Converges in measure then a subsequence converges almost everywhere 
Let $(X,\mathcal{M},\mu)$ be a measure space. Let $\{f_n\}_{n\in\mathbb{N}}$ a sequence of measurable functions which converge in measure to a function $f$. Prove that exists a subsequence $\{f_{n_k}\}_{k\in\mathbb{N}}$ such that converges almost everywhere to $f$.

Attempt: (Since I not used the Borel-Cantelli Lemma I'm not sure if it's right, I would appreciate a lot if you check)
Since $f_n$ converges in measure to $f$ we have the following:
$$\forall\epsilon>0\ \ \forall\delta>0\ \ \exists\, n_o\in\mathbb{N}\, \big(n\geq n_o \Rightarrow \mu(x\in X : |f(x)-f_n(x)|\geq\epsilon)<\delta\big)$$
For each $m\in\mathbb{N}$ we define $\epsilon_m=\frac{1}{m}$ and the following subsequence of $f_n$. Taking $\delta_n=\frac{1}{n}$ exist $n_{m,n}\in\mathbb{N}$ such that $\mu(x\in X : |f(x)-f_{\displaystyle n_{m,n}}(x)|\geq\frac{1}{m})<\frac{1}{n}$.
This defines for each $m$ a subsequence $\{f_{\displaystyle n_{m,n}}\}_{n\in\mathbb{N}}$ of $f_n$. I taked $f_{\displaystyle n_{k,k}}=f_{n_k}$ then for each $k\in\mathbb{N}$ we have $\mu(x\in X: |f(x)-f_{n_k}(x)|\geq\frac{1}{k})<\frac{1}{k}$ so $f_{n_k}$ converges almost everywhere to $f$. I was quite informal in the last step, but this idea I think that works. How being more precise writting this last step, or is my idea completely wrong?
 A: This got a bit long for a comment, but it continues our discussion from the comments above.
Up to your last step you have just used the definition of convergence in measure, and that is fine. But in the last step, you've hidden the main difficulty of the problem in your assertion, without proof, that $f_{n_k}$ converges a.e. to $f$.
In fact, your condition on $f_{n_k}$ is not enough to guarantee this.
Look at the measure space $[0,1]$ with Lebesgue measure. Let $f_k$ be the characteristic function of the interval $I_k$, where $I_k$ is an interval of length $1/k$ which move left along $[0,1]$ and "wrap around" when they reach the end, i.e.
$$ I_1 = [0,1] $$
$$ I_2 = [0,1/2] $$
$$ I_3 = [1/2,5/6] $$
$$ I_4 = [5/6,1] \cup [0,1/12] $$
$$ I_5 = [1/12,17/60] $$
etc.
(I hope you get the idea; try drawing a picture.)
Then 
$$ \mu(\{x: |f_k(x) - 0| \geq 1/k\}) \leq 1/k $$
because each $f_k$ is $0$ except on an set of measure $1/k$.
On the other hand $\{f_k\}$ does not converge pointwise to zero anywhere, much less almost everywhere. (This is because the harmonic series diverges, and so any $x\in[0,1]$ is in some $I_k$ for infinitely many values of $k$.)
So my functions $f_k$ satisfy the exact same assumption as your $f_{n_k}$ do, but they do not converge pointwise almost everywhere.
So you need to refine your argument; the idea is to choose your $f_{n_k}$ better and then apply Borel-Cantelli.
