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$X_n$ converges to $X$ in $L^1$, then $\limsup_H|EX_n1_H-EX1_H|=0$. I want to prove it, is the following proof right?

$$\lim|EX_n1_H-EX1_H|=\lim|E(X_n-X)1_H|=|\lim E(X_n-X)1_H|\\=|E\lim(X_n-X)1_H|=0$$ It's true for all $H$. So, $\limsup|EX_n1_H-EX1_H|=0$

And I also confuse how I can get $|=|\lim E(X_n-X)1_H|=|E\lim(X_n-X)1_H|=0$ by dominated convergence theorem.

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And I am also struggling with the proof of necessity. I don't konw hot to use sup. –  user30795 Jul 9 '12 at 12:51
    
Excellent. Well done. –  Did Jul 15 '12 at 5:14

1 Answer 1

up vote 3 down vote accepted

For every $H$, $|E(X_n1_H)-E(X1_H)|\leqslant E(|X_n-X|\cdot1_H)\leqslant E(|X_n-X|)=\|X_n-X\|_1$ hence $\sup\limits_H|E(X_n1_H)-E(X1_H)|\leqslant\|X_n-X\|_p$ for every $p\geqslant1$.

As a consequence, if $X_n\to X$ in $L^p$, then $\lim\limits_{n\to\infty}\sup\limits_H|E(X_n1_H)-E(X1_H)|=0$.

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Sorry,I am also struggling with the proof of necessity. I don't konw hot to use sup. –  user30795 Jul 9 '12 at 12:20
    
I do not know what it is exactly that you call the proof of necessity, but it seems to be another question anyway. Hence post another question (and this will have the side effect that we will know what the question is, really). –  Did Jul 9 '12 at 14:23
    
$X_n$ converges to $X$ in $L^p$, if and only if $\limsup_H|EX_n1_H-EX1_H|=0$. –  user30795 Jul 10 '12 at 0:14
    
Then this does not hold. –  Did Jul 10 '12 at 5:28

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