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Let $(\Omega, \mathbb{A},P)$ be a probability triple and $X_n$ be a sequence of random variables. $X_n$ converges almost surely if and only if $Prob(w \in \Omega:X_n(w) \to X(w) \ as \ n \ \to \ \infty )=1$.

How do we know that this set is measurable?

Thanks.

Christian

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    The assumption of measurability of that set is part of the definition.2012-11-01
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    @JonasMeyer I do not understand your comment. What are you alluding to?2012-11-01
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    @did: Looking at it now, probably I misunderstood. It looked like the first part is a statement of the definition of almost sure convergence to me. If that had been the case, the appearance of the statement "$Prob(w \in \Omega:X_n(w) \to X(w) \ as \ n \ \to \ \infty )=1$" would presume that $\{w \in \Omega:X_n(w) \to X(w) \ as \ n \ \to \ \infty \}$ is something at which "$Prob$" can be meaningfully evaluated, i.e., is measurable. Apparently I am confused.2012-11-01
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    You're not confused (Although I sense a hint of sarcasm).2012-11-02

1 Answers 1

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If every $X_n$ is measurable and the event $[X_n\to X]$ is almost sure, then $X$ is measurable${}^*$ since $$ [X_n\to X]\cap[X\leqslant x]=[X_n\to X]\cap\bigcap_{n\geqslant1}\bigcap_{k\geqslant1}\bigcup_{i\geqslant k}[X_i\leqslant x+\tfrac1n]. $$ If $X$ and every $X_n$ are measurable, the set $[X_n\to X]$ is measurable since $$ [X_n\to X]=\bigcap_{n\geqslant1}\bigcup_{k\geqslant1}\bigcap_{i\geqslant k}\,[|X_i-X|\leqslant\tfrac1n]. $$ Edit: ${}^*$ This assumes that the sigma-algebra $\mathcal A$ is complete, that is, that $\mathcal A$ contains every subset of every $A$ in $\mathcal A$ such that $\mathbb P(A)=0$. Otherwise, assume that $[X_n\to X]=\Omega$.

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    Can you please tell me why the set within the unions and intersections is measurable...?2012-11-01
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    If $Y:\Omega\to\mathbb R$ is a measurable function, then every subset $[Y\leqslant y]=Y^{-1}((-\infty,y])$ of $\Omega$ is measurable because each subset $(-\infty,y]$ of $\mathbb R$ is.2012-11-01
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    That's the definition of measurability of Y. But why is the absolute value of the difference of two RVs measurable?2012-11-01
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    Because this is a measurable function of the measurable pair $(X_i,X)$. (By the way, which textbook are you following?)2012-11-01
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    I only recently started studying measure-theory... Foundations of Modern Probability by Kallenberg. Don't you have to prove that it is a measurable function?2012-11-01
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    Surely you might have heard that $(x',x)\mapsto x'-x$ is measurable.2012-11-01
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    That's not obvious to me. Again, I just started studying measure theory.2012-11-01
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    So I had guessed. Hint: this function is continuous.2012-11-01
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    Continuous functions are measurable...2012-11-01