Continuity of parameter dependent integral (source needed)
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.