Let $f$ be an absolutely continuous function with $f' \in L^1$.
Is it then true that $\lim_{h \downarrow 0}\int_0^t \frac{f(x+h)-f(x)}{h}dx= \int_0^t f'(s) ds$?
Here $t>0$ is finite.
What is problematic for me is to find an upper bound for the integrand. Obviously the right side exists, but is there a way to justify this?
By Lebesgue's fundamental theorem, if $f$ is absolutely continuous on $[0,t]$ then $f'$ is integrable and
$$\tag{FTC}\int_0^t f'(x) \, dx = f(t) - f(0)$$
We also have,
$$\int_0^t [f(x+h) - f(x)] \, dx = \int_h^{t + h}f(x) \, dx - \int_0^{t }f(x) \, dx = \int_t^{t + h}f(x) \, dx - \int_0^{h }f(x) \, dx, $$
and, since absolute continuity implies continuity of $f$,
$$\lim_{h \to 0+}\frac{1}{h}\int_0^t [f(x+h) - f(x)] \, dx = \lim_{h \to 0+}\frac{1}{h}\int_h^{t + h}f(x) \, dx - \lim_{h \to 0+}\frac{1}{h}\int_0^{h }f(x) \, dx = f(t) - f(0) \\ = \int_0^t f'(x) \, dx$$