Almost Sure Convergence and Lacunary Sequences Is there an example of a sequence $X_n$ of random variables so that for every lacunary sequence $n_k$ it holds that $X_{n_k}$ converges almost surely to $0$, but $X_n$ does not converge almost surely to $0$?
A sequence $n_k$ is lacunary when there exists a $\lambda > 1$ so that $n_{k+1} > \lambda n_k$ for all $k$.
 A: The probability space is $[0,1]$with Lebesgue measure.
Let
$$ X_{2^n + m} = \cases{I_{[m/n^2,(m+1)/n^2]} & if $0 \le m < n^2$ \\ 0 & otherwise.}$$
Clearly $X_n$ diverges everywhere.  If $n_k$ is lacunary, then there exists a fixed number $M$ (related to $\log_2 \lambda$) such that at most $M$ of the $n_k$ lie in any $[2^n, 2^{n+1})$, and the set where each of these is non-zero has measure at most $\frac 1{n^2}$.  So using the Borel-Cantelli Lemma we see that $X_{n_k} \to 0$ a.s.
You could also make the $X_n$ independent, but with the same distribution.  Then you can show that $X_n$ diverges using the second Borel-Cantelli Lemma.
A: As the accepted answer makes clear, the Borel Cantelli lemma makes this equivalent to the much easier question of finding a sequence $p_k\ge 0$ that is not summable but so that every lacunary subsequence is summable.
For instance, take $p_t$ to be a decreasing function with $\sum_{k=1}^{\infty}p_{k}=\infty$, like $p_t = 1/t$ for $t\in \mathbb{R}_{+}$.  Let $X_n$ be a sequence of independent Bernoulli $(p_n)$ random variables.  Then $\sum_{k=0}^\infty \mathbb{P}(X_k> 0) = \sum_{k=0}^\infty\frac{1}{k} = \infty$, so almost surely, this sequence will be $1$ infinitely often (similarly, it's going to be $0$ infinitely often too).  Therefore, with probability $1$, it does not converge.  On the other hand, for any lacunary sequence $n_k$, there will be some $\lambda > 1$ so that $n_k > \lambda^k n_1$.  Therefore,
$$
\sum_{k=1}^{\infty}\mathbb{P}(X_{n_{k}} > 0) = \sum_{k=1}^{\infty}p_{n_{k}}\le \sum_{k=1}^{\infty}p_{\lambda^{k}n_{1}} = \sum_{k=1}^{\infty}\frac{1}{n_{1}\lambda^{k}} < \infty
$$
and so the probability that $X_{n_{k}} > 0$ infinitely often is $0$ by Borel Cantelli, and so the sequence converges to $0$ almost surely.
