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172 lines
8.9 KiB
Markdown
172 lines
8.9 KiB
Markdown
---
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lecture: 15
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date: 2025-03-06
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---
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# Lebesgue's Monotone Convergence Theorem
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Say $X$ has a [[Measure|measure]] $\mu$, and let $f_{n} : X \to [0, \infty]$ be [[Measurable|measurable]] and $f_{1} \leq f_{2} \leq f_{3} \leq \dots$.
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Then $\int f_{m} \, d\mu \to \int \lim_{ n \to \infty } f_{n} \, d\mu$ as $m \to \infty$.
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> [!note]-
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> **Left Integral:**
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> $\int f \, d\mu = \sup_{0 \leq s \leq f} \int s \, d\mu$
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>
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> **Right Integral:**
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> $\lim_{ m \to \infty } \int f_{m} \, d\mu = \int \lim_{ m \to \infty } f_{m} \, d\mu$
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## Proof
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Note that $f \equiv \lim_{ n \to \infty }f_{n} : X \to [0, \infty]$ is a [[Measurable|measurable]] function as
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> [!note]-
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> - $\equiv$ is [[Pointwise|pointwise]]
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> - To make $[0, \infty]$, you can do "$[0, a\rangle, \langle a, b \rangle, \langle b, \infty]$"
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$f^{-1}([0, a\rangle) = \cap_{n=1}^{\infty} \underbrace{f_{n}^{-1}([0, a \rangle)}_{\text{Measurable}}$
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> [!note]- More on the right, measurable, function
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> $x \in X$ such that
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> $f(x) \lt a$
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> $\implies f_{n}(x) \leq f(x) < a\ \forall n$
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> $f_{n}(x) \lt a\ \forall n$
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> $\implies f(x) < a$
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Let $b \equiv \lim_{ n \to \infty } \int f_{n} \, d\mu \leq \int f \, d\mu$ as $f_{n} \leq f$.
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Let $0 \leq s \leq f$, $s$ [[Measurable|measurable]] [[Simple Function|simple function]], and $c \in \langle 0, 1 \rangle$.
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Let $A_{n} = \{ x \in X \, | \, c \times s(x) \leq f_{n}(x) \} = (\underbrace{f_{n} - cs}_{measurable function})^{-1}(\underbrace{[0, \infty]}_{open})$.
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Then $A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]], and $\cup_{n} A_{n} \overbrace{=}^{\text{(*)}} X$
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> [!note] Continuing (\*)
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> Say $x \in X$.
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> If $f(x) = 0$, then $x \in A_{1}$.
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> If $f(x) \gt 0$, then $c \times s(x) \lt f(x)$, so $c \times s(x) \lt f_{n}(x)$ for some $n$, and $x \in A_{n}$.
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>
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> By the previous two lemmas, we have
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> $b \geq \lim_{ n \to \infty } \int_{A_{n}} f_{n} \, d\mu \geq \lim_{ n \to \infty } c \times \int_{A_{n}} s \, d\mu$
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> > [!note]- Note on the $A_{n}$
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> >
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> > $$\int_{A} \subset \int_{X}$$
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> > $$A \subset X$$
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>
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> $= c \lim_{ n \to \infty } \int_{A_{n}} s \, d\mu \overbrace{=}^{\text{2 lemmas}} c \times \int_{\cup A_{n}} s \, d\mu = c \times \int s \, d\mu$, so $b \geq c \times \int f \, d\mu$ and $b \geq \int f \, d\mu$
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>
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> > [!info]- Reminder of the two lemmas
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> > 1. $A \mapsto \int_{A} s \, d\mu$ [[Measure|measure]] ($s = 1 \implies \int_{A} s \, d\mu = \mu(A)$)
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> > 2. For any measure $\nu$ and $A_{1} \subset A_{2} \subset \dots$ [[Measurable|measurable]] $\implies \nu(\cup A_{n}) = \lim_{ n \to \infty } \nu(A_{n})$
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QED.
