Continuity of parameter dependent integral (source needed) I am looking for a reference from a book for the result of continuity of an integral (found in https://www.encyclopediaofmath.org/index.php/Parameter-dependent_integral):

Let $D \subset \mathbb{R}^n$ and $t \in (0,T)$, and let $f(x,t)$ be a function that is continuous in $t$ for almost every $x \in D$, and let $|f(x,y)| \leq g(x)$ where $g$ is integrable. Then 
  $$J(t) = \int_{D}f(x,t)\,dx$$
  is continuous with respect to $t$.

I tried the references on the site but they require continuity of $t \mapsto f(x,t)$ for every $x$, but I need it for almost every $x$ only.
 A: It's rather straightforward to reduce it to the assumption that $t\mapsto f(x,t)$ is continuous for all $x$, one just needs to replace $f$ with $h(x,t) = f(x,t)\cdot \chi_{D\setminus N}(x)$, where $N$ is a null set such that $t\mapsto f(x,t)$ is continuous for all $x\in D\setminus N$. $J(t)$ isn't changed at all.
However, you ask for a citable reference, so:


*

*Jürgen Elstrodt, Maß- und Integrationstheorie, Springer Verlag 1996, Chapter IV, Theorem 5.6


gives a slightly more general formulation. (I don't know if the book has been translated, it should have been, it's a good book.)
The translation is mine:

5.6 Theorem (Continuous dependency of an integral on a parameter):
  Let $T$ a metric space and $f\colon T\times X\to \mathbb{K}$ satisfy
  
  
*
  
*$x \mapsto f(t,x) \in \mathcal{L}^1$ for all $t\in T$.
  
*For $\mu$-almost all $x\in X$, $t\mapsto f(t,x)$ is continuous in $t_0 \in T$.
  
*There exists a neighbourhood $U$ of $t_0$ and an integrable function $g \colon X \to [0,\infty]$ such that for all $t \in U$: $$ \lvert f(t,\,\cdot\,)\rvert \leqslant g\quad \mu-\text{a.e.}^\dagger $$
Then the function $F \colon T \to \mathbb{K}$, $$ F(t) := \int_X f(t,x) d\mu(x)\quad (t\in T) $$ is continuous in the point $t_0\in T$, and also the map $\Phi \colon T\to \mathcal{L}^1,\; \Phi(t) := f(t,\,\cdot\,) \in \mathcal{L}^1\quad (t\in T)$ is continuous in $t_0\in T$
$ (\dagger) $ The union of the null sets $ N_t = \{\lvert f(t,\,\cdot\,)\rvert > g\}\quad (t\in U) $ need not be a null set.     

A: Let me give a direct proof.
Fix $ t\in (0, T) $, and let $ \{t_n\} $ be a sequence in $ (0, T) $ converging to $ t $. Then we have $$ \lvert f(x,t_n)\rvert\leqslant g(x)\quad\text{for every }\ n $$
where $ g $ is integrable. Since $ f(x,t) $ is continuous in $ t $ for almost every $ x\in D $, denote the set where $ f(x,t) $ is not continuous in $ t $ by $ 
E $ and $ \mu(E)=0 $ where $ \mu $ denote the Lebesgue measure. Hence $$ \lim_{t_n\to t}f(x, t_n)\to f(x, t)\quad\text{for} \ x\in D\setminus E .$$
Using the Dominated Convergence Theorem, we deduce that $ f(x,t) $ is integrable on $ D\setminus E $ and $$ \lim_{n\to\infty}\int_{D\setminus E}f(x,t_n)dx=\int_{D\setminus E}f(x,t)dx $$
Since $ \mu(E)=0 $, of course we have
$$ \lim_{n\to\infty}\int_{D}f(x,t_n)dx=\int_{D}f(x,t)dx .$$
Then
\begin{align}
\lvert J(t)-J(t_n)\rvert &=\left| \int_D f(x,t)dx-\int_D f(x,t_n)dx \right|
\end{align}
Let $ n\to\infty $, we have $ J(t_n)\to J(t) $. By the definition of Heine continuity we know that $ J(t) $ is continuous with respect to $ t $.
