Convergence in measure implies convergence almost everywhere (on a countable set!) Here is an interesting problem from "Real Analysis for Graduate Students" by Richard Bass (which is an amazing book, by the way). 

Suppose $(X, \mathcal{A}, \mu)$ is a measure space, and $X$ is a countable set. Prove that if $f_n$ is a sequence of measurable functions converging to $f$ in measure, then $f_n$ also converges to $f$ a.e. (almost everywhere).

This is Exercise 10.9 in page 79. I tried applying the theorem that if $f_n\to f$ in measure, then $f_n$ has a subsequence $\{f_{n_k}\}$ that converges to $f$ a.e. But I am having trouble using the fact that $X$ is a countable set. I would appreciate a hint in the right direction. 
 A: Assume that $X=\{x_j,j\in\mathbb N\}$ and that $\mu\{x_j\}>0$ for each $j$ and $f=0$, $f_n\geqslant 0$. We have to show that for each $j$, $f_n(x_j)\to 0$ as $n$ goes to infinity.  
Fix $j$ and $\varepsilon\gt 0$; for $n$ large enough, we have $\mu\{x,f_n(x)\geqslant\varepsilon\}\lt\mu\{x_j\}$, hence $f_n(x_j)\lt\varepsilon$ for these $n$.
In general, it may be possible that $\{x_j\}$ is not measurable. However, the $\sigma$-algebra on $X$ is generated by a countable partition and each $f_n$ is constant on each element of the partition, hence we can without loss of generality assume that $X=\{x_j,j\in\mathbb N\}$ endowed with the powerset.
This shows the convergence at the points $x_j$ which have positive measure. In a general measure space, we can show that if $f_n\to 0$ in measure and $\{x_0\}  $ has positive measure, then $f_n(x_0)\to 0$. But this does not guarantee the almost everywhere convergence because we cannot remove (as in the case where $X$ is countable) the points $x$ such that the measure of $\{x\}$ is $0$. 
