---
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
Defined in the lecture here: [[Fatou's Lemma]]
> [!info] Definition
> Have [[Measure|measure]] $\mu$ on $X$, and $f_{n} : X \to [0, \infty]$ [[Measurable|measurable]]. Then $\int \lim_{ n \to \infty } \inf f_{n} \, d\mu \le \lim_{ n \to \infty } \inf \int f_{n} \, d\mu$
## 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]]
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# 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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---
Exercise part of the session.

These exercises are from Exercise 8.
# Question 3
Prove $\mu(A) \leq \mu(B)$ when $A \subset B$
## Proof
Have $B = A \cup (\underbrace{A^{\complement} \cap B}_{B \setminus A})$, so
> [!example]- What this set looks like
> ![[Drawing 2025-03-06 13.17.23.excalidraw.dark.svg]]
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$\mu(B) = \mu(A) + \underbrace{\mu(B \setminus A)}_{\geq 0}$
$\implies \mu(B) >+ \mu(A)$.
# Question 4
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]].

$N \equiv \{ A \subset Y \, | \, f^{-1}(A) \in M \}$
prove that $N$ is a $\sigma$[[Sigma-Algebra|-algebra]].
## Proof
1. Have $\emptyset \in N$ since $f^{-1}(\emptyset) = \emptyset \in M$.
2. If $A \in N$, then $f^{-1}(A) \in M$, so $f^{-1}(A)^{\complement} \in M = f^{-1}(A^{\complement}) \implies A^{\complement} \in N$.
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 $\cup A_{n} \in N$.

# A question I don't know the number of
Say $f : X \to Y$ and they are [[Topological Space|topological spaces]].
Show that if $f$ is [[Measurable|measurable]], then $f^{-1}(A)$ is [[Borel Measurable|Borel measurable]] for any [[Borel Sets|Borel set]] $A \subset Y$.
## Proof
Consider $N = \{ A \subset Y \, | \, f^{-1}(A)\ \text{Borel} \}$. This is a $\sigma$[[Sigma-Algebra|-algebra]].
> [!example]-
> ![[Drawing 2025-03-06 13.41.17.excalidraw.dark.svg]]
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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]].

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)
# Question 5
$X$ with $\sigma$[[Sigma-Algebra|-algebra]] $M$.
$f: X \to [0, \infty]$ is [[Measurable|measurable]] $\iff f^{-1}(\langle a, \infty]) \in M,\ \forall a \gt 0$.
## Proof
### $\Rightarrow$
"Obvious".
### $\Leftarrow$
In $[0, \infty]$ [[Ball|balls]] are $[0, a\rangle$, $\langle a, \infty ]$, $[0, \infty]$, or $\langle a, b \rangle$ for $a, b \in \mathbb{R}$.

Any open set in $[0, \infty]$ is a [[Countable|countable]] union of these "building blocks".

Checking they belong to $M$:
- $[0, a]$
- $[0, b\rangle \cap \langle a, \infty = \langle a, b \rangle$
- $[0, a \rangle = \cup_{n=1}^{\infty}\left[ 0, a-\frac{1}{n} \right]$
QED.

$f_{n} : X \to [0, \infty]$ is [[Measurable|measurable]]
$f \equiv \sup f_{n}$ is [[Measurable|measurable]]
> [!note] Can also use [[Infimum|infimum]] instead
> $\inf f_{n} = -\sup(-f_{n})$

Now need to check if $f^{-1}(\langle a, \infty ])$ is [[Measurable|measurable]] and is in $M$

$f^{-1}(\langle a, \infty ]) = \{ x \in X \, | \, f(x) \gt a \}$
$= \cup_{n}f_{n}^{-1}(\langle a, \infty ]) \in M$
Why is the set and the union both the same?
$\cup_{n} \{ x \in X \, | \, f_{n}(x) \gt a \} = \{ x \in X \, | \, f_{n}(x) \gt a\ \forall n \}$