Take $\Omega=(0,1]$ and let $\mathcal{A}$ be the $\sigma$-algebra generated by all the one-point sets $\{x\}$, $x\in\Omega$.
I've come to the conclusion that $\mathcal{A}$ must then be the set of all $A\subset\Omega$ such that either $A$ or $A^{c}$ is at most countable. I know want to show that there exists a probability measure $P$ on $(\Omega,\mathcal{A})$ such that $P\{x\}=0$ for all $x\in(0,1]$. My idea is to use the Lebesgue measure and know that a single point has Lebesgue measure 0 this works? I'm hoping I'm on the right path here?
Take $X:(0,1]\rightarrow\mathbb{R}$ an integrable random variable. Then I also need to determine $\mathbb{E}[X\mid \mathcal{A}]$. Here is where i get stuck. Could anyone help me understand what this should be?
Best Answer
As you say, $\mathcal{A}$ is the set of subsets $A\subseteq \Omega$ such that $A$ or $A^c$ is at most countable. The measure $P:\mathcal{A}\rightarrow [0,1]$ is: $P(A)=0$ if $A$ is at most countable and $P(A)=1$ if $A^c$ is at most countable.
If $X:\Omega\rightarrow\mathbb{R}$ is measurable then I claim that there exists $r\in\mathbb{R}$ such that $X^{-1}(r)$ is uncountable. In fact, suppose that $X^{-1}(r)$ is at most countable for all $r$. Then $X(\Omega)$ is uncountable and therefore there exists $a\in \mathbb{R}$ such that $X(\Omega)\cap (-\infty,a)$ and $X(\Omega)\cap (a,\infty)$ are uncountable. If $A=X^{-1}((-\infty,a))$ then $A^c=X^{-1}((a,\infty))$ and it is clear that both $A$ and $A^c$ are uncountable, which is a contradiction.
So, let $X:\Omega\rightarrow\mathbb{R}$ measurable and let $r\in\mathbb{R}$ such that $A=X^{-1}(r)$ is uncountable. Then $P(A^c)=0$ and therefore $$E[X,\mathcal{A}]=\int_{\Omega} XdP=\int_AXdP=rP(A)=r$$