# Necessity of uniformity in "almost uniform convergence $\implies$ convergence a.e"

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.

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.

• Thanks for confirming my suspicion. A followup question: am I right that uniform convergence almost everywhere implies almost uniform convergence (because we can choose $A$ such that $\mu(A^c) = 0$, so it's trivially satisfied), while the converse is false, because in general, just because we have uniform convergence on each $A_k$, we may not have uniform convergence on $\bigcup A_k$ (an explicit example doesn't come to mind immediately, but I'm sure I can think of something in $\Bbb{R}$). Jul 31, 2020 at 23:03
• @user580918 You are right on what's weaker than what. Compare with the Taylor series of $(1+x^2)^{-1}$, which converges uniformly on compact subsets of $(-1,1)$ and pointwise on the interval, but for all $\delta$ and $n$ there is some $\varepsilon>0$ and $m>n$ such that $\lvert f-f_k\rvert>\delta$ on $[\varepsilon,1)$ for all $k> m$.
– user239203
Jul 31, 2020 at 23:29