Let $f:I\times E\to \mathbb R$ where $I$ is an interval. Suppose
$x\longmapsto f(x,t)$ is integrable for all $t$.
$t\longmapsto f(x,t)$ is derivable for a.e $x$, we denote the derivative $\frac{\partial f}{\partial t}$.
There is $\varphi:E\to \mathbb R^+$ in $L^1(E)$ s.t. $$\left|\frac{\partial f}{\partial t}(x,t)\right|\leq \varphi(x)$$ for a.e. $x$.
Set $$F(t)=\int_E f(x,t)dx.$$ Prove that $$F'(t)=\int_E \frac{\partial f}{\partial t}(x,t)dx,$$ for all $t$.
Attempts
We made the proof when $I$ is an interval of the form $[a,b]$ and $f$ is $\mathcal C^1([a,b])$. So the theorem is just a consequence of mean value theorem. I would like to prove this more general theorem. I did as follow :
Fix $t\in I$. We suppose first $I$ bounded. Since $$\lim_{h\to 0}\frac{f(x,t+h)-f(x,t)}{h}=\frac{\partial f}{\partial t}(x,t),$$ there is $\delta>0$ s.t. $$|\frac{f(x,t+h)-f(x,t)}{h}|\leq \left|\frac{\partial f}{\partial t}(x,t)\right|+1\leq \varphi(x)+1\in L^1(I).$$
Using DCT, we get $$\lim_{h\to 0 }\frac{F(t+h)-F(t)}{h}=\int_E\lim_{h\to 0}\frac{f(x,t+h)-f(x,t)}{h}dx=\int_E\frac{\partial f}{\partial t}(x,t)dx.$$
Q1) Is it correct ?
Q2) How can I do when $I$ is not supposed bounded ?
