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lecture	date
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

A \mapsto \int_{A} s \, d\mu Measure (s = 1 \implies \int_{A} s \, d\mu = \mu(A))

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

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

[!info] What is \lim\inf? Definition of Infimum (it is basically the opposite of a 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 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|)