I have read some results about integrals of the form $$\int_D f(x,t)\,dt$$ for instance, the dominated convergence theorem and MCT. Also I see results for $F(x) = \int_0^\infty f(x,t)\,dt$ being continuous at $x_0$, provided $\int_0^t f(x,\eta)\,d\eta$ has a limit as $t \to \infty$ which is uniform in some open set around $x_0$. Also, under somewhat more stringent conditions, $\lim_{x\to\infty}\int_0^\infty f(x, \eta) \,d\eta = \int_0^\infty \lim_{x\to\infty} f(x, \eta) \,d\eta$. And of course if we fix $x_0$, of course $\int_0^t f(x_0,\eta)\,d\eta$ is continuous in $t$ by the continuity of the integral.
Suppose on the other hand we fix $t_0$. When is $F(x)=\int_0^{t_0} f(x,\eta)\,d\eta$ continuous at $x_0$? Since this should be easier than the improper integral mentioned above, it certainly seems strong enough that $f(x,\eta)$ converges uniformly as $t \to t_0$ in a neighborhood around $x_0$. But can we get by with less?
Is it then an easy extension to show when $$F(x,t)=\int_0^tf(x,\eta)\,d\eta,$$
is a continuous multivariable function of $(x,t)$? What restrictions do we need on $f(x, \eta)$?
I apologize if these are easy consequences of the other theorems, and I'm just not seeing it. Also I apologize if these are canonical results that I don't know. A google search didn't help me to answer these questions.
Assume that $f$ is defined on $X\times I$, where $X$ is a metric space and $I$ is an interval of $\mathbb R$ containing $0$. Here is a reasonably general set of conditions ensuring that the function $F$ is well-defined and continuous:
(o) $f(x,t)$ is measurable with respect to $t$, for each fixed $x\in X$;
(i) $f(x,t)$ is continuous with respect to $x$, for each fixed $t\in I$;
(ii) for any compact set $K\subset X$ and any compact interval $J\subset I$, there is an integrable function $g:J\to \mathbb R_+$ such that $\vert f(x,\eta)\vert\leq g(\eta)$ whenever $(x,\eta)\in K\times J$
Note that these conditions are satisfied if $f(x,t)$ is continuous wrt $x$ and $f$ is locally bounded on $X\times I$; in particular, this holds if $f$ is $jointly$ continuous on $X\times I$.
Conditions (o) and (ii) with $K=\{ x\}$ and $J=[0,t]$ implies that $F(x,t)$ is well-defined for any $(x,t)$.
To check continuity at some given point $(x,t)$, let us fix a sequence $(x_n,t_n)\subset X\times I$ converging to $(x,t)$. Then $K=\{ x\}\cup \{ x_n;\; n\in\mathbb N\}$ is a compact subset of $X$, and one can find a compact interval $J\subset I$ containing $\{ t\}\cup\{ t_n;\; n\in\mathbb N\}\cup\{ 0\}$. Let $g:J\to\mathbb R_+$ be chosen according to (ii). Assuming for example $t_n\geq t$, one can write \begin{eqnarray*} \left\vert F(x_n,t_n)-F(x,t)\right\vert&=&\left\vert \int_0^{t_n} f(x_n,\eta)\, d\eta-\int_0^t f(x,\eta)\, d\eta\right\vert\\ &=&\left\vert \int_t^{t_n} f(x_n,\eta)\, d\eta+\int_0^t \left(f(x_n,\eta)-f(x,\eta) \right)\, d\eta\right\vert\\ &\leq& \int_t^{t_n} g(\eta)\, d\eta +\int_J \vert f(x_n,\eta)-f(x,\eta) \vert\, d\eta \end{eqnarray*}
The second integral in the right-hand side goes to $0$ by (i) the dominated convergence theorem; and the first one as well by dominated convergence again, because it can be written as $\int_J \mathbb 1_{(t,t_n)}(\eta)g(\eta)\, d\eta$. This shows that $F$ is continuous at $(x,t)$.