# When does mean convergence implies almost sure convergence?

Suppose I have a sequence $(X_n)$ of random variables such that $X_n\to X$ in mean $(L_1)$.

What are the conditions that I need to impose so that I can affirm that $X_n\to X$ almost surely?

I know that Lebesgue's Dominated Convergence Theorem can be utilized to show that a sequence of random variables that converges almost surely converges in mean. But for this I do not know any conditions.

• Convergence in $L_1$ implies convergence in probability, and there are conditions ensuring that convergence in probability implies almost sure convergence. – user52227 Jun 6 '18 at 15:48
• Convergence in $L_1$ implies almost sure convergence of a subsequence of the $X_n$. So perhaps try to find conditions such that almost sure convergence of a subsequence implies almost sure convergence of the entire sequence. – Math1000 Jun 6 '18 at 16:01
• @Math1000 Is this true? I know that convergence in probability implies convergence almost sure of a subsequence, but I didn't know that the same is true for convergence in $L_1$ – Gabriel Jun 6 '18 at 18:16
• Convergence in $L_1$ implies convergence in probability, so... – Math1000 Jun 6 '18 at 19:50

One simple sufficient condition is $\sum E|X_n -X| <\infty$. This condition implies $\sum |X_n -X| <\infty$ almost surely,which implies $X_n \to X$ almost surely.