Let $f(t,s)$ be a (jointly) continuous bounded function on $(a,b) \times (a,b) \subseteq \mathbb{R}^2$. Suppose further than for each fixed $s \in (a,b)$, the continuous function $t \mapsto f(t,s)$ is differentiable (Lebesgue) almost everywhere on $(a,b)$ and that $\partial_t f(t,s) \in L^1(a,b)_t$.
I would like to know whether the continuous function $t \mapsto \int_a^b f(t,s) ds$ is differentiable for almost all $t \in (a,b)$ with derivative equal to $\int_a^b \partial_t f(t, s) ds$ (note there is some subtlety here about showing $s \mapsto \partial_t f(t,s)$ is measurable for almost all $t$).
The issue is that when looking at the quantity
$$ \frac{1}{h} \int_a^b f(t+h,s) - f(t,s) ds,$$
we want to use a statement like "for almost all $s$, $\frac{f(t+h,s) - f(t,s)}{h}$ converges to $\partial_t f(t,s)$ as $h \to 0$," but the only thing we can assume is sort of the opposite (for each fixed $s$, $\frac{f(t+h,s) - f(t,s)}{h} \to \partial_tf(t,s)$ for almost all $t$).
Note that if each $t \mapsto f(t,s)$ is differentiable on all of $(a,b)$, then by dominated convergence, there is no issue (apart from showing $s \mapsto \partial_t f(t,s)$ is measurable for each $t$).
The reason I care about proving this proposition is because I am trying to establish the variation of parameters formula for Sturm Liouville problems having coefficients that are $L^1$ functions. So, in this situation, we can only expect our solutions to be differentiable almost everywhere. The function $f(t,s)$ is really the the primary fundamental matrix in disguise (i.e., for each $s$, $t \mapsto f(t,s)$ is the solution of the ODE satisfying the initial condition $f(s,s)= 1$).
Edit: It seems that the variation of parameters formula I was looking at (Theorem 1.3.1 in Zettl's book Sturm-Liouville Theory) is not completely accurate. If you look at Theorem 3.1 of Coddington-Levinson, there is a simpler version of the formula which still solves the ODE (and therefore must be the unique solution) and in which the $t$-dependence appears outside the integral only. So then there is no issue. It would still be interesting to know whether the above assertion is true, but I no longer need it for the application I had in mind.
The statement is wrong.
Indeed, let $a=0$, $b=1$, and $$ f(t,s) = c(t-s)\mathbf{1}_{t\ge s}, $$ where $c$ is the Cantor staircase. Then, $f$ is jointly continuous, and for any $s\in [0,1]$, $\frac{\partial}{\partial t} f(t,s) = 0$ for almost all $t\in [0,1]$. However, for any $t\in (0,1]$, $$ \left(\int_0^1 f(t,s)ds\right)'_t = \left(\int_0^t c(t-s)ds\right)'_t= \left(\int_0^{t} c(u)du\right)'_t\\= c(t) > 0 = \int_0^1 \frac{\partial}{\partial t} f(t,s)ds. $$
As @pseudocydonia comments, certain stronger assumption like absolute continuity is needed.