A well-known theorem states that if $\frac{d}{dx} a_n(x)$ is uniformly convergent then the limit $a_n(x)\rightarrow a(x)$ can be differentiated term by term, i.e. $\frac{d}{dx}a(x) = \lim_{n\rightarrow \infty}\frac{d}{dx}a_n(x)$.
Now suppose $f(x)=\lim_{n\rightarrow \infty} f_n(x)$ converges on $\mathbb{R}$ and is differentiable. Suppose $ f_n'(x)$ also converges pointwise everywhere. Without any uniform convergence assumptions, must we have $f'= \lim_{n\to\infty} f_n'(x)$ a.e.?
I've seen an example here where they differ at one point.
I think the answer is yes since pointwise convergence of continuous functions implies uniform convergence on a set of measure as large as we want, so they at least must agree on a large set. I tried to modify the proof given in Rudin but its not clear where we can swap "uniform convergence" for "existence of $g$ derivative"
Best Answer
Thanks to Dave Renfro (whose posts on StackExchange are legend). In the link
he points us to this paper:
Darji, Udayan B. Limits of differentiable functions. Proc. Amer. Math. Soc. 124 (1996), no. 1, 129-134.
Abstract. Suppose that $\{f_n\}$ is a sequence of differentiable functions defined on $[0,1]$ which converges uniformly to some differentiable function $f$, and $f_n'$ converges pointwise to some function $g$. Let $M= \{x: f'(x)\not = g(x)\}$. In this paper we characterize such sets $M$ under various hypotheses. It follows from one of our characterizations that $M$ can be the entire interval $ [0,1]$.
I cannot resist quoting Dave's account of Darji's research (by the way we all call Darji "Darji").