"almost uniformly" means that for any $\epsilon>0$, there exists a set $E$, $\mu(E)<\epsilon$ such that $f_n\to f$ uniformly on $E^c$. The name may be a little misleading. By Egorov's theorem, if $\mu(X)<\infty$, then a.s. convergence implies almost uniform convergence.
What you're asked to show that almost uniform convergence does not imply the stronger uniform convergence almost everywhere.
A simple example for this is the sequence $f_n=x^n$ on $[0,1]$. This converges pointwise to the function $0$ on $(0,1)$ and to $1$ at $0$, that is, it converges a.e. to the constant function $f=0$. The sequence clearly converges almost uniformly to $f$, because it converges uniforly to $0$ on every interval of the form $[0,1-\epsilon]$.
What about uniform convergence except on a set of measure zero ?
Let $E$ have zero measure. Observe that for any $m$, there exists a point in the interval $[1-1/m,1-1/(m+1)]$ not in $E$ (othwise $E$ would have positive measure). Therefore (using monotonicity of $f_nx^n$):
$$\sup_{x \in [0,1]\backslash E} |f_n (x) - f(x)|\ge (1-1/m)^n-0,$$
for any $m$. Letting $m\to \infty$, we see that
$$\sup_{x\in [0,1]\backslash E} |f_n (x)-f(x)|\ge 1-0=1.$$
(It is of course equal to $1$).
Thus, no matter which measure zero set $E$ we choose, we will never converge uniformly to the (a.e.) limit $f=0$ outside $E$.
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.
Best Answer
Not necessarily.
Consider a Smith-Volterra-Cantor set, a compact and nowhere dense set $C \subset [0,1]$ with positive Lebesgue measure. Let $C_n$ be the finite union of closed intervals remaining at stage $n$ of the construction, so that $C_1 \supset C_2 \supset \dots$ and $C = \bigcap_n C_n$.
Set $f_n = 1_{C_n}$ and $f = 1_C$ to the corresponding indicator functions, so we have $f_n \to f$ pointwise everywhere. Now I claim that for each $x \in C$, we do not have $f_n \to f$ locally uniformly at $x$. Let $U$ be any neighborhood of $x$ and let $n$ be arbitrary. Since $x \in C_n$, which is a union of closed intervals, we have $x \in I \subset C_n$ for some closed interval $I$. But $C$ is nowhere dense, so there exists some $y \in (I \setminus C) \cap U$. Hence we have $y \in U$, $f_n(y) = 1$, and $f(y) = 0$, so $\sup_{z \in U} |f_n(z) - f(z)| = 1$, and so the sequence $f_n$ does not converge to $f$ uniformly on $U$. This holds for all $x \in C$, and $C$ has positive measure.