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Let $(\Omega, \mathcal F, P)$ be a probability space. If $X_n \leq C$ for each $n\geq 1$ and $X_n \to X$ almost everywhere, prove that $X_n \to X$ in $L^p$.

So far I have just showed that $X, X_n \in L^p$. For the $L^p$ convergence I have just used the DCT to write $\int_{\Omega}X_n dP \to \int_{\Omega}X dP$, but don't know what else to do.

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    $\begingroup$ This is false. If you change $X_n \leq C$ to $|X_n| \leq C$ then it becomes true. $\endgroup$ Jul 24 at 7:18
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Hint:

As stated by the comment below the question, it is not true unless you have $|X_n| \le C$ instead.

If we assume that, apply the DCT to $$\int_\Omega |X_n-X|^p dP$$

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    $\begingroup$ As stated by OP the result is false. $\endgroup$ Jul 24 at 7:18
  • $\begingroup$ Ho, yes you are correct, I have edited. $\endgroup$
    – nicomezi
    Jul 24 at 7:20
  • $\begingroup$ @nicomezi: Thanks. I got it (when $|X_n| \leq C$), but what is a counterexample for the case $X_n \leq C$? $\endgroup$
    – Keio203
    Jul 24 at 7:34
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    $\begingroup$ I dont have the example in mind (I think you can find it quite easily on some books) but the problem here is that $X_n$ can go to $- \infty$. $\endgroup$
    – nicomezi
    Jul 24 at 7:42
  • $\begingroup$ I think I have it. Define $P(X_n=0)=1-1/n$ and $P(X_n=-n)=1/n$ and $X=0$ a.s. . $X_n$ converges a.s. to $0$ but $X_n$ does not converge to $X$ in $L^p$ since $E([X_n-X|^p)=-n^p\times1/n$. $\endgroup$
    – nicomezi
    Jul 24 at 7:48

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