Let $(X,\mathcal{A},\mu)$ be a measure space and $(f_n)_{n \in \mathbb{N}}$ such that $f_n \to f$ in measure, i.e.
$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$
for any $\varepsilon >0$. Setting $\varepsilon=2^{-k}$, $k \in \mathbb{N}$, we can choose $n_k$ such that
$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$
for all $n \geq n_k$. Without loss of generality, $n_{k+1} \geq n_k$ for all $k \in \mathbb{N}$. Set
$$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\}.$$
As $$\sum_{k \geq 1} \mu(A_k) \leq \sum_{k=1}^{\infty} 2^{-k} < \infty,$$ the Borel-Cantelli lemma yields
$$\mu \left( \limsup_{k \to \infty} A_k \right) =0.$$
It is not difficult to see that this implies
$$\lim_{k \to \infty} f_{n_k}(x) =f(x)$$
$\mu$-almost everywhere.
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 $0$.
Best Answer
Assume $\{f_n\}$ converges pointwise a.e. to $f$. Fix $\varepsilon > 0$ and define: $$ E_N = \{x \in X \mid \exists n > N : |f_n(x) - f(x)| > \varepsilon\} $$
We have $E_1 \supset E_2 \supset \cdots$ and $\mu(E_1) \le \mu(X) < \infty$. Furthermore, since $\{f_n\}$ converges pointwise a.e. to $f$, we have $\mu\left(\bigcap_{N \in \mathbb{N}} E_N\right) = 0$.
Hence, $\lim_{N \to \infty} \mu(E_N) = 0$ and we can find an $N$ for which $\mu(E_N) < \varepsilon$.
It's clear that: $$ \forall n > N : \{x \in X \mid |f_n(x) - f(x)| > \varepsilon\} \subset E_N $$
Hence, $\{f_n\}$ converges to $f$ in measure.