Suppose I have a sequence of random variables $\{X_n\}_{n \in \mathbf{N}}$ such that for every subsequence there exists a further subsequence that converges almost surely to $X$. Can I prove that $X_n \to X$ almost surely?
The confusion I have is that if $X_n$ converges in probability to $X$, then I have the statement mentioned above. On the other hand, from the real analysis we know one trick to prove convergence of a sequence is via the subsequence argument for deterministic sequences.
However, convergence in probability is weaker than the almost sure convergence.
Best Answer
I think the answer is no.
Consider $\Omega = [0,1]$ with uniform measure, $\{X_n\} = \{Y_{n,j}: 0 \leq j \leq 2^n - 1\}$, where $Y_{n,j}$ is $1$ if $x \in I_{n,j} = [\frac{j}{2^n}, \frac{j + 1}{2^n}]$ and $0$ otherwise. Let $X = 0$.
I claim that this sequence has the subsequence property you claimed. Say we have a sequence $\{Y_{n_i, j_i}\}$. Then we can take the subsequence with distinct $n_i$, so that $$\sum_{i} \mathbb{P}(Y_{n_i, j_i} \neq X) < \infty.$$ Thus by the Borel-Cantelli lemma, $Y_{n_i, j_i}$ converges almost surely to $X$. But obviously $X_n$ does not converge almost surely to $X$.