real analysis - Could convergence in measure imply converge almost everywhere for the entire sequence and not just sub-sequence here?

This doesn't seem to have been asked on this site before. I've been self-studying measure theory and came across this problem.

Given a sequence of measurable functions $\{f_n\}_{n \in \mathbb{N}}$ such that for all $\epsilon \gt 0$

$$\sum_{n=1}^\infty \mu (\{x: |f_n (x)| \gt \epsilon \}) \lt \infty$$

Prove that $f_n \to 0$ a.e.

I've given the problem a try and clearly the sum being finite implies the summand converges to 0 which satisfies the definition of convergence in measure. Also, I know that there exists a subsequence $f_{n_j}$ converging a.e to 0. In fact, were this a finite measure space, I could prove the problem statement. However, in this more general setting, I'm unable to prove the statement.

Any help would be appreciated. This is my first time posting here so I hope I've met all conventions.

Answer

Looking at @user39756 's answer, I believe I have the answer. My hunch is that this is basically Borel-Cantelli I.

Borel Cantelli I states that:

Given measurable sets $\{E_n\}_{n \in \mathbb{N}}$ with $\sum_{n=1}^\infty \mu(E_n) \lt \infty$, we have $\mu(\limsup_{n \to \infty} E_n) = 0$.

Recall $\limsup_{n \to \infty} E_n = \bigcap_{i=1}^\infty \bigcup_{j=i}^\infty E_j$

Defining our appropriate measurable sets for this problem.

For $m \in \mathbb{N}$, let $A_{m,n} = \{x : |f_n(x)| \gt \frac{1}{m}\}$

Thus define $$A := \bigcup_{m \in \mathbb{N}} \bigcap_{i=1}^\infty \bigcup_{j=i}^\infty A_{m,j} = \bigcup_{m \in \mathbb{N}}\limsup_{n \to \infty} A_{m,n} = \{x : |\lim_{n\to \infty} f_n(x)| \gt 0 \} = \{x : \lim_{n\to \infty} f_n(x) \neq 0 \}$$

We want $\mu (A) = 0$, that is, $f_n \to 0$ a.e.

Now, $0 \le \mu (A) = \mu (\bigcup_{m \in \mathbb{N}}\limsup_{n \to \infty} A_{m,n}) \le \sum_{m \in \mathbb{N}} \mu(\limsup_{n \to \infty} A_{m,n}) = \sum_{m \in \mathbb{N}} 0 = 0$.

The above follow from subadditivity, and the fact that for each $m \in \mathbb{N}$, $\mu(\limsup_{n \to \infty} A_{m,n}) = 0$ by the finiteness of the sum in the problem statement for arbitrary $\epsilon \gt 0$.

Thus, $\mu(A) = 0$.

Would appreciate any feedback.