Lecture 13 - Measure Theory

Sigma Algebra

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

Measure

(See Measure)

Example

$M=\wp (X)$ is a $\sigma$-algebra with the counting measure $\mu$ given by
$\mu (A)=\{\begin{array}{ll}|A|,& \text{when}A\text{is finite}\\ \infty ,& \text{when}A\text{is infinite}\end{array}$
${\mu }^{\prime }(A)=0,\forall A$

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 of Borel Sets

${\cap }_{n=1}^{\infty }\left[-\frac{1}{n},\frac{1}{n}\right]=\{0\}$

Proposition

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

Proof

Say $f$, $g$ are measurable and $a,b\in \mathbb{C}$.

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

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 measurable.

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

Note

$f+g=\mathrm{Re}(f+g)+i\mathrm{Im}(f+g)=\mathrm{Re}f+\mathrm{Re}g+i(\mathrm{Im}f+\mathrm{Im}g)$
$(a,b)\mapsto x+ib$ is continuous.

Let $h:x\to {\mathbb{R}}^2$; $h(x)=(f(x),g(x))$.
Claim that $h$ is 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 $(s,t)\stackrel{p}{\mapsto }s+t$ and $(s,t)\stackrel{q}{\mapsto }s\times t$, so $p\circ h=f+g$ and $q\circ h=f\times g$.
(All that needs to be proven now is that $h$ is measurable)

Show the claim: note that open $V\subset {\mathbb{R}}^2$ 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 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.

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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 $\cap E_i\ne \varnothing$

Proof

Net argument

$X=\mathbb{R}$
$E_n=\left[-\frac{1}{n},\frac{1}{n}\right]$
$[-1,1]\supset \left[-\frac{1}{2},\frac{1}{2}\right]\supset \left[-\frac{1}{3},\frac{1}{3}\right]\supset \dots$
${\cap }_n[-\frac{1}{n},\frac{1}{n}]=\{\varnothing \}\ne \varnothing$

$I=\{E_i|i=1,\dots ,\infty \}$
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 $x_{{E_i}_j}\in A\forall j\ge k$.
So $A\cap {E_i}_j\ne \varnothing \forall j\ge k$.

If $x\notin \cap E_i$, then $\exists$ neighbourhood
$B$ of $x$ such that $B\cap (\cap E_i)=\varnothing$.
Then $B\cap E_i=\varnothing \forall i\ge m$.
Pick $B=A$ and get a contradiction.

Subcover argument

$V_n=E_n^{\complement }$ open in Hausdorff space if $\cap E_i=\varnothing$, then $\{V_n\}$ will be an open cover of $E_1$ that has no finite subcover, contradicting that $E_1$ is compact.

$\cup V_n=\cup E_n^C=(\cap E_n{)}^{\complement }={\varnothing }^{\complement }=X$

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

$\iota :(X,\stackrel{\text{Compact Hausdorff}}{\overbrace{\tau }})\to (X,\stackrel{\text{Weaker}}{\overbrace{\tau }})$ (${\tau }^{\prime }\subset \tau$)
$\iota (x)=x$, so $\iota$ is bijective

This is a continuous map since ${\tau }^{\prime }\subset \tau$.

$\iota$ 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 $(X,{\tau }^{\prime })$ as $\iota$ is continuous, so $\iota (E)$ is closed in $(X,{\tau }^{\prime })$ as it is Hausdorff. So in this case ${\tau }^{\prime }=\tau$.

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

$⟨c,d⟩=\cup V_{\frac{1}{n}}^a$
$V_n=\{n\}$

$\mathbb{R}={\cup }_{a\in \mathbb{R}}{V}_a$
$V_a=\{a\}$
(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 $⟹X$ 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 $A$ with $x\in A$ and $A\cap \overline{\{x_n\}}=\varnothing$. But $\exists m$ such that $x\in V_m$…