14/10/24

Reference: Lecture notes 3.1, 3.2

Lecture 5 Measure spaces.pdf

The properties of measurable sets motivate the following abstract notion.

Definition. Let $X$ be a set and let $P(X)$ be its power set. A $\sigma$-algebra is a subset $\mathcal{A}\subset P(X)$ with the following properties

  1. (Non-emptiness) $X \in \mathcal{A}$.
  2. (Closure under complement) If $A\in\mathcal{A}$ then $A^c=X\setminus A\in\mathcal{A}$.

A measurable space is a pair $(X,\mathcal{A})$ where $X$ is a set and $\mathcal{A}$ is a $\sigma$-algebra.

  1. (Closure under countable union) If $A_1,A_2,\ldots\in\mathcal{A}$, then $\bigcup_{n=1}^\infty A_n \in \mathcal{A}$.

Remark. The first condition is equivalent to $\mathcal{A}$ being non-empty, as if there exists any$A\in\mathcal{A}$, then from the other two conditions we get $X=A\cup A^c \in \mathcal{A}.$ The last condition can be replaced by countable intersections using

$\bigcap_{n=1}^\infty A_n = (\bigcup_{n=1}^\infty A_n^c)^c$

(an element is not in the intersection if and only if it does not belong to one of the sets).

Examples

  1. $(X,P(X))$ is a measurable space for every non-empty set $X$.
  2. Let $\Sigma\subset P(\R)$ be the set of measurable sets. We have proved that $\Sigma$ is a $\sigma$-algebra, so $(\R,\Sigma)$ is a measurable space.
  3. If $X$ is uncountable, then $\mathcal{A}=\{A\subset X : A \text{ or } A^c \text{ is countable} \}$ is a $\sigma$-algebra.
  4. If $X$ is countable, then $\mathcal{A}=\{A\subset X \ | \ A \text{ or } A^c \text{ is finite}\}$ contains $X$, is closed under complement, and closed under finite unions, but it is not a $\sigma$-algebra.

To specify a $\sigma$-algebra, we do not need to list all sets contained in it. It often suffices to specify a smaller collection of sets.

Proposition. Given a subset $\mathcal{G} \subset P(X)$, there exists a unique $\sigma$-algebra $\sigma(\mathcal{G})\subset P(X)$ which is the smallest $\sigma$-algebra containing $\mathcal{G}$, that is: if $\mathcal{A}\subset P(X)$ is a $\sigma$-algebra containing $\mathcal{G}$, then $\sigma(G)\subset\mathcal{A}.$ We call $\sigma(G)$ the $\sigma$-algebra generated by $\mathcal{G}$.

Proof. Let $\mathfrak{S}$ be the set of all $\sigma$-algebras in $P(X)$ containing $\mathcal{G}$. There is at least one such $\sigma$-algebra, namely all of $P(X)$, so $\mathfrak{S}$ is non-empty. Set