Does uniform integrability imply almost sure convergence of a subsequence?

Question

True or false? If $X_n$ is uniformly integrable then there exists a subsequence $X_{n_k}$ such that $X_{n_k}$ converges a.s. to a random variable $X$ as $k→\infty$.

My attempt so far:

Since $X_n$ is UI, given $\varepsilon>0\ \exists\delta>0$ s.t. $\mathbb{E}[|X_n|\mathbb{1}_{|X_n|>K}]<\varepsilon$ whenever $\mathbb{P}({|X_n|>K})<\delta$. So choose $K$ s.t. $\delta=\frac{1}{2^n}$. Then $\limsup\limits_{n\rightarrow\infty}\mathbb{P}({|X_n|>K})=0$. Thus $|X_n|\leq K$. Then by the Bolzano-Weierstrass theorem, $X_n$ has a convergent subsequence.

Any help would be appreciated, thanks in advance.

Answer 1

Not really. Please note that if $(X_n)_n$ are independent, Rademacher random variables (that is $\mathbb P(X_n = 1) =\mathbb P(X_n=-1) = \frac{1}{2}$), then family $(X_n)_n$ is uniformly integrable (for example since this family is dominated by constant random variable $Y=1$). Moreover, for any subsequence $(X_{n_k})_k$ we have $$ \sum_n \mathbb P(X_{n_k} = 1) = \sum_n \frac{1}{2} =\infty, $$ hence by Borell Cantelli, we have $\mathbb P(\limsup_k \{X_{n_k}=1\}) = 1$. Similarly, $\mathbb P(\limsup_k\{X_{n_k}=-1\})=1$, so $X_{n_k}$ cannot converge almost surely (in fact the convergence can occur only on the set of measure $0$).

Answer 2

In your argument, the chosen $K$ depends a priori on $n$. Moreover, even if $K$ can be chosen independently, the application of the Bolzano-Weierstrass theorem does not give the wanted result because the subsequence may depend on $\omega$. An explicit example is given by Domini Kutek.

Note that Dunford-Pettis theorem show that uniform integrability of a sequence is equivalent to the fact that every subsequence has a weakly convergent subsequence.