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I need to prove the following:

If $\alpha$ is a probability measure and ${X_n \to X} $ a.e then ${X_n \to X}$ in measure. Show that the opposite may not be true.

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    Another way is to use Egoroff's theorem, or tu use the fact that $X_n \to X$ in probability iff $\frac{|X_n - X|}{1+|X_n - X|}$ goes to $0$ in $L^1$ together with dominated convergence theorem.2012-10-17
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    @Deven Ware : I don't understand what $X = \cap_N \cup_{n \ge N} X_n$ means when $X_n$ and $X$ are random variables. Can you give further details ?2012-10-17
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    @Ahriman, yes sorry I read this as a measure space when I saw the problem last night, I don't know anything about probability, so I've retracted my comment2012-10-18

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