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Convergence in probability implies convergence on a subsequence almost surely.

But this means we fix a subsequence, such that $X_{n_k}$ converges for almost every $\omega$, right? The subsequence we pick does not depend on the $\omega$ right?

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Right.${}{}{}{}$ –  Michael Greinecker Oct 27 '12 at 20:14
    
Thanks very much. –  Lost1 Oct 27 '12 at 20:21
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1 Answer 1

up vote 5 down vote accepted

Yes, we can take a sequence $\{n_k\}$ which works for almost all $\omega$. To see that, fix a subsequence $\{n_k\}$ such that for each $k$, $$P(|X_{n_k}-X|>2^{-k})\leq 2^{-k}.$$ This one can be constructed by induction. We have $$P(\limsup_{k\to+\infty}\{|X_{n_k}-X|>2^{-k}\})=0,$$ by a similar argument of Borel-Cantelli theorem proof, using the fact that $\sum_{j\geq k}2^{—j}\to 0$ when $k\to +\infty$. This proves convergence almost everywhere of $\{X_{n_k}\}$ to $X$.

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