I wanted to check that the proof I wrote for the following question is correct.
Let $H = [a,b] \times [c,d] \subset \mathbb{R}^2$, and suppose $f: H \rightarrow \mathbb{R}$ is continuous and $g:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable. Show that: $$F(y) = \int_a^bf(x,y)g(x)dx$$ is continuous
Pf:
Let $M(y)=\max_{x\in[a,b]}f(x,y)$. We note here that $M$ is an increasing monotonic function from $d$ to $c$. Suppose that $|p-q| < \delta$. Further assume $p>q$. Analyze the difference $|F(p)-F(q)|$: $$|F(p)-F(q)| = |\int_a^bf(x,p)g(x)dx - \int_a^bf(x,q)g(x)dx| \\ \leq |M(p)-M(q)|\int_a^bg(x)dx$$
$M$ is monotonic so the difference $|M(p)-M(q)|$ can be made arbitrarily small. $g$ is Riemann Integrable so $|M(p)-M(q)|\int_a^bg(x)dx$ can be made arbitrarily small. Hence $F$ is continuous.
I'm worried about my reasoning with $M$, I think it might be flimsy. In particular, $M$ may have jumps - I don't think this is true though because $f$ is continuous. I'm not sure how to argue here though.
Your function $M$ is not an increasing monotonic function: there is an $f$ such that $y_{1} < y_{2}$ that it could not be the case that $f(x,y_{1}) \geq f(x,y_{2})$ for all $x\in [a,b]$. For instance, if $f(x,y) = -xy$ and $0 < a < b$, then $f(x,1) \geq f(x,2)$. Another way to this is to consider $H=[0,1]^{2}$ and a function $f$ that only attains its global max at $(0.5,0.5)$. Then if you move away from $y=0.5$, $M(y)$ must decrease.
To prove this result, you need to use uniform continuity and consider $|F(p)-F(q)| \leq \int_{a}^{b} |f(x,p)-f(x,q)||g(x)|$.
Another way to see it is to consider the sequence of functions $f_{n}(x) = f(x,y_{n})g(x)$ and prove it converges uniformly to $f(x,y)g(x)$ if $y_{n}\to y$.