Proving pointwise convergence almost everywhere I would like to prove the following statement:
Let $(X, F, \mu)$ be a measure space, not necessarily finite. Suppose that for every $\epsilon > 0$ there exists a natural number $N$ such that $ \mu(\bigcup_{n = N}^{\infty} \{x \in X : |(f_n(x) - f(x)| > \epsilon \}) < \epsilon $.
Then $f_n \to f$ pointwise almost everywhere.
My idea was to prove the statement by contradiction, i.e. assuming the negation of pointwise convergence almost everywhere:
Suppose there exists an $\epsilon'$ such that for all $N$ there exists an $n \geq N$ with $|f_n(x) - f(x)| > \epsilon$ for almost every $x \in X$.
By assumption we know that for any $\epsilon$, in particular for $\epsilon'$, we can find an $N_0$ such that for all $n \geq N_0$ we have $|f_n(x) - f(x)| > \epsilon'$ for almost every $x \in X$.
I thought this would imply that
$\mu(\bigcup_{n = N_0}^{\infty} \{x \in X : |(f_n(x) - f(x)| > \epsilon' \}) = \mu(X)$,
but even if that were the case, I do not see a way to reach a contradiction as I cannot assume that $\mu(X) > \epsilon$, I've probably made a mistake, but I cannot see where.
Is there any way I could complete this proof, or is there a better way?
Thank you in advance.
 A: Since OP asks for an alternative way, I give a constructive approach.  I prefer this since we see the truth along the proof.  Moreover, the skill of converting an uncountable union into a countable one is often re-used.
For any fixed $\epsilon' > 0$,
\begin{align}
& \{x \in X : f_n(x) \not\to f_n(x) \} \\
=& \{x \in X : \exists \epsilon > 0, \forall N \in \Bbb{N}, \exists n \ge N, |f_n(x) - f(x)| > \epsilon \} \\
=& \bigcup_{\epsilon > 0} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon \}\\
=& \bigcup_{k \in \Bbb{N}} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}.
\end{align}
Show that the last equality is true:


*

*$\subseteq$: take $k$ large enough so that $\epsilon > \epsilon'/2^k$

*$\supseteq$: for any $k \in \Bbb{N}$, choose $\epsilon$ sufficiently small so that $\epsilon'/2^k > \epsilon$
For each $k \in \Bbb{N}$, invoke the given condition (with $\epsilon = \epsilon'/2^k$) to find $N_k \in \Bbb{N}$ so that
$$\mu\left(\bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}\right) < \epsilon'/2^k.$$
It's not hard to check that
$$\bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \subseteq \bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \},$$
from which we get our desired conclusion
\begin{align}
& \mu\left(\bigcup_{\epsilon > 0} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon \}\right) \\
=& \mu\left(\bigcup_{k \in \Bbb{N}} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}\right) \\
\le& \sum_{k \in \Bbb{N}} \mu\left( \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \right) \\
\le& \sum_{k \in \Bbb{N}} \mu\left( \bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \right) \\
\le& \sum_{k \in \Bbb{N}} \epsilon'/2^k = \epsilon'.
\end{align}
Since $\epsilon' > 0$ is arbitrary, we conclude that $\mu(\{f_n \not\to f\}) = 0$, i.e. $f_n \to f$ a.e.
