Let $f(x,t)\in L^\infty(\Omega\times (0,T))$ $\Omega$ is a bounded domain in $\mathbb{R}^d$ and extend $f$ to zero oustide this domain. Consider for $h>0$ the integral $$F_h(t):= \int_\Omega \frac{f(x,t+h)-f(x,t)}{h}$$ By Fubini's theorem the integral exists for almost every $t\in(0,T)$. I want to know/show whether this integral is continuous in $h$. Note that the domain of integration is not $\Omega \times (0,T)$
What have I tried so far: I chose a sequence $h_n\to h$ and tried to estimate $$|F_h(t)-F_{h_n}(t)|\leq \int_\Omega |\frac{f(x,t+h)-f(x,t)}{h} - \frac{f(x,t+h_n)-f(x,t)}{h_n}|$$ Then I estimated the integrand $$|\frac{f(x,t+h)-f(x,t)}{h} - \frac{f(x,t+h_n)-f(x,t)}{h_n}|\leq |f(x,t)/h_n-f(x,t)/h| + |f(x,t+h)/h-f(x,t+h_n)/h_n|$$
Probably not the best estimate. The problem I find in any attempt i tried is that in general $|f(x,t+h)-f(x,t+h_n)|$ does not need converge to zero since $f$ is not continuous. Think of $h=1, t=0, h_n<1$ and $f=0$ on $[0,1)$ and $f(1)=1$. Of course this occurs in one single point. Can this occur on a non Null set? Why or Why not?
You can get a counterexample without even thinking about the $x$ dependence - if $f$ does not depend on $x$, then the integral collapses to $\mu(\Omega)$ and you're left with the question of whether the finite difference quotients of a function in $L^{\infty}((0,T))$ are continuous.
Consider $f(x,t) = \chi_{\{t \ge 1/2\}}$, $\chi$ the characteristic function. Then for every $t \in (0,1/2)$ we have
$$ F_h (t) = \mu(\Omega) \frac{1}{h} \chi_{\{ h \ge 1/2 - t \}} $$
which is piecewise continuous but jumps from $0$ to $\mu(\Omega) \frac{2}{1-2t}>0$ at $h=\frac{1}{2} - t$; so $F_h(t)$ is not continuous in $h$ (and nor is any function equal to it almost everywhere).