [0, \infty]$ is $\int_{A} f \, d\mu \equiv \sup_{0 \leq s \leq f} \int_{A} s \, d\mu$
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Cut out $A$ when $A = X$.
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$\int f \, d\mu \equiv \int f(x) \, d\mu(x)$
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Behaves nice under limits
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$\int \lim f_{n} \, d\mu = \lim \int f_{n} \, d\mu$
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---
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# Lemma
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Given [[Measure|measure]] $\mu$, and
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$A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]] sets.
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> [!info]- What this subset of a subset of a subset... looks like
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> ![[Drawing 2025-03-03 11.49.17.excalidraw.dark.svg]]
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%%[[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]]%%
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Then $\mu(A_{n}) \to \mu(\cup_{m = 1}^{\infty} A_{m})$ as $n \to \infty$.
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## Proof
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Consider $B_{n} = A_{n} \setminus A_{n-1}$ with $A_{0} \equiv \emptyset$.
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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 $n \neq m$.
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Hence $\mu(\cup A_{m}) = \mu(\cup B_{m}) = \Sigma_{m = 1}^{\infty} \mu (B_{m}) = \lim_{ N \to \infty } \Sigma_{m=1}^{N} \mu (B_{m})$.
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# Lemma
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$X$, $\mu$, $s : X \to [0, \infty]$ [[Measurable|measurable]] [[Simple Function|simple function]].
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Then $A \mapsto \int_{A} s \, d\mu$ defines a [[Measure|measure]] on $X$.
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## Proof
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$A = \emptyset \implies$ measure is $0$. Given [[Continuous|continuous]] disjoint union $\cup B_{n}$ with $B_{n}$ [[Measure|measure]], then
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$$
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\int_{\cup B_{n}} s \, d\mu = \Sigma_{i} a_{i} \mu(\underbrace{A_{i} \cap (\cup B_{n})}_{\cup(A_{i} \cap B_{n})}) = \Sigma_{i} a_{i} \Sigma_{n}^{\infty} \mu (A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \Sigma_{i} a_{i} \mu(A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \int_{B_{n}} s \, d\mu
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$$
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QED |