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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.

  • 0
    Excellent. Well done.2012-07-15

1 Answers 1

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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 $\lim\limits_{n\to\infty}\sup\limits_H|E(X_n1_H)-E(X1_H)|=0$.

  • 0
    Then this does not hold.2012-07-10