# Almost sure downward convergence with some conditions implies convergence in $L^1$

If $X_n\downarrow X$ a.s., each $X_n$ is integrable and $inf_n E[X_n] > -\infty$, then $X_n \rightarrow X$ in $L^1$.

As far as I know, "$X_n\downarrow X$ a.s." means that for every n, $X_n > X$ almost everywhere, and $lim_{n \rightarrow \infty} X_n = X$ almost everywhere.

I know that almost sure convergence implies convergence in probability, but any inequality I can find relating probabilities to expectations is facing the wrong direction to get $E[|X_n - X|] = 0$ (that is, the expectation is usually the larger quantity). I'd like to use some kind of dominated convergence. Since X is smaller than all of the $X_n$'s, my best hope is to get some Y in $L^1$ so that $|X_n - X| < Y$, but I'm having trouble finding a good candidate Y. Every $X_n$ is integrable and each $X_n$ dominates X almost surely, which suggests (though I'm not sure if it's rigorously true) that X must also be integrable, so (I think) $|X_n - X|$ should also be integrable. But their sup might not be; otherwise I would say something like $|X_n - X| \leq sup_n |X_n - X|$ (I'm not sure if this is quite the right inequality either; hence "something like"). I'm just not clear how knowing something about the inf helps here.

• Do you know the monotone convergence theorem? – saz Oct 5 '14 at 16:53
• The version of that in my book seems to require that the X_n go up to X almost surely, rather than down to it. – Xindaris Oct 5 '14 at 18:49
• Apply the monotone convergence theorem to $(-X_n) \uparrow (-X)$. – saz Oct 5 '14 at 18:59
• I'm not sure how to do monotone for that, but I can definitely make that work for the dominated theorem I know. Thank you! – Xindaris Oct 5 '14 at 19:48