ACIT4330-Page/content/Lectures/Lecture 15.md

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15	2025-03-06

Lebesgue's Monotone Convergence Theorem

Say X has a Measure \mu, and let f_{n} : X \to [0, \infty] be 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 function as

[!note]-

• \equiv is 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 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, 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 (s = 1 \implies \int_{A} s \, d\mu = \mu(A))
2. For any measure \nu and A_{1} \subset A_{2} \subset \dots 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 \mu on X, and f_{n} : X \to [0, \infty] 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 functions. QED

Lebesgue's Dominated Convergence Theorem

(Also defined Lebesgue's Dominated Convergence Theorem, 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.md, and the Drawing 2025-03-06 11.57.37.excalidraw.light.svg%%

Definition

Given 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 f : X \to [0, \infty].

Then \exists Measurable Simple Function 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

Proof

Define h_{n} : [0, \infty] \to [0, \infty \rangle by ! %%Drawing 2025-03-06 12.14.05.excalidraw.md, and the Drawing 2025-03-06 12.14.05.excalidraw.light.svg%% ! %%Drawing 2025-03-06 12.16.07.excalidraw.md, and the Drawing 2025-03-06 12.16.07.excalidraw.light.svg%% Continue like this. ! %%Drawing 2025-03-06 12.23.12.excalidraw.md, and the Drawing 2025-03-06 12.23.12.excalidraw.light.svg%%

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.md, and the Drawing 2025-03-06 12.24.31.excalidraw.light.svg%%

Set s_{n} = h_{n} \circ f.

! %%Drawing 2025-03-06 12.25.26.excalidraw.md, and the Drawing 2025-03-06 12.25.26.excalidraw.light.svg%%

---

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.md, and the Drawing 2025-03-06 13.17.23.excalidraw.light.svg%%

\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 and f : X \to Y set. Then the collection N of subsets A \subset Y such that f^{-1}(A) Measurable, is a $\sigma$Sigma-Algebra.

N \equiv \{ A \subset Y \, | \, f^{-1}(A) \in M \} prove that N is a $\sigma$Sigma-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. Show that if f is Measurable, then f^{-1}(A) is Borel Measurable for any Borel Sets A \subset Y.

Proof

Consider N = \{ A \subset Y \, | \, f^{-1}(A)\ \text{Borel} \}. This is a $\sigma$Sigma-Algebra.

[!example]- ! %%Drawing 2025-03-06 13.41.17.excalidraw.md, and the Drawing 2025-03-06 13.41.17.excalidraw.light.svg%%

It contains all the Open Sets in Y since f is Measurable and then f^{-1}(V) is Borel Measurable for V Open Sets.

Hence N contains all the Borel Sets in Y. If A is Borel Sets, then A \in N, so f^{-1}(A) is Borel Sets. (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 M. f: X \to [0, \infty] is Measurable \iff f^{-1}(\langle a, \infty]) \in M,\ \forall a \gt 0.

Proof

\Rightarrow

"Obvious".

\Leftarrow

In [0, \infty] Ball 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 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 f \equiv \sup f_{n} is Measurable

[!note] Can also use Infimum instead \inf f_{n} = -\sup(-f_{n})

Now need to check if f^{-1}(\langle a, \infty ]) is 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 \}