---
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 $0 \leq s \leq f$, $s$ [[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 $\cup_{n} A_{n} \overbrace{=}^{\text{(*)}} X$
> [!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.
# 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]]
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![[Drawing 2025-03-06 12.16.07.excalidraw.dark.svg]]
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Continue like this.
![[Drawing 2025-03-06 12.23.12.excalidraw.dark.svg]]
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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]]
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Set $s_{n} = h_{n} \circ f$.

![[Drawing 2025-03-06 12.25.26.excalidraw.dark.svg]]
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