Quartz sync: Mar 6, 2025, 1:07 PM

.gitignore

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+content/Excalidraw/**/*.md

content/Definitions/Functions/Infimum.md

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+# Definition
+Say we have $A_{m} = \{ x_{n}\, | \, n \geq m \}$
+Then $\inf A =$ greatest lower bound of $A$.
+> [!note] What is the "lower bound"?
+> $c \lt a,\ \forall a \in A$
+
+The infimum is denoted by $\inf$.

content/Definitions/Functions/Supremum.md

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+# Definition
+Say we have $A_{m} = \{ x_{n}\, | \, n \geq m \}$
+Then $\sup A$ is the least upper bound of $A$.
+
+The **supremum** is denoted by $\sup$.

content/Definitions/Measure Theory/Lebesgue's Dominated Convergence Theorem.md

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+# Definition
+Let $g$ be a real function on $X$.
+
+Define $g^{+} = \max \{ g, 0 \}$, $g^{-} = -\min \{ g, 0 \}$.
+
+Then $g = g^{+} - g^{-}$ and $g^{\pm} \geq 0$.
+
+> [!example]-
+> ![[Drawing 2025-03-06 11.57.37.excalidraw.dark.svg]]
+%%[[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]]%%

content/Definitions/Measure Theory/Lebesgue's Monotone Convergence Theorem.md

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+# Definition
+Say $X$ has a [[Measure|measure]] $\mu$, and let $f_{n} : X \to [0, \infty]$ be [[Measurable|measurable]] and $f_{1} \leq f_{2} \leq f_{3} \leq \dots$. 
+
+Then $\int f_{m} \, d\mu \to \int \lim_{ n \to \infty } f_{n} \, d\mu$  as $m \to \infty$.

content/Lectures/Lecture 15.md

@@ -27,7 +27,7 @@ $f^{-1}([0, a\rangle) = \cap_{n=1}^{\infty} \underbrace{f_{n}^{-1}([0, a \rangle
 > $\implies f(x) < a$
 
 Let $b \equiv \lim_{ n \to \infty } \int f_{n} \, d\mu \leq \int f \, d\mu$ as $f_{n} \leq f$.
-Let $0 \leq s \leq f$, $s$ [[Measure|measure]] simple, and $c \in \langle 0, 1 \rangle$.
+Let $0 \leq s \leq f$, $s$ [[Measurable|measurable]] [[Simple Function|simple function]], and $c \in \langle 0, 1 \rangle$.
 Let $A_{n} = \{ x \in X \, | \, c \times s(x) \leq f_{n}(x) \} = (\underbrace{f_{n} - cs}_{measurable function})^{-1}(\underbrace{[0, \infty]}_{open})$.
 
 Then $A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]], and $\cup_{n} A_{n} \overbrace{=}^{\text{(*)}} X$
@@ -49,3 +49,64 @@ Then $A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]]
 > > 2. For any measure $\nu$ and $A_{1} \subset A_{2} \subset \dots$ [[Measurable|measurable]] $\implies \nu(\cup A_{n}) = \lim_{ n \to \infty } \nu(A_{n})$
 
 QED.
+# Corollary - Fatou's Lemma
+Have [[Measure|measure]] $\mu$ on $X$, and $f_{n} : X \to [0, \infty]$ [[Measurable|measurable]]. Then $\int \lim_{ n \to \infty } \inf f_{n} \, d\mu \le \lim_{ n \to \infty } \inf \int f_{n} \, d\mu$
+> [!info] What is $\lim\inf$?
+> Definition of [[Infimum|infimum]] (it is basically the opposite of a [[Supremum|supremum]]).
+> 
+> $\{ x_{n} \} \subset [0, \infty]$
+> $\lim_{ n \to \infty }\inf x_{n} = \sup_{m}\inf_{n \geq m} x_{n}$
+> 
+> $\inf_{n \geq m} = y_{m} \leq y_{m+1} \leq \dots$
+
+## Proof
+Use [[Lebesgue's Monotone Convergence Theorem]] on $g_{m} = \inf_{n \geq m} f_{n}$.
+$g_{1} \leq g_{2} \leq \dots$ are [[Measurable|measurable]] functions.
+QED
+# Lebesgue's Dominated Convergence Theorem
+(Also defined [[Lebesgue's Dominated Convergence Theorem|here]], it's the same thing)
+
+Let $g$ be a real function on $X$.
+
+Define $g^{+} = \max \{ g, 0 \}$, $g^{-} = -\min \{ g, 0 \}$.
+
+Then $g = g^{+} - g^{-}$ and $g^{\pm} \geq 0$.
+> [!example]-
+> ![[Drawing 2025-03-06 11.57.37.excalidraw.dark.svg]]
+%%[[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]]%%
+# Definition
+Given [[Measure|measure]] $\mu$ on $X$.
+Define $L'(\mu) = \left\{  f : X \to \mathbb{C}\ \text{measurable and}\ \int |f| \, d\mu \lt \infty  \right\}$.
+
+Define integral for $f \in L'(\mu)$ by $\int f \, d\mu \equiv \int (\mathrm{Re}f)^{+} \, d\mu - \int (\mathrm{Re}f)^{-} \, d\mu + i \int (\mathrm{Im} f)^{+} \, d\mu - i \int (\mathrm{Im} f)^{-} \, d\mu$.
+
+Use $f = \mathrm{Re} f + i \mathrm{Im} f = (\mathrm{Re} f)^{+} - (\mathrm{Re} f)^{-} + i((\mathrm{Im} f)^{+} - (\mathrm{Im} f)^{-})$.
+
+The integral definition makes sense as each integral on the RHS is finite.
+($(\mathrm{Re} f)^{+} \leq |f|$)
+## Lemma
+Given [[Measure|measure]] $f : X \to [0, \infty]$.
+
+Then $\exists$ [[Measurable|measurable]] [[Simple Function|simple functions]] $s_{n}$ such that
+1. $0 \leq s_{1} \leq s_{2} \leq \dots \leq f$
+2. $\lim_{ n \to \infty } s_{n} = f$ [[Pointwise|pointwise]]
+### Proof
+Define $h_{n} : [0, \infty] \to [0, \infty \rangle$ by
+![[Drawing 2025-03-06 12.14.05.excalidraw.dark.svg]]
+%%[[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]]%%
+![[Drawing 2025-03-06 12.16.07.excalidraw.dark.svg]]
+%%[[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]]%%
+Continue like this.
+![[Drawing 2025-03-06 12.23.12.excalidraw.dark.svg]]
+%%[[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]]%%
+
+Have $0\leq h_{1} \leq h_{2} \leq \dots \leq h_{n} \to l\ \text{as}\ n \to \infty$.
+
+![[Drawing 2025-03-06 12.24.31.excalidraw.dark.svg]]
+%%[[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]]%%
+
+Set $s_{n} = h_{n} \circ f$.
+
+![[Drawing 2025-03-06 12.25.26.excalidraw.dark.svg]]
+%%[[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]]%%
+