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Show that $X_n$ converges to $X$ in $L^1$, if and only if $\limsup_H|EX_n1_H-EX1_H|=0$. I am struggling with the proof of the necessatity part.

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    I think it should be $\lim_{n\to \infty} \sup_H |EX_n1_H-EX1_H|=0$ and $H$ comes from the associated sigma algebra.2012-07-11

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Here are some steps of a proof:

  1. If $H=[X_n\gt X]$, then $\mathrm E(X_n1_H)-\mathrm E(X1_H)=\mathrm E((X_n-X)^+)$.
  2. If $H=[X_n\lt X]$, then $\mathrm E(X_n1_H)-\mathrm E(X1_H)=-\mathrm E((X_n-X)^-)$.
  3. Show that $|X_n-X|=(X_n-X)^++(X_n-X)^-$.
  4. Deduce that $\mathrm E(|X_n-X|)\leqslant2\cdot\sup\limits_H|\mathrm E(X_n1_H)-\mathrm E(X1_H)|.$

As a consequence, if $\lim\limits_{n\to\infty}\left(\sup\limits_H|\mathrm E(X_n1_H)-\mathrm E(X1_H)|\right)=0$, then $\lim\limits_{n\to\infty}\mathrm E(|X_n-X|)=0$, that is, $X_n\to X$ in $L^1$.