ACIT4330-Page/content/Lectures/Lecture 13 - Measure Theory.md

lecture, date

lecture	date
13	2025-02-27

Sigma Algebra

Defined in the lecture: Sigma-Algebra (See also Sigma-Algebra#Measurable)

Measure

(See Sigma-Algebra#Measure)

[!example] M = \wp(X) is a $\sigma$-algebra with the counting measure \mu given by \mu(A) = \begin{cases}|A|, & \text{when}\ A\ \text{is finite}\\ \infty, & \text{when}\ A\ \text{is infinite}\end{cases} \mu'(A) = 0,\ \forall A

Given a collection N of subsets of any set X, the intersection M of all $\sigma$-algebras containing N is a $\sigma$-algebra; the one generated by N.

Borel Sets

Defined in the lecture: Borel Sets

[!example] Example of Borel Sets \cap^{\infty}_{n=1}\left[ -\frac{1}{n}, \frac{1}{n} \right] = \{ 0 \}

Proposition

Say X has a $\sigma$-algebra. Then all Borel Measurable functions X \to \mathbb{C} is an algebra under Pointwise operations.

Proof

Say f, g are Borel Measurable and a, b \in \mathbb{C}.

Note that k : \mathbb{C} \to \mathbb{C} given by b \mapsto ab is Continuous then (af)(x) = (\underbrace{k}_{\text{continuous}} \circ \underbrace{f}_{\text{measurable}})(x) (= k(\underbrace{f(x)}_{b}) = ab = a \times f(x) = (af)(x))

Since \mathrm{Re}, \mathrm{Im} : \mathbb{C} \to \mathbb{R} by \mathrm{Re}(a + ib) = a and \mathrm{Im}(a + ib) = b are Continuous the real and imaginary parts of f are Borel Measurable.

So to show that f+g, f \times g are Borel Measurable, we may assume that f and g are real.

Note

f + g = \mathrm{Re}(f+g) + i \mathrm{Im}(f+g) = \mathrm{Re} f + \mathrm{Re} g + i (\mathrm{Im} f + \mathrm{Im} g)

(a, b) \mapsto x + i b is Continuous.

Let h : x \to \mathbb{R}^2; h(x) = (f(x), g(x)). Claim that h is Borel Measurable. Then use that f + g, f \times g are compositions of h with the Continuous functions \mathbb{R}^2 \to \mathbb{R} given by (s, t) \xmapsto{p} s + t and (s, t) \xmapsto{q} s \times t, so p \circ h = f + g and q \circ h = f \times g. (All that needs to be proven now is that h is Borel Measurable)

Show the claim: note that Open Sets V \subset \mathbb{R}^2 is a Countable union of rectangles I \times J for segments I, J \subset \mathbb{R}.

Also note that (taking the inverse image of the rectangle) h^{-1}(I \times J) = f^{-1}(I) \cap g^{-1}(J) is Borel Measurable. Then so is h^{V} = h^{-1}(\cup_{n}(I_{n} \times J_{n})) = \cup_{n}h^{-1}(I_{n} \times J_{n}).

QED.

---

Exercise Part of the session...

Question 3 (Exercise 6)

Show that a sequence E_{1} \supset E_{2} \supset E_{3} \supset \dots of a Compact non-empty subsets of a Hausdorff space has a non-empty intersection \cap E_{i} \neq \emptyset

Proof

Net argument

X = \mathbb{R} E_{n} = \left[ -\frac{1}{n}, \frac{1}{n} \right] [-1,1] \supset \left[ -\frac{1}{2}, \frac{1}{2} \right] \supset \left[ -\frac{1}{3}, \frac{1}{3} \right] \supset \dots \cap_{n} [-\frac{1}{n}, \frac{1}{n}] = \{ \emptyset \} \neq \emptyset

I = \{ E_{i} | i=1,\dots,\infty \} Upward filtered ordered set under reverse inclusion.

Defined by axiom of choice X_{E_{i}} \in E_{i}, so we get a net \{ x_{E_{i}} \}_{E_{i} \in I} \subset E_{i}

Then it has a convergent subnet \{ x_{{E_{i}}_{j}} \}_{j}, say x_{{E_{i}}_{j}} \to x \in E_{1}.

Claim x \in \cap E_{i}: If A is a neighbourhood of x in E_{1}, then x_{{E_{i}}_{j}} \in A\ \forall j \geq k. So A \cap {E_{i}}_{j} \neq \emptyset\ \forall j \geq k.

!Excalidraw/ACIT4330/Lecture 13/Drawing 2025-02-27 13.19.24.excalidraw

If x \notin \cap E_{i}, then \exists neighbourhood B of x such that B \cap (\cap E_{i}) = \emptyset. Then B \cap E_{i} = \emptyset\ \forall i \geq m. Pick B=A and get a contradiction.

Subcover argument

V_{n} = E_{n}^{\complement} Open Sets in Hausdorff space if \cap E_{i} = \emptyset, then \{ V_{n} \} will be an Open Cover of E_{1} that has no finite Subcover, contradicting that E_{1} is compact.

\cup V_{n} = \cup E_{n}^{C} = (\cap E_{n})^{\complement} = \emptyset^{\complement} = X

Question 2 (Exercise 7)

Show that Compact Hausdorff spaces are rigid; if X is Compact Hausdorff it has no weaker or stronger Topology that is Compact Hausdorff.

Proof

\iota : (X, \overbrace{\tau}^{\text{Compact Hausdorff}}) \to (X, \overbrace{\tau}^{\text{Weaker}}) (\tau' \subset \tau) \iota(x) = x, so \iota is Bijective

This is a Continuous map since \tau' \subset \tau.

\iota is also an Open Map, because it takes closed sets to closed sets, because if E is closed in (X, \tau), then it is Compact, and \iota(E) is Compact in (X, \tau') as \iota is Continuous, so \iota(E) is closed in (X, \tau') as it is Hausdorff. So in this case \tau' = \tau.

Question 1 (Exercise 7)

A Topological Space is second Countable if we have a sequence \{ V_{n} \} of Open Sets such that any open set is a union of some of these $V_{n}$s.

Proof

\langle c, d \rangle = \cup V_{\frac{1}{n}}^{a} V_{n} = \{ n \}

\mathbb{R} = \cup_{a \in \mathbb{R}}V_{a} V_{a} = \{ a \} (which is not Countable)

Instead: I \times J Any open A \subset \mathbb{R} Countable union of I.

X is separable if it has a dense sequence \{ x_{n} \} = X.

Claim: X selectable countable \implies X separable

Axiom of choice def $x_{n} \in V_{n}. Then \overline{\{ x_{n} \}} = X. If \overline{\{ x_{n} \}} \neq X.

x \in \overline{\{ x_{n} \}}^{\complement} \exists Open Sets A with x \in A and A \cap \overline{\{ x_{n} \}} = \emptyset. But \exists m such that x \in V_{m}.......