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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. –  Ahriman Oct 17 '12 at 6:44
    
@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 ? –  Ahriman Oct 17 '12 at 6:55
    
@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 comment –  Deven Ware Oct 18 '12 at 3:43

1 Answer 1

For simple, let $Y_n:=|X_n-X|$. We have $$P\left(\bigcap_{l=1}^{+\infty}\bigcup_{n\geq 1}\bigcap_{j\geq n}\{Y_j\leq l^{—1}\}\right)=1$$ by hypothesis, hence $$P\left(\bigcup_{l=1}^{+\infty}\bigcap_{n\geq 1}\bigcup_{j\geq n}\{Y_j>l^{—1}\}\right)= 0.$$ This gives that for each $l\geq 1$, $P(\limsup_{j\to +\infty}\{Y_j>l^{-1}\})=0$. This implies $\limsup_{j\to +\infty}P(\{Y_j>l^{-1}\})=0$ for each $l$, what we wanted.

For a counter-example showing the converse doesn't hold, see this thread.

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