Measure Theory Conjecture While I was doing some math here, I made this conjecture.
Let $f_n:X\rightarrow \mathbb{R}$ be a sequence of measurable functions from the measure space $(X,\mathcal{A},\mu)$ to the measurable space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, and let $\mu:\mathcal{A}\rightarrow[0,\infty]$ be a positive measure. Consider that for all $\varepsilon > 0$, the sets $A_n = \{x\in X: \ |f_n(x)| > \varepsilon\}$ are such that $$\limsup_{n\to\infty}\mu(A_n) = 0.$$
Then its true that $f_n\to0$, $\mu$-almost everywhere.
The thing is, I think it is true but I don't know how to prove it. You can consider $\mu$ as a probability measure if it makes more easier to prove.
Thank you.
 A: The property you have named is called "convergence in measure":

We say that ${f_n}$ converges to ${f}$ in measure if, for every ${\epsilon > 0}$, the measures $\mu( \{ x \in X: |f_n(x) - f(x)| \geq \epsilon \} )$ converge to zero as $n \rightarrow \infty$.  (source)

In particular, assuming you mean $A_n = \{x \in X \; : \; |f_n(x)| > \epsilon\}$, your conjecture can be restated as

If $f_n : X \to \mathbb{R}$ are functions such that $f_n \to 0$ in measure, must it be true that $f_n \to 0$ $\mu$-almost everywhere?

(Note that you use $\limsup$ and the definition of convergence in measure uses $\lim$, but the two are equivalent since the measure is nonnegative.)
In general your conjecture is false.  However, a weaker statement is true: there must exist a subsequence of $f_n$ which converges to $0$ $\mu$-almost everywhere.
Counterexample to your conjecture as stated
The idea is to make $f_n$ equal to $1$ on a smaller and smaller set, so that $\mu(A_n) \to 0$, but to make those smaller and smaller sets jump around enough so that $f_n(x)$ is $1$ infinitely often for any $x$.
Consider $X = \mathbb{R}$ with Lebesgue measure, and define following sequence of functions:
$$
\chi_{[0,1]},
\chi_{[0,1/2]}, \chi_{[1/2,1]},
\chi_{[0,1/3]}, \chi_{[1/3,2/3]}, \chi_{[2/3,1]},
\chi_{[0,1/4]}, \ldots
$$
Then for any $\epsilon > 0$, $\mu(A_n) \to 0$ because it is the sequence $1, 1/2, 1/2, 1/3, 1/3, 1/3, \ldots$.  So $f_n \to 0$ in measure.  However, $f_n$ does not approach $0$ for any $x \in [0,1]$, so $f_n$ does not converge to the zero function almost everywhere.
Proof that some subsequence of $\boldsymbol{f}_\boldsymbol{n}$ converges almost everywhere to $\boldsymbol{0}$
Proof.
Let $f_n \to f$ in measure; we want to show a subsequence of $f_n$ converges to $f$ pointwise almost everywhere.
For any $k$, let $E_{n,k}$ be the set of points where $\left|f_n - f\right| > \frac{1}{k}$.
For a fixed $k$, convergence in measure implies $\mu(E_{n,k}) \to 0$.
For each $i$, choose an integer $n_i$ so that $n_i$ is an increasing sequence of integers, and $\mu(E_{n_i, i}) < \frac{1}{2^i}$.
Notice that this means $\mu(E_{n_i, k}) < \frac{1}{2^i}$ for $k \le i$ since $E_{n_i, k} \subset E_{n_i, i}$ when $k \le i$.
Now consider the set $E$ of points $x$ where $f_{n_i}(x) \not \to f(x)$.
We have
$f_{n_i}(x) \not \to f(x)$ if and only if there is some $k$ such that $\left|f_{n_i}(x) - f(x) \right| > \frac1k$ infinitely often.
Therefore,
$$
E = \bigcup_{k=1}^\infty \bigcap_{N=1}^\infty \bigcup_{i=N}^\infty E_{n_i,k}
$$
So
\begin{align*}
\mu(E)
&\le \sum_{k =1}^\infty \mu \left( \bigcap_{N=1}^\infty \bigcup_{i=N}^\infty E_{n_i,k} \right) \\
&\le \sum_{k =1}^\infty \lim_{N \to \infty} \mu \left( \bigcup_{i=N}^\infty E_{n_i,k} \right) \\
&\le \sum_{k=1}^\infty \lim_{N \to \infty} \sum_{i=N}^\infty \mu(E_{n_i,k}) \\
&= \sum_{k=1}^\infty {\lim_{N \to \infty \atop N \ge k}} \sum_{i=N}^\infty \mu(E_{n_i,k}) \\
&\le \sum_{k=1}^\infty {\lim_{N \to \infty \atop N \ge k}} \sum_{i=N}^\infty \mu(E_{n_i,i}) \quad \quad \text{(since } k \le i \text{)} \\
&\le \sum_{k=1}^\infty {\lim_{N \to \infty \atop N \ge k}} \sum_{i=N}^\infty \frac{1}{2^{i}} \\
&= \sum_{k=1}^\infty (0) = 0\\
\end{align*}
Therefore, $f_{n_i} \to f$ pointwise outside the set $E$ of measure $0$.
$\square$
