How can you show that if a sequence $(x_n)$ converges in $L^2$ to $X$, and converges almost surely to $Y$, then $X$ and $Y$ are almost surely the same. Can you give me a hint?
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Yes, But I find it really hard for me to finish the prove after triangle inequality. – Lee Nov 10 '14 at 22:31
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Convergence in $L^2$ implies convergence in probability... – Daniele A Nov 10 '14 at 22:58
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so, convergence in probability means every sub sequence has a further sub sequence that convergence almost surely to X, then what should I do? that's all I have done. Thanks! – Lee Nov 10 '14 at 23:08
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Yes! But this subsequence converges almost surely to $Y$ as well. – Daniele A Nov 10 '14 at 23:09
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That is where I have problem with, can you give me some more details? Thank you very much! – Lee Nov 10 '14 at 23:44
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You're welcome! Since $x_n$ converges almost surely to $Y$, any subsequence converges almost surely to $Y$ as well. Tell me if you want more details. – Daniele A Nov 10 '14 at 23:49
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As Daniele A said, the equality $X=Y$ a.s. follows from the fact that a sequence that converges to $X$ in $L^2$ has a subsequence converging to $X$ a.s. Since the same subsequence converges to $Y$ a.s., the result follows.
A more general statement can be found in Pointwise a.e. convergence and weak convergence in Lp
