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.
This is one of those things that is helpfully studied using an example. A very nice example for this issue is the "wandering block". Informally, the wandering block is the sequence of indicator functions of $[0,1],[0,1/2],[1/2,1],[0,1/4],[1/4,1/2],[1/2,3/4],[3/4,1]$, etc. More explicitly, it is the sequence $g_n(x)$ which comes about from enumerating the "triangular array" $f_{j,k}(x)=\chi_{[j2^{-k},(j+1)2^{-k}]}(x)$, where $k=0,1,\dots$ and $j=0,1,\dots,2^{k}-1$.
The sequence $g_n$ converges in measure to the zero function. You can see this as follows. Given $n$, write $g_n=f_{j,k}$, then $m(\{ x : |g_n(x)| \geq \varepsilon \})=2^{-k}$ for any given $\varepsilon \in (0,1)$. Since $k \to \infty$ as $n \to \infty$, this measure goes to $0$ as $n \to \infty$.
On the other hand, the sequence $g_n(x)$ does not converge at any individual point, because any given point is in infinitely many of these intervals and also not in infinitely many of these intervals. Thus the sequence $g_n(x)$ contains infinitely many $1$s and infinitely many $0$s, and so it cannot converge.
On an infinite measure space, there is an example for the other direction: $f_n(x)=\chi_{[n,n+1]}(x)$ on the line converges pointwise to $0$ but does not converge in measure, since $m(\{ x : |f_n(x)| \geq \varepsilon \})=1$ for $\varepsilon \in (0,1)$. A corollary of Egorov's theorem says that this is impossible on a finite measure space.
On a related note, the wandering block example also shows a nice, explicit example of the theorem "if $f_n$ converges in measure then a subsequence converges almost everywhere". Here, for any fixed $j$, the sequence $h_k=f_{j,k}$ (defined for sufficiently large $k$ that this makes sense) converges almost everywhere.
Best Answer
If the convergence in measure is rapid enough, then a.e. convergence will follow. Suppose $(f_n)$ converges in measure on the measure space $(E,\mathcal E,\mu)$. If $\mu(\{x\in E: |f_n(x)-f(x)|>\epsilon\})$ converges to $0$ so rapidly that $\sum_n\mu(\{x\in E: |f_n(x)-f(x)|>\epsilon\})<\infty$ for each $\epsilon>0$ then $\mu(\{x\in E: \limsup_n|f_n(x)-f(x)|>\epsilon\})=0$ for each $\epsilon>0$ (Borel-Cantelli lemma), so $f_n\to f$ $\mu$-a.e.