Almost sure convergence of random variables with $\mathbb{E}[X_n]=O(1/\log n)$ Let $(X_n: n\ge 1)$ be a sequence of (not necessarily independent) random variables such that $|X_{n+1}-X_n|\le 1$ for all $n$, and 
$$
\mathbb{E}[X_n]=O\left(\frac{1}{\log n}\right) \,\,\,\text{ and }\,\,\,\text{Var}[X_n]=O\left(\frac{1}{n}\right)
$$
as $n\to \infty$. 

Question. Is it true that $X_n\to 0$ almost surely?

 A: Here is a classic counter-example: Define $(X_n)_{n\geq 1}$ as follows:


*

*Consider the probability space $\Omega = [0, 1)$ equipped with the Borel $\sigma$-algebra and the Lebesgue measure restricted to $\Omega$.

*Consider the intervals of the form $I_{m,k} = [k/2^m, (k+1)/2^m)$ for $m \geq 0$ and $0 \leq k < 2^m$. Then we enumerate all such $I_{m, k}$'s in lexicographical order to obtain the single-indexed sequence $(I_n)_{n\geq 1}$. That is, $(I_1, I_2, I_3, I_4, \cdots) = (I_{0,0}, I_{1,0}, I_{1,1}, I_{2,0}, \cdots)$.

*Define $(X_n)_{n\geq 1}$ by $ X_n = \mathbf{1}_{I_n} $. This is a Bernoulli RV with parameter $\mathbf{P}(I_n)$.
Then it is clear that $|X_{n+1} - X_n| \leq 1$. Also, by noting that $\mathbf{P}(I_n) = 2^{-\lfloor \log_2 n \rfloor} \leq \frac{2}{n}$, we get
$$ \mathbf{E}[|X_n|]
= \mathbf{P}(I_{n})
\leq \frac{2}{n}
\qquad \text{and} \qquad
\mathbf{Var}(X_n)
\leq \mathbf{P}(I_{n})
\leq \frac{2}{n}. $$
On the other hand, $X_n(\omega)$ takes each value in $\{0, 1\}$ infinitely often for every $\omega \in \Omega$. So $X_n$ cannot converge almost surely.
