$(X,\mathscr{A}, \mu)$ measure space and $(f_n)_{n \in \mathbb{N}}$ sequence of increasing nonnegative measurable functions such that $f_{n}$ converges in measure to $f$. Prove $f_n$ converges to $f$ almost everywhere.
Attempt For a fixed $N \in \mathbb{N}$ and for all sufficiently large $n$ it holds $\mu(\{|f_n(x)−f(x)|\geq \frac{1}{k}\})<\frac{1}{2^N}$
Let $A_{n}=\{x \in X : |f_{n}(x)-f(x)|\geq \frac{1}{k}\}$. If $f_n \leq f_{n+1}(x) \leq \dots \leq f$ for all $n\in \mathbb{N}$ then $|f_{n}(x)-f(x)|\geq|f_{n+1}(x)-f(x)|$, so $A_{n} \supset A_{n+1}$.
Then, $A= \bigcap_{n \in \mathbb{N}}A_{n}$. We see $\mu(A) \leq \mu(A_{n})\leq \frac{1}{2^n} \forall n \in \mathbb{N}$ , so $\mu(A)=0$.
If $x \notin A$ then exists $n_0 \in \mathbb{N}$ such that $x \notin A_{n}$ for all $n \geq n_{0}$ then $|f_{n}(x)-f(x)|<\frac{1}{k}$.
Then $f_{n} \rightarrow f$ almost everywhere.
Could you help me if this prove is correct? or how can I proceed to solve this problem?