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# Corollary - Fatou's Lemma
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Defined in the lecture here: [[Fatou's Lemma]]
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> [!info] Definition
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> Have [[Measure|measure]] $\mu$ on $X$, and $f_{n} : X \to [0, \infty]$ [[Measurable|measurable]]. Then $\int \lim_{ n \to \infty } \inf f_{n} \, d\mu \le \lim_{ n \to \infty } \inf \int f_{n} \, d\mu$
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## Proof
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Use [[Lebesgue's Monotone Convergence Theorem]] on $g_{m} = \inf_{n \geq m} f_{n}$.
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$g_{1} \leq g_{2} \leq \dots$ are [[Measurable|measurable]] functions.
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QED
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# Lebesgue's Dominated Convergence Theorem
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(Also defined [[Lebesgue's Dominated Convergence Theorem|here]], it's the same thing)
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Let $g$ be a real function on $X$.
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Define $g^{+} = \max \{ g, 0 \}$, $g^{-} = -\min \{ g, 0 \}$.
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Then $g = g^{+} - g^{-}$ and $g^{\pm} \geq 0$.
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> [!example]-
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> ![[Drawing 2025-03-06 11.57.37.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 11.57.37.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 11.57.37.excalidraw.light.svg|light exported image]]%%
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# Definition
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Given [[Measure|measure]] $\mu$ on $X$.
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Define $L'(\mu) = \left\{ f : X \to \mathbb{C}\ \text{measurable and}\ \int |f| \, d\mu \lt \infty \right\}$.
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Define integral for $f \in L'(\mu)$ by $\int f \, d\mu \equiv \int (\mathrm{Re}f)^{+} \, d\mu - \int (\mathrm{Re}f)^{-} \, d\mu + i \int (\mathrm{Im} f)^{+} \, d\mu - i \int (\mathrm{Im} f)^{-} \, d\mu$.
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Use $f = \mathrm{Re} f + i \mathrm{Im} f = (\mathrm{Re} f)^{+} - (\mathrm{Re} f)^{-} + i((\mathrm{Im} f)^{+} - (\mathrm{Im} f)^{-})$.
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The integral definition makes sense as each integral on the RHS is finite.
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($(\mathrm{Re} f)^{+} \leq |f|$)
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## Lemma
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Given [[Measure|measure]] $f : X \to [0, \infty]$.
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Then $\exists$ [[Measurable|measurable]] [[Simple Function|simple functions]] $s_{n}$ such that
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1. $0 \leq s_{1} \leq s_{2} \leq \dots \leq f$
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2. $\lim_{ n \to \infty } s_{n} = f$ [[Pointwise|pointwise]]
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### Proof
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Define $h_{n} : [0, \infty] \to [0, \infty \rangle$ by
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![[Drawing 2025-03-06 12.14.05.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 12.14.05.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 12.14.05.excalidraw.light.svg|light exported image]]%%
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![[Drawing 2025-03-06 12.16.07.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 12.16.07.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 12.16.07.excalidraw.light.svg|light exported image]]%%
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Continue like this.
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![[Drawing 2025-03-06 12.23.12.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 12.23.12.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 12.23.12.excalidraw.light.svg|light exported image]]%%
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Have $0\leq h_{1} \leq h_{2} \leq \dots \leq h_{n} \to l\ \text{as}\ n \to \infty$.
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![[Drawing 2025-03-06 12.24.31.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 12.24.31.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 12.24.31.excalidraw.light.svg|light exported image]]%%
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Set $s_{n} = h_{n} \circ f$.
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![[Drawing 2025-03-06 12.25.26.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 12.25.26.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 12.25.26.excalidraw.light.svg|light exported image]]%%
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---
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Exercise part of the session.
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These exercises are from Exercise 8.
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# Question 3
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Prove $\mu(A) \leq \mu(B)$ when $A \subset B$
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## Proof
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Have $B = A \cup (\underbrace{A^{\complement} \cap B}_{B \setminus A})$, so
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> [!example]- What this set looks like
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> ![[Drawing 2025-03-06 13.17.23.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 13.17.23.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 13.17.23.excalidraw.light.svg|light exported image]]%%
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$\mu(B) = \mu(A) + \underbrace{\mu(B \setminus A)}_{\geq 0}$
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$\implies \mu(B) >+ \mu(A)$.
