$(\Omega,\mathscr{F},\mu)$ measure space. We have a map $f : \Omega \times \mathbb{R} \rightarrow \mathbb{R}$ with these three properties:
a) For all $x \in \mathbb{R}$ : $f(x,\star) : \Omega \rightarrow \mathbb{R}$ is integrable.
b) For all $\omega \in \Omega$: $f(\star, \omega) : \mathbb{R}\rightarrow \mathbb{R}$ is differentiable.
c) There exists an integrable function $g$, so that for all $x \in \mathbb{R}$ : $|\frac{\partial}{\partial x} f(x, \star) | \le g$ $\mu$-almost everywhere.
Show that:
$F: \mathbb{R}\rightarrow \mathbb{R}$ is differentiable.
$$F(x) := \int_{\Omega} f(x,\omega) d\mu(\omega).$$
And show that for all $x \in \mathbb{R}$:
$$F '(x) = \int_{\Omega} \frac{\partial}{\partial x} f(x,\omega) d\mu(\omega)$$
So what I have tried:
I'm pretty sure, that we have to use the dominated convergence theorem for Lebesgue integration. First of all we could start with $\lim_{x \rightarrow x_o} \frac{F(x)-F(x_o)}{x-x_o}$. Then we get $\lim_{x \rightarrow x_o} \frac{\int_{\Omega} f(x,\omega) d\mu(\omega) - \int_{\Omega} f(x_o,\omega) d\mu(\omega)}{x-x_o}$. But what do I now? Maybe: $\lim_{x \rightarrow x_o} \int_{\Omega} \frac{ f(x,\omega) dµ(\omega) - f(x_o,\omega) d\mu(\omega)}{x-x_0}$. Is this right? Could I use now the dominated convergence theorem? Then I could use the fact that $f$ is differentiable? But I'm not sure.
Thank you for your reply.
You already have $$\frac{F(x) - F(x_0)}{x-x_0} = \int_{\Omega} \frac{ f(x,\omega) - f(x_0,\omega)}{x-x_0} d\mu(\omega)$$
Additionally you have that $$\left|\frac{f(x,\omega) - f(x_0,\omega)}{x-x_0}\right| = \left|\frac{\partial}{\partial x}\ f(\xi,\omega)\right| \le g(\omega)$$ $\mu$-almost everywhere for a $\xi \in (x,x_0)$ hence you can use dominated convergence and you are done…