Let $F:[a,b]\times [a,b] \rightarrow \mathbb{R}$ and $f:[a,b] \rightarrow \mathbb{R}$ be two continuous functions, let $g:[a,b] \rightarrow \mathbb{R}$ defined as:
$$ g(x) = \int_a^b F(x,y) f(y) dy$$
Show that $g$ is continuous
My attempt
Since $F$ is continuous and it is defined on a compact subset of $\mathbb{R}^2$ then $F$ is uniformly continuous, so it is easy to see that for each $x,y \in [a,b]$ the functions $F_x, F_y:[a,b] \rightarrow \mathbb{R}$ defined as: $$ F_x(y) = F(x,y)$$ $$F_y(x) = F(x,y)$$
are also uniformly continuous. We can write $g$ as:
$$ g(x) = \int_a^b F_y(x) f(y) dy$$
Note that: $$|g(x_1)-g(x_2)| = \left|\int_a^b F_y(x_1) f(y) dy - \int_a^b F_y(x_2) f(y) dy \right| = \left|\int_a^b (F_y(x_1)-F_y(x_2)) f(y) dy \right|$$
Let $\epsilon > 0$, there exists $\delta >0$ such that if $|x_1-x_2| < \delta$ then
$|F_y(x_1)-F_y(x_2)| < \dfrac{\epsilon}{b-a}$ $\Rightarrow \int_a^b |F_y(x_1)-F_y(x_2)|dy < \epsilon$
So we have:
$$\left|\int_a^b F_y(x_1)-F_y(x_2) dy \right| \leq \int_a^b |F_y(x_1)-F_y(x_2)| dy < \epsilon$$
But the left side of the inequality is not exactly $|g(x_1)-g(x_2)|$, so I'm having trouble trying to find an expression to prove the continuity of $g$. Is there an easy way to prove it?
$$\begin{align}\vert g(x_1)-g(x_2)\vert &= \left\vert\int_a^b F(x_1,y)f(y)dy-\int_a^b F(x_2,y)f(y)dy \right\vert\\ &= \left\vert\int_a^b (F(x_1,y)-F(x_2,y))f(y)dy \right\vert\\ &\leq \int_a^b\vert F(x_1,y)-F(x_2,y)\vert \vert f(y)\vert dy \\ &\leq \sup_{a<y<b}\vert F(x_1,y)-F(x_2,y)\vert \int_a^b\vert f(y)\vert dy\end{align}$$
The integral is finite as $f$ is continuous. Can you finish it from here?