Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 26K (Probability and Measure)

A sigma-algebra $\mathcal{F}$ on $X$ contains $\varnothing$, is closed under complements in $X$, and is closed under countable unions. A measure on the measurable space $(X,\mathcal{F})$ is a functionsuch that $\mu (\varnothing )=0$ and, for every pairwise disjoint sequence $(A_n)$,A probability measure is a measure normalized by

The Caratheodory extension theorem states that a premeasure ${\mu }_0$ on an algebra of subsets of $X$ extends to a measure on the generated sigma-algebra. If ${\mu }_0$ is sigma-finite, this extension is unique. In particular, a finite premeasure has a unique extension.

First, $\mathcal{B}\subseteq \mathcal{C}$ by taking $B=A$. If $A\in \mathcal{C}$, thenso $\mathcal{C}$ is closed under complements. Now let $A_i\in \mathcal{C}$ and put $A={\bigcup }_i{A}_i$. Because $\mu (X)<\infty$, continuity from below of a measure gives an $N$ such thatChoose $B_i\in \mathcal{B}$ with $\mu (A_i△B_i)<ϵ/(2N)$ and let $B={\bigcup }_{i=1}^N{B}_i\in \mathcal{B}$. The union bound for symmetric differences givesThus $\mathcal{C}$ is a sigma-algebra containing $\mathcal{B}$, and

Finiteness is essential. Take $X=\mathbb{N}$ with counting measure and let $\mathcal{B}$ be the finite-cofinite Boolean algebra of sets. Then $\sigma (\mathcal{B})=\mathcal{P}(\mathbb{N})$. For $ϵ<1$, the condition $\mu (A△B)<ϵ$ forces $A=B$, so $\mathcal{C}=\mathcal{B}$. Any infinite coinfinite subset of $\mathbb{N}$ belongs to $\sigma (\mathcal{B})$ but not to $\mathcal{C}$.

Yes. The regularity of Lebesgue measure says that for every Lebesgue measurable set $A\subseteq [0,1]$ and every $ϵ>0$, there is an open set $O\supseteq A$ withWrite $O\cap [0,1]$ as a countable disjoint union of intervals. Since its total measure is at most one, a finite subunion $B$ can be chosen withThen $B\in \mathcal{B}$ and $m(A△B)<ϵ$. Hence every Lebesgue measurable set lies in $\mathcal{C}$, while $\mathcal{C}\subseteq {\mathcal{L}}_{[0,1]}$ by definition. Therefore