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.
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.
Best Answer
If $(f_n:n\in\mathbb{N})$ is a sequence in $(X,\mathcal{M},\mu)$ and if it converges in measure to $f$, then $(f_n:n\in\mathbb{N})$ is a Cauchy sequence with respect to convergence in measure $\mu$. Indeed, since $f_n\rightarrow f$ in measure, for every $\epsilon>0$ there exists $N_{\epsilon}>0$ such that $\mu\{x:|f_n(x)-f(x)|\geq\frac{\epsilon}{2}\}<\frac{\epsilon}{2}$ for $n\geq N_{\epsilon}$. Then for $m,n\geq N_{\epsilon}$ we have $$\mu\{x:|f_m(x)-f_n(x)|\geq\epsilon\}\leq\mu\{x:|f_m(x)-f(x)|\geq\frac{\epsilon}{2}\}+ \mu\{x:|f_n(x)-f(x)|\geq\frac{\epsilon}{2}\}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. $$