# Almost everywhere convergence and $L^p$ probability

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.

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

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 $$\int_\Omega |X_n-X|^p dP$$
• @nicomezi: Thanks. I got it (when $|X_n| \leq C$), but what is a counterexample for the case $X_n \leq C$? Jul 24 at 7:34
• 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 $- \infty$. Jul 24 at 7:42
• 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 $E([X_n-X|^p)=-n^p\times1/n$. Jul 24 at 7:48