I already kept myself busy with a proof concerning the Theorem of Lebesgue and differentiation of a parameter integral. Unfortunately I did not get an answer there yet.
Now my task is nearly the same - but this time dealing with continuity.
Consider the measruable space $(\Omega,\mathcal{A},\mu)$. Let $\left\{f_t\colon\Omega\to\overline{\mathbb{R}}\right\}_{t\in B}$ be a family of measruable function, induced by a subset $B\subset E$ of a metric space $E$, so that for a $g\in\mathfrak{L}_{\mu}^1$ it is $\sup\limits_{t\in B}\lvert f_t\rvert\leq g$ almost surely. Moreover assume that the function $t\mapsto f_{t}(\omega)$ is continious in $t_0\in B$ for $\omega\in\Omega$ almost surely. Show, that $t\mapsto\varphi(t):=\int_A f_t\, d\mu, B\to\mathbb{R},$ is continious in $t_0$ for all $A\in\mathcal{A}$.
Proof
Consider any $A\in\mathcal{A}$.
Let $(t_n)_{n\in\mathbb{N}}$ be a sequence in $B$ with $\lim_{n\to\infty}t_n=t_0$.
Set $$ f_n:=f_{t_{n}}, n\in\mathbb{N},~~~~~M_n:=\left\{f_{t_n}\mbox{ is not continious in }t_0\right\}, n\in\mathbb{N},~~~~~M:=\bigcup_{n=1}^{\infty}M_n. $$
$M_n, n\in\mathbb{N},$ and $M$ are zero sets, i.e. $$ \mu(M_n)=\mu(M)=0. $$
My intention is to apply the Theorem of Lebesgue.
It is $f_{t_n}\to f_{t_0}$ alomost surely. That would be fine in order to apply Lebesgue.
However, in order to apply Lebesgue, it is furthermore necessary that $f_{t_0}$ is measurable. Because the $f_n$ are measurable this would be the case if the limit $\lim_{n\to\infty}f_{t_n}(\omega)$ would exist pointwise. But this is not the case here, because the limit only exists almost surely!
In order to solve this problem I decided to consider
$g_n:=1_{A\setminus M}f_{t_n}$,
because the $g_n$ are measurable and hence here the pointwise limit exists and is $1_{A\setminus M}f_{t_0}$, so this time the limit is indeed measurable.
Because of $$ \lvert g_n\rvert\leq\lvert f_n\rvert\leq\sup_{t\in B}\lvert f_t\rvert\leq g\text{ almost surely} $$
I now can apply the Theorem of Lebesgue, getting $$ \lim_{n\to\infty}\int_A g_n\, d\mu=\int_A 1_{A\setminus M}f_{t_0}\, d\mu $$
and because of $g_n=f_n$ almost surely and $1_{A\setminus M}f_{t_0}=f_{t_0}$ almost surely, it is $$ \lim_{n\to\infty}\int_A f_{t_n}\, d\mu=\lim_{n\to\infty}\int_A g_n\, d\, d\mu=\int_A 1_{A\setminus M}f_{t_0}\, d\mu=\int_A f_{t_0}\, d\mu. $$
And that was to show. $\diamond$
Please tell me, if my proof is allright. I am really excited to know if my efforts are right!
Sincerely yours,
math12