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+---
+lecture: 15
+date: 2025-03-06
+---
+# Lebesgue's Monotone Convergence Theorem
+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 $m \to \infty$.
+> [!note]-
+> **Left Integral:**
+> $\int f \, d\mu = \sup_{0 \leq s \leq f} \int s \, d\mu$
+> 
+> **Right Integral:**
+> $\lim_{ m \to \infty } \int f_{m} \, d\mu = \int \lim_{ m \to \infty } f_{m} \, d\mu$
+## Proof
+Note that $f \equiv \lim_{ n \to \infty }f_{n} : X \to [0, \infty]$ is a [[Measurable|measurable]] function as
+> [!note]-
+> - $\equiv$ is [[Pointwise|pointwise]]
+> - To make $[0, \infty]$, you can do "$[0, a\rangle, \langle a, b \rangle, \langle b, \infty]$"
+
+$f^{-1}([0, a\rangle) = \cap_{n=1}^{\infty} \underbrace{f_{n}^{-1}([0, a \rangle)}_{\text{Measurable}}$
+> [!note]- More on the right, measurable, function
+> $x \in X$ such that
+> $f(x) \lt a$
+> $\implies f_{n}(x) \leq f(x) < a\ \forall n$
+> $f_{n}(x) \lt a\ \forall n$
+> $\implies f(x) < a$
+
+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 $A_{n} = \{ x \in X \, | \, c \times s(x) \leq f_{n}(x) \} = (\underbrace{f_{n} - cs}_{measurable function})^{-1}(\underbrace{[0, \infty]}_{open})$.
+
+Then $A_{1} \subset A_{2} \subset A_{3} \subset \dots$ [[Measurable|measurable]], and $\cup_{n} A_{n} \overbrace{=}^{\text{(*)}} X$
+> [!note] Continuing (\*)
+> Say $x \in X$.
+> If $f(x) = 0$, then $x \in A_{1}$.
+> 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 $x \in A_{n}$.
+> 
+> By the previous two lemmas, we have
+> $b \geq \lim_{ n \to \infty } \int_{A_{n}} f_{n} \, d\mu \geq \lim_{ n \to \infty } c \times \int_{A_{n}} s \, d\mu$
+> > [!note]- Note on the $A_{n}$
+> > $$\int_{A} \subset \int_{X}$$
+> > $$A \subset X$$
+> 
+> $= c \lim_{ n \to \infty } \int_{A_{n}} s \, d\mu \overbrace{=}^{\text{2 lemmas}} c \times \int_{\cup A_{n}} s \, d\mu = c \times \int s \, d\mu$, so $b \geq c \times \int f \, d\mu$ and $b \geq \int f \, d\mu$
+> 
+> > [!info]- Reminder of the two lemmas
+> > 1. $A \mapsto \int_{A} s \, d\mu$ [[Measure|measure]] ($s = 1 \implies \int_{A} s \, d\mu = \mu(A)$)
+> > 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.
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diff --git a/content/index.md b/content/index.md
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@@ -22,3 +22,4 @@ title: ACIT4330 Lecture Notes
 # Measure Theory
 - [[Lecture 13 - Measure Theory]]
 - [[Lecture 14]]
+- [[Lecture 15]]