6

Suppose that we are given a probability space $(\Omega, \mathcal{F}, \mathsf P)$ and an increasing sequence of $$\mathcal{F}_1\subset \ldots \subset\mathcal{F}_n\subset \mathcal{F}_{n+1} \subset \ldots \subset \mathcal{F}$$ of $\sigma$-algebras. Assume that $\mathsf P|_{\mathcal{F}_n}$ is atomless for each $n$. Is so $\mathsf P$ restricted to the $\sigma$-algebra generated by the union of $\mathcal{F}_n$ ($n\in \mathbb{N}$)?

Pist
  • 61

2 Answers2

4

Actually, something much stronger is true: If $P$ is a measure on a $\sigma$-algebra $\mathcal{F}$ and $\mathcal{G}\subseteq\mathcal{F}$ is a $\sigma$-subalgebra on which $P$ is atomless and $\sigma$-finite, then $P$ is atomless on all of $\mathcal{F}$. To show this, suppose $A\in \mathcal{F}$ is an atom. Then for any $B\in \mathcal{G}$, either $P(A\cap B)=0$ or $P(A\cap B)=P(A)$. Let $\mathcal{I}$ be the collection of all $B\in\mathcal{G}$ such that $P(A\cap B)=0$. Then $\mathcal{I}$ is a closed under countable unions, and must contain all $B$ such that $P(B)<P(A)$. But since $P$ is atomless and $\sigma$-finite on $\mathcal{G}$, every element of $\mathcal{G}$ can be written as a union of countably many sets $B\in\mathcal{G}$ such that $P(B)<P(A)$. It follows that $\mathcal{I}$ is all of $\mathcal{G}$. But clearly $\Omega\not\in \mathcal{I}$, so this is a contradiction.

Eric Wofsey
  • 342,377
  • 1
    Beautiful proof! Let me just add a little bit more details about how $G\in\mathcal G$ can be written as a union of countably many sets each probability less that $P(A)$. Since $P|\mathcal G$ is $\sigma$-finite, there exists $(H_n){n\in\mathbb N}\in\mathcal G^{\mathbb N}$ such that $P(H_n)<\infty$ for each $n\in\mathbb N$ and $\Omega=\bigcup{n\in\mathbb N} H_n$. For each $n\in\mathbb N$ such that $P(H_n)>0$, the lack of atoms ensures that one can decompose $H_n$ into finitely many disjoint set $B_n^1,\ldots B_n^{k_n}$ from $\mathcal G$ ($k_n\in\mathbb N$) with measure less than $P(A)$ each. – triple_sec Jul 10 '15 at 04:22
  • Now one can take $$G=\bigcup_{n\in\mathbb N}(H_n\cap G)=\bigcup_{n\in\mathbb N}\bigcup_{m=1}^{k_n}(B_n^m\cap G).$$ Clearly, the union is countable and involves elements of $\mathcal G$. Also, for each $n\in\mathbb N$ and $m\in{1,\ldots,k_n}$, one has $$P(B_n^m\cap G)\leq P(B_n^m)<P(A).$$ [If $P(H_n)=0$ for some $n\in\mathbb N$, then simply take $k_n=1$ and $B_n^1=H_n$.] – triple_sec Jul 10 '15 at 04:27
  • 1
    @triple_sec: Why does the lack of atoms ensure that one can decompose $H_n$ into finitely many disjoint sets of measure less than $P(A)$ each? – shalin Jul 10 '15 at 14:05
  • 2
    @Shalop http://math.stackexchange.com/questions/254728/simpler-proof-non-atomic-measures –  Jul 10 '15 at 14:33
  • All right, thanks. – shalin Jul 11 '15 at 01:36
-2

An atom is an indivisible set in the sigma algebra with positive measure.

Consider $\mathcal{F}_n$ a sigma algebra of subsets of the space $[0,1]^2$.

The construction is a bit artificial. so I will divide it into steps.

1) consider a the sets $[0,1] \cup (n,n+1)\backslash E_n$ as the basic sets of your sigma algebra $\mathcal{F}_n$ ($E_n \subset [n-1,n+1]$ is a denumerable set)

Note that there are no atoms in $\mathcal{F}_n$

2) set P([0,1]) = 1

Note that $[0,1]\in \sigma(\mathcal{F}_1 \cup \mathcal{F}_2)$ $[0,1]$ is indivisible and has positive measure. Therefore it is an Atom.

Conrado Costa
  • 6,239