Let $(a,b)\subset\mathbb{R}$ be an interval and let $\left\{f_t\colon\Omega\to\mathbb{R}\right\}_{t\in (a,b)}$ be a family of measurable functions on the measurable space $(\Omega,\mathcal{A},\mu)$, which are differentiable in $t_0\in (a,b)$ almost surely and for which there exists a $g\in\mathfrak{L}_{\mu}^1$ with $\lvert f_t-f_{t_0}\rvert\leq g\lvert t-t_0\rvert$ almost surely for all t in a neighboorhood $U$ of $t_0$. Show that for all $A\in\mathcal{A}$ it is $$ \left. \frac{d}{dt}\int_A f_t\, d\mu\right|_{t=t_0}=\int_A\left. \frac{d}{dt} f_t\right|_{t=t_0}\, d\mu. $$
Good evening!
Here is my proof, would you please say me if it is okay or if there are mistakes?
Proof
Consider any $A\in\mathcal{A}$. Without loss of generality consider a sequence $(t_n)_{n\in\mathbb{N}}\subset U\cap (a,b)$ with $t_n\to t_0$ and $t_n\neq t_0\forall~n\in\mathbb{N}$. Define $$ f_n:=\frac{f_{t_n}-f_{t_0}}{t_n-t_0},~~M_n:=\left\{f_{t_n}\text{ not differentiable in }t_0\right\},~~M:=\bigcup_{n=1}^{\infty}M_n. $$ It is $\mu(M_n)=\mu(M)=0$.
Moreover define $$ g_n:=1_{A\setminus M}f_n, n\geq 1. $$ $g_n$ is measurable, because $f_n$ and $1_{A\setminus M}$ are measurable. Additionally, it is pointwise $$ g_n\to 1_{A\setminus M}\left. \frac{d}{dt}f_t\right|_{t=t_0} $$ and $1_{A\setminus M}\left. \frac{d}{dt}f_t\right|_{t=t_0}$ is measurable, because $g_n$ is measurable and the pointwise limit exists.
(By the way: This is the reason why I defined $g_n$ and did not use $f_n$, because $f_n\to\left. \frac{d}{dt}f_t\right|_{t=t_0}$ almost surely (what would be okay for the Theorem of Lebesgue), but the limit of $f_n$ exists only almost surely and so $\left. \frac{d}{dt}f_t\right|_{t=t_0}$ is not measurable - but this is needed to apply Lebesgue.)
So all is given to use Lebesgue's Theorem.
$$ \lim_{n\to\infty}\int_A g_n\, d\mu=\int_A1_{A\setminus M}\left. \frac{d}{dt}f_t\right|_{t=t_0}\, d\mu. $$
Now, because of $$ g_n=f_n\text{ almost surely },~~1_{A\setminus M}\left. \frac{d}{dt}f_t\right|_{t=t_0}=\left. \frac{d}{dt} f_t\right|_{t=t_0}\text{ almost surely}, $$ it is $$ \lim_{n\to\infty}\int_A f_n\, d\mu=\lim_{n\to\infty}\int_A g_n\, d\mu=\int_A1_{A\setminus M}\left. \frac{d}{dt}f_t\right|_{t=t_0}\, d\mu=\int_A\left. \frac{d}{dt}f_t\right|_{t=t_0}\, d\mu. $$ And because of the linearity of the Lebesgue-integral it is finally $$ \lim_{n\to\infty}\int_A f_n\, d\mu=\left. \frac{d}{dt}\int_A f_t\, d\mu\right|_{t=t_0}. $$
Please say me if my proof is allright. Would be very kind!
Sinccerely yours,
math12