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I want to prove that:

If f is a continuous function, and ${X_n \to X} $ a.e then ${f(X_n) \to f(X)}$ a.e

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    I think it would be helpful to include the space, sigma-algebra and measure under consideration.2012-10-17
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    If $X_n(\omega)\to X(\omega)$ for $n\to\infty$ then by continuity $f(X_n(\omega))\to f(X(\omega))$ for $n\to\infty$.2012-10-17
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    @Lord_Farin What for?2012-10-19
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    Because it seems hard to believe that things can be said about arbitrary measure spaces built on arbitrary topologies. At least some assumptions need to be made. The sigma-algebra may have nothing to do with the topology.2012-10-19
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    @Lord_Farin The sigma-algebra and the measure indeed have nothing to do with a topology. Random variables $X$ usually are measurable functions $X:\Omega\to\mathbb R^n$ where $\Omega$ is endowed with some sigma-algebra $\mathcal F$ and $\mathbb R^n$ is endowed with the Borel sigma-algebra $\mathcal B(\mathbb R^n)$. Hence at least the sigma-algebra $\mathcal F$ of the source-set $\Omega$ is irrelevant. Note also that the result holds irrespectively of the probability measure on $(\Omega,\mathcal F)$. (Unrelated: please use @.)2012-10-20
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    @did: Confusion arose from me being more into abstract measure theory. The only reference to the fact that we are dealing with probability theory is that there is a tag "probability". Furthermore, your point is disproved by assuming the Borel $\sigma$-algebra, which arguably relies on the Euclidean topology on $\Bbb R$, in which (topological) sense the function is continuous. There's too many gaps of information left for the reader to decide - apparently you had less problems with guessing how to fill these gaps. Even that $X$ is a RV is not mentioned... Gaps gaps gaps.2012-10-20
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    @Lord_Farin Agreed. (And I might re-use your *Gaps gaps gaps*, which I like very much. That is, if you don't mind.)2012-10-20
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    @did Please go ahead :).2012-10-20

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As Stefan commented, the 'event' (set of $\omega$'s) that $X_n\to X$, is included in the 'event' when $f(X_n)\to f(X)$, just because of continuity.

So (using $\mu$ for the measure now), $$0 = \mu(X_n \not\to X) \ge \mu(f(X_n)\not\to f(X)) \ge 0$$

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    The set $\{X_n\not\to X\}$ need not be measureable.2012-10-17
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    @StefanHansen Sorry but this is not true: if $X$ and every $X_n$ are measurable, then $[X_n\to X]$ and $[X_n\not\to X]$ are measurable.2012-10-19
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    @did: Yep, that is true. But it is nowhere mentioned that $X$ and $X_n$ are measurable.2012-10-20
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    @StefanHansen Hmmm... One can presume that the question is asked about *random variables* $X_n$ and $X$, otherwise I fail to see the point.2012-10-20
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    @did: I see your point. It's just that I've been taught that convergence a.e. does not require measurability (i.e. $\{X_n\not\to X\}$ is a, not necessarily measurable, null-set). So I guess my point is that if the OP shares this definition, then he has to be careful when writing $\mu(X_n\not\to X)$.2012-10-20
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    @StefanHansen True. But as soon as $X_n$ and $X$ are all measurable, then the set $[X_n\not\to X]$ is "necessarily" measurable. If this point is clear, everything is fine. :-)2012-10-20
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    @did: Indeed it is :)2012-10-20