# Martingale and uniform random variable

I'm just wondering if I'm understanding any of this correct. Let $Y\sim \mathrm{Unif}(0,1)$ and if $\mathcal{F}_n$ a filtration such that $\mathcal{F}_{\infty}$ is generated by nullsets. Then let $X_n = \mathbb{E}[Y\mid\mathcal{F}_n]$. Is it then true that $\mathbb{P}(X_{\infty}\in (a,b))=0$, or even $\mathbb{P}(X_{\infty}=Y)=0$?

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Are you in the same class as MathGirl? – Did Oct 31 '12 at 11:11

The hypothesis about $\mathcal F_\infty$ means that every event $A$ in $\mathcal F_\infty$ is such that $\mathbb P(A)=0$ or $1$. Now, for every Borel set $B$, the event $[X_\infty\in B]$ belongs to $\mathcal F_\infty$ hence $\mathbb P(X_\infty\in B)=0$ or $1$ (but not only $0$). Finally, all this means that there exists some nonrandom $x$ such that $\mathbb P(X_\infty=x)=1$. For every random variable $Y$ with a nonatomic distribution (such as $Y$ uniform on $(0,1)$), one knows that $\mathbb P(Y=x)=0$ hence $\mathbb P(X_\infty=Y)=0$.

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Thank you, great! – DinkyDoe Oct 31 '12 at 10:03