I've seen a lot of related questions, but no one ever consider the convergence in measure.
Suppose you have a sequence of functions $k_n$ that are $C^1$ on $[0,1]$. Suppose that:
- $k_n$ converge in measure to a function $k$
- $k'_n$ converge in measure to a function $h$
Call $E$ the set of points of differentiability for $k$. Is it true that $k'(x) = h(x)$ almost everywhere in $E$?
Notice that in the case $k\in C^1$, we can rephrase the problem as $$ k_m\to 0,\quad k'_m\to h \implies h=0 \quad a.e.? $$
From what I've seen, I expect the existence of a counterexample
Thanks to the properties of the convergence in measure, you can consider equivalently the convergence almost everywhere. Moreover, if $k_m'$ are equibounded, then the claim is true thanks to Lebesgue's Dominated Convergence Theorem, so a possible counterexample must have $k_m'$ not equibounded.
Here's an idea that should work for a counterexample. Define $f:\mathbb R\to [0,1)$ as follows: On $[0,1)$ set $f(x) = x,$ and then extend $f$ to all of $\mathbb R$ by making $f$ periodic with period $1.$ Now set $f_n(x) = f(nx)/n, n=1,2,\dots$ Then on $[0,1],f_n \to 0$ uniformly. Furthermore, for each $n,$ $f_n$ is differentiable on $(0,1/n)\cup(1/n,2/n)\cup\cdots \cup ((n-1)/n,1),$ with $f_n'=1$ there. Thus we have $f_n\to 0$ in measure, and $f_n'\to 1$ in measure.
Of course these $f_n$ are not in $C^1[0,1],$ nor even in $C[0,1].$ But surely for each $n$ there is $g_n\in C^1[0,1]$ close enough to $f_n$ to make this work. In fact, there exist smooth $g_n$ such that $0\le g_n \le 1/n,$ with $g_n'=1$ except for a set $E_n$ of small measure. How small? $m(E_n)\to 0$ is enough to give $g_n'\to 1$ in measure.