Let $(X,\mathcal{A}, \mu)$ be a measure space, and $E$ a Banach space (for this discussion, a metric space suffices I guess). We say a sequence of functions $f_n:X \to E$ is $\mu$-almost uniformly convergent if for every $\delta>0$, there is a measurable set $A\in \mathcal{A}$ with $\mu(A^c) < \delta$ and such that the restricted sequence $\{f_n|_{A}\}$ is uniformly convergent.
It is then a common theorem that "almost uniform convergence implies convergence a.e", and the proof goes like this:
For each $k\in \Bbb{N}$, we set $\delta_k = \frac{1}{k}$ for example, and obtain a corresponding measurable set $A_k$ as per the definition. Put $A:= \bigcup A_k$, then it's easy to see that $\mu(A^c) = 0$. Since $\{f_n|_{A_k}\}$ is uniformly convergent, it is pointwise convergent, and hence we can define $f:X\to E$ by
\begin{align}
f(x):=
\begin{cases}
\lim\limits_{n\to \infty}f_n(x) & \text{if $x\in A$}\\\\
0 & \text{otherwise}
\end{cases}
\end{align}
Then, $f_n \to f$ pointwise on $A$, which completes the proof (because $\mu(A^c) = 0$).
My question is whether the uniformity assumption is actually necessary, because based on the proof it seems that it's not needed, but every source I read always adds in this seemingly extra hypothesis (maybe they just want to give a sufficient condition?). So, I would just like some verification, to make sure I'm not overlooking something obvious.
Best Answer
Uniformity is not needed, but while there is a difference between almost uniform convergence and uniform convergence almost everywhere, there is no material reason to introduce the terminology almost pointwise convergence: a sequence $\{f_n\}_{n\in\Bbb N}$ such that for all $\delta>0$ there is some $A$ such that $\mu(X\setminus A)<\delta$ and $\{\left.f_n\right\rvert_A\}_{n\in\Bbb N}$ converges pointwise is just almost-everywhere convergent, and the seemingly weaker condition is not easier to state, nor does it look significantly easier to check.