yesterday I asked a question whether for $f$ absolutely continuous with $f' \in L^1$:$$\lim_{h \downarrow 0} \int_0^t \frac{f(s+h)-f(s)}{h}= \int_0^t f'(s)ds$$
In fact what I needed though was that for some continuous function $g$
$$\lim_{h \downarrow 0} \int_0^t g(s)\frac{f(s+h)-f(s)}{h}= \int_0^t g(s)f'(s)ds.$$
The thread is here: click me.
I was hoping that the proof of the first statement could help me find one for the second one, but in fact, it did not, because the presents of $g$ is an actual problem apparently.
Since $f$ is absolutely continuous, you have that $f(s+h)-f(s)=\int_s^{s+h}f'(t)\,dt$ so you can write $$\int_0^t \left(g(s)\frac{f(s+h)-f(s)}{h} -g(s)f'(s)\right)\,ds=\int_0^t g(s)\left(\frac{1}{h}\int_s^{s+h}(f'(t)-f'(s))\,dt \right)\,ds.$$ Now the function $\varphi(x)=\chi_{[-1,0]}(x)$ is a mollifier. Hence $\varphi_h(x)=\frac1h\chi_{[-1,0]}(x/h)=\frac1h\chi_{[-h,0]}(x)$ and $(\varphi_h* f')(s)=\int_{\mathbb{R}}\varphi_h(s-t)f'(t)\,dt=\frac1h \int_s^{s+h}f'(t)\,dt$. So you have $$\int_0^t \left(g(s)\frac{f(s+h)-f(s)}{h} -g(s)f'(s)\right)\,ds=\int_0^t g(s)((\varphi_h* f')(s)-f'(s))\,ds.$$ By the properties of mollifiers you have that $\varphi_h* f'\to f'$ in $L^1$ as $h\to 0$ and so $$\int_0^t |g(s)((\varphi_h* f')(s)-f'(s))|\,ds\le \Vert g\Vert_\infty\int_0^t |(\varphi_h* f')(s)-f'(s)|\,ds\to 0$$