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# Question 4
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Show that if $X$ has a $\sigma$[[Sigma-Algebra|-algebra]] and $f : X \to Y$ set. Then the collection $N$ of subsets $A \subset Y$ such that $f^{-1}(A)$ [[Measurable|measurable]], is a $\sigma$[[Sigma-Algebra|-algebra]].
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$N \equiv \{ A \subset Y \, | \, f^{-1}(A) \in M \}$
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prove that $N$ is a $\sigma$[[Sigma-Algebra|-algebra]].
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## Proof
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1. Have $\emptyset \in N$ since $f^{-1}(\emptyset) = \emptyset \in M$.
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2. If $A \in N$, then $f^{-1}(A) \in M$, so $f^{-1}(A)^{\complement} \in M = f^{-1}(A^{\complement}) \implies A^{\complement} \in N$.
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3. If $A_{n} \in N$, then $f^{-1}(A_{n} \in M)$, so $f^{-1}(\cup A_{n}) = \cup f^{-1}(A_{n}) \in M$, so $\cup A_{n} \in N$.
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# A question I don't know the number of
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Say $f : X \to Y$ and they are [[Topological Space|topological spaces]].
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Show that if $f$ is [[Measurable|measurable]], then $f^{-1}(A)$ is [[Borel Measurable|Borel measurable]] for any [[Borel Sets|Borel set]] $A \subset Y$.
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## Proof
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Consider $N = \{ A \subset Y \, | \, f^{-1}(A)\ \text{Borel} \}$. This is a $\sigma$[[Sigma-Algebra|-algebra]].
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> [!example]-
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> ![[Drawing 2025-03-06 13.41.17.excalidraw.dark.svg]]
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%%[[Drawing 2025-03-06 13.41.17.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 13.41.17.excalidraw.light.svg|light exported image]]%%
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It contains all the [[Open Sets|open sets]] in $Y$ since $f$ is [[Measurable|measurable]] and then $f^{-1}(V)$ is [[Borel Measurable|Borel measurable]] for $V$ [[Open Sets|open]].
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Hence $N$ contains all the [[Borel Sets|Borel sets]] in $Y$. If $A$ is [[Borel Sets|Borel]], then $A \in N$, so $f^{-1}(A)$ is [[Borel Sets|Borel]]. (**Note**: not sure if on this if the "Borel"s are about them being Borel sets or Borel measurable)
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# Question 5
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$X$ with $\sigma$[[Sigma-Algebra|-algebra]] $M$.
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$f: X \to [0, \infty]$ is [[Measurable|measurable]] $\iff f^{-1}(\langle a, \infty]) \in M,\ \forall a \gt 0$.
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## Proof
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### $\Rightarrow$
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"Obvious".
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### $\Leftarrow$
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In $[0, \infty]$ [[Ball|balls]] are $[0, a\rangle$, $\langle a, \infty ]$, $[0, \infty]$, or $\langle a, b \rangle$ for $a, b \in \mathbb{R}$.
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Any open set in $[0, \infty]$ is a [[Countable|countable]] union of these "building blocks".
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Checking they belong to $M$:
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- $[0, a]$
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- $[0, b\rangle \cap \langle a, \infty = \langle a, b \rangle$
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- $[0, a \rangle = \cup_{n=1}^{\infty}\left[ 0, a-\frac{1}{n} \right]$
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QED.
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$f_{n} : X \to [0, \infty]$ is [[Measurable|measurable]]
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$f \equiv \sup f_{n}$ is [[Measurable|measurable]]
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> [!note] Can also use [[Infimum|infimum]] instead
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> $\inf f_{n} = -\sup(-f_{n})$
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Now need to check if $f^{-1}(\langle a, \infty ])$ is [[Measurable|measurable]] and is in $M$
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$f^{-1}(\langle a, \infty ]) = \{ x \in X \, | \, f(x) \gt a \}$
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$= \cup_{n}f_{n}^{-1}(\langle a, \infty ]) \in M$
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Why is the set and the union both the same?
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$\cup_{n} \{ x \in X \, | \, f_{n}(x) \gt a \} = \{ x \in X \, | \, f_{n}(x) \gt a\ \forall n \}$
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