Let $(X,\mathcal{A},\mu)$ be a measure space and $(f_n)$ a sequence of real-valued functions on $X$ which is Cauchy in measure; that is, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ such that for all $m,n\geq N$ we have $\mu(\{x\in X \ | \ |f_n(x)-f_m(x)|\geq \epsilon\})<\epsilon$. Prove that there is a function $f$ on $X$ to which $(f_n)$ converges to $f$ in measure.
(Convergence in measure: For any $\epsilon>0,\displaystyle\lim_{n \to \infty} \mu(\{x\in X \ | \ |f_n(x)-f(x)|\geq \epsilon\})=0.$)
(This is not a homework problem.) I'm finding this very difficult, probably because the definitions are so complicated. I'm not sure how to approach it. Usually when one tries to prove 'Cauchy implies convergent'-type statements it's just a routine application of the completeness of $\mathbb{R}$. One approach may be to establish to existence of a Cauchy subsequence for each $x$ but that's more of a guess rather than an idea inspired by understanding. It's just difficult to get a strong enough intuitive understanding for what's going on to solve it. Fixing an $x$ and constructing $f(x)$ individually is not easy as it seems to depend on all the other $f(x)$ as well (measure being a more global property). Any small hints would be appreciated.
Edit: Not sure if this is common thing on StackExchange but I wanted to make a copy of this question myself in order to get a hint for the problem instead of seeing the answer (and spoiling the problem).