If $f$ is continuous on rectangle $[a,b]\times[c,d]$, and if $g\in R$ on $[a,b]$, then the function $F$ defined by equation $$F(y)=\int_a^bg(x)f(x,y)dx$$ is continuous on $[c,d]$.
Proof:
If $G(x)=\int_a^xg(t)dt$, then by Theorem 1, $F(y)=\int_a^bf(x,y)dG(x)$.
Now by Theorem 2 , $F(y)=\int_a^bg(x)f(x,y)dx$ is continuous on $[c,d]$.
Theorem 1: Suppose $f\in R(\alpha)$ and $g\in R(\alpha)$ on $[a,b]$ where $\alpha$ is increasing on $[a,b]$. We define $F(x)=\int_a^xf(t)d\alpha (t)$ and $G(x)=\int_a^xg(t)d\alpha (t)$ if $x\in [a,b]$. Then $f\in R(G),g\in R(F), fg\in R(\alpha )$ on $[a,b]$ and $\int_a^bf(x)g(x)d\alpha (x)=\int_a^bf(x)dG(x)=\int_a^bg(x)dF(x)$.
Theorem 2: Let $f$ be a continuous function in each point $(x,y)$ of rectangle $[a,b]\times[c,d]$. Suppose $\alpha$ is bounded variation on $[a,b]$ and let $F$ be the function defined on $[c,d]$ by $F(y)=\int_a^bf(x,y)d\alpha (x)$. Then $F$ is continuous on $[c,d]$.
How the proof apply theorem 1? I don't see it.
When using theorem 2, how do we know that g is bounded variation?
In applying Theorem 1, let $\alpha(t) = t$. Since $\alpha$ is increasing, the theorem guarantees that $f \in R(G)$ with
$$\int_a^b f(x,y) g(x) \, dx = \int_a^b f(x,y) g(x) \, d\alpha(x) = \int_a^b f(x,y) \, dG(x),$$ where $$G(x) = \int_a^x g(t) \, d\alpha(t) = \int_a^x g(t) \, dt.$$
In applying Theorem 2, you need to satisfy the hypothesis that $G$ be of bounded variation. Note that for any Riemann integrable function $g$ the function
$$G(x) = \int_a^x g(t) \, dt$$
is of bounded variation since $g$ can be decomposed as $g = g^+ - g^-$ where $g^+ \geqslant 0$ is the positive part and $g^- \geqslant 0$ is the negative part.
Thus,
$$G(x) = \int_a^x g^+(t) \, dt - \int_a^x g^-(t) \, dt$$
is the difference of two non-decreasing functions and of bounded variation by the Jordan decomposition theorem.