Let $X_1, X_2, \cdots$ be a sequence of $p$-integrable $\mathbb{R^d}$ valued random variables. Assume that $X_n$ converges $0$ weakly, then can we say that $r(\omega) = liminf\{ |X_1 (\omega)|, |X_2 (\omega)|, \cdots\}$ is $0$ for almost every $\omega ?$

The classical example of a weakly convergent but not strongly convergent sequence is those of orthogonal basis, yet it is not a counterexample for the above claim. I could not prove the claim but intuitively I believe it holds, at least for $\mathbb{R^d}$ valued random vectors.

If needed one can assume that the sequence $(X_n)_n$ is bounded in $p$-norm, I do not feel like this is necessary.


Weak convergence together with boundedness of second moments implies convergence of expectations. By Fatou's Lemma we get $E \lim \inf |X_n| \leq \lim \inf E|X_n|=0$ since $E|X_n| \to 0$. This implies that $ \lim \inf |X_n|=0$ almost surely.

  • $\begingroup$ I don't see how weak convergence together with boundedness of second moments implies convergence of expectations. Are you using Vitaly type theorem here? For every $Y$ $q$-integrable, we have $E[Y X_n] \rightarrow E[Y \, 0] = 0,$ and there is some $W$ $p$-integrable such that $|X_n| \leq W$ for all $n.$ I dont see how we proceed. Could you provide more details please? $\endgroup$ Jan 4 at 9:52
  • $\begingroup$ @vekinpirna Boundeness of second moments implies uniform integrabilty. And $Y_n \to 0$ weakly, $(Y_n)$ uniform integrable implies $EY_n \to 0$. (one quick way of proving this to replace weak convergence by almost sure convergence using Skorohod Representation Theorem), but you can do it without this theorem also). $\endgroup$ Jan 4 at 10:00
  • $\begingroup$ Thank you so much for your response. I see that Skorohod Representation Theorem gives some conditions under which random variables which converge in distribution can lead to almost sure convergence on some larger probability space. Do the notions of weak convergence and convergence in distribution coincide in this case? $\endgroup$ Jan 4 at 10:11
  • $\begingroup$ We are only intersetd in the real random variables $|X_n|$. And $X_n \to 0$ weakly implies $|X_n| \to 0$ weakly, so the usual Skorohod Theorem applies. @vekinpirna $\endgroup$ Jan 4 at 10:15
  • $\begingroup$ Is it obvious that $X_n \rightarrow 0$ weakly implies $|X_n| \rightarrow 0$ weakly? I will take my time working on that. $\endgroup$ Jan 4 at 10:20

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