ACIT4330-Page/content/Lectures/Lecture 14.md

lecture, date

lecture	date
14	2025-03-03

Recap

Going over Lecture 13 - Measure Theory

• Definition of Sigma-Algebra
  • Concept of Measurable function
  • If F \subset \wp(X) the $\sigma$-algebra generated by F is \cap_{F \subset M} M $\sigma$-algebra
  • New?: Borel Sigma-Algebra
• Definition of Pointwise

Measure \mu on X with M: \mu : M \to [0, \infty] such that \mu(0) = 0 \mu(\cup_{n=1}^\infty A_{n}) = \Sigma_{n=1}^\infty \mu(A_{n}) For A_{n} \in M pairwise disjoint

Example M = \wp(X) define measure: \mu(A) = \begin{cases}\#A & \text{When}\ A\ \text{is finite}\\ \infty & \text{When}\ n\ \text{is infinite}\end{cases}

Simple Function

A simple function on X is a function s : X \to \mathbb{R} of the form s = \Sigma_{i=1}^{n} a_{i} \times X_{a_{i}} for pairwise disjoint A_{i} \subset X and distinct real numbers a_{i}

X_{A}(x) = \begin{cases}1 & \text{If}\ x \in A\\ 0 & \text{If}\ x \notin A\end{cases}

Note

If X has M then: s is Measurable \iff all A_{i} are Measurable

A_{i} = s^{-1}(\{ a_{i} \}) = s^{-1}(\mathbb{R} \setminus \{ a_{i} \})^{\complement}

If we have a Measure \mu on X, define \int_{A} s \, d\mu \equiv \Sigma_{i=1}^{n} a_{i} \times \mu (A \cap A_{i}) for A \in M (they are all \in [0, \infty]).

Lebesgue Integral

Defines Lebesgue Integral in the lecture

f : X \to [0, \infty] is \int_{A} f \, d\mu \equiv \sup_{0 \leq s \leq f} \int_{A} s \, d\mu

Cut out A when A = X. \int f \, d\mu \equiv \int f(x) \, d\mu(x) Behaves nice under limits \int \lim f_{n} \, d\mu = \lim \int f_{n} \, d\mu

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Lemma

Given Measure \mu, and A_{1} \subset A_{2} \subset A_{3} \subset \dots Measurable sets.

[!info]- What this subset of a subset of a subset... looks like ! %%Drawing 2025-03-03 11.49.17.excalidraw.md, and the Drawing 2025-03-03 11.49.17.excalidraw.light.svg%%

Then \mu(A_{n}) \to \mu(\cup_{m = 1}^{\infty} A_{m}) as n \to \infty.

Proof

Consider B_{n} = A_{n} \setminus A_{n-1} with A_{0} \equiv \emptyset. So A_{n} = B_{1} \cup \dots \cup B_{n}, and \cup B_{n} = \cup A_{n}, and B_{n} \cap B_{m} = \emptyset when n \neq m. Hence \mu(\cup A_{m}) = \mu(\cup B_{m}) = \Sigma_{m = 1}^{\infty} \mu (B_{m}) = \lim_{ N \to \infty } \Sigma_{m=1}^{N} \mu (B_{m}).

Lemma

X, \mu, s : X \to [0, \infty] Measurable Simple Function. Then A \mapsto \int_{A} s \, d\mu defines a Measure on X.

Proof

A = \emptyset \implies measure is 0. Given Continuous disjoint union \cup B_{n} with B_{n} Measure, then

\int_{\cup B_{n}} s \, d\mu = \Sigma_{i} a_{i} \mu(\underbrace{A_{i} \cap (\cup B_{n})}_{\cup(A_{i} \cap B_{n})}) = \Sigma_{i} a_{i} \Sigma_{n}^{\infty} \mu (A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \Sigma_{i} a_{i} \mu(A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \int_{B_{n}} s \, d\mu 

QED