@Davide Giraudo's answer works quite well with the probability space, which has finite measure. And I am going to show a more general result, i.e.
$\textbf{Claim:}$ if the measurable space $(\Omega,\mathcal{F},\mu)$ has finite measure($\mu(\Omega)<\infty$), then the a.e. convergence implies convergence in measure.
which is suggested by @Ashok. Also, I will introduce another mode of convergence to help people understand the relationship between these modes of convergence(since I think it is hard to help the question owner after 8 years...).
Key observation is that
$X_n$ converges a.e. to $X$ is equivalent to(by definition)
$$\mu(\mathop{\cap}\limits_{m=1}^{\infty}\mathop{\cup}\limits_{n=m}^{\infty}\{|X_n-X|\geq\epsilon\})=0,$$
while converges in measure is equivalent to(by definition as well)
$$\lim\limits_{n\rightarrow\infty}\mu(\{|X_n-X|\geq \epsilon\})=0.$$
They are both true under the following statement:
$$\lim\limits_{m\rightarrow\infty}\mu(\mathop{\cup}\limits_{n=m}^{\infty}\{|X_n-X|\geq \epsilon\})=0,$$
in all measurable space. And this kind of convergence is called $\textbf{almost uniform convergence}$, or a.u. convergence in short, since you can read from the above formula that for any $\epsilon>0$, there is some measurable set $A\in\mathcal{F}$ such that $\mu(A)<\epsilon$ and $X_n$ converges uniformly on $A$.
(Well, it may not be direct to "read" from it, but can be proved with moderate effort, and is a very good exercise for everyone :) Or you can go to Egorov's theorem for the proof--we only need a part of it to show the equivalence. However, I still recommend you to try on yourself.
Now, you can find: if the measure space is finite, then a.e. convergence is equivalent to a.u. convergence, since you can now pull the intersection out to be a limit, then implies convergence in measure. Q.E.D.
In general measure spaces, a.e. convergence and convergence in measure can be totally irrelevant:
Let $X_n(x)=1(|x|>n)$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, then it converges a.e. but not in measure(Lebesgue measure);
Let $X_n(x)=1(x\in[\frac{i}{2^k},\frac{i+1}{2^k}))$ be the famous "typewritter" function, where $n=2^k+i$, then it converges in measure but not a.e.