If $J_1=[a,b], J_2=[c,d]$, and if $f$ is continous on $J_1\times J_2$ to $\mathbb{R}$ and $g$ is Riemann integrable on $J_1$, then the function $F$, defined on $J_2$ by $$F(t)=\int_a^bf(x,t)g(x)dx$$ is continuous on $J_2$.
My approach:
I know have to show that for $t_0\in [c,d]$ $\forall$, $\exists \delta > 0$, such that for all $t$, $|t-t_0|<\delta\quad$ then $\quad |\int_a^bf(x,t)g(x)dx-\int_a^bf(x,t_0)g(x)dx|<\epsilon$
I can see that the continuity of $F$ comes from the continuity of $f$. And that $g$ is Riemann Stieltjes integrable with respect to $x$.
I've been struggling with this proof for a while. Any suggestions would be great!
$f$ being continuous on a compact set ($J_{1} \times J_{2}$) implies uniform continuity.
$$ \begin{aligned} | F(t) - F(t_0) | & = \left| \int_{a}^{b} (f(x,t)-f(x,t_0)) g(x) \text{ d}x \right| \\ & \leqslant \int_{a}^{b} \underbrace{|f(x,t) - f(x,t_0)|}_{< \ \varepsilon \text{ by uniform continuity}} |g(x)| \text{ d}x \\ & \leqslant \varepsilon \int_{a}^{b} |g(x)| \text{d}x \end{aligned} $$
$g$ being Riemann integrable implies that $g$ is bounded (and in particular $|g|$ is bounded too).