Continuity of Integral of Two Variables If $f:[a,b] \times Y \mapsto\mathbb{R}$ is continuous, $Y$ a generic metric space, then why is $$F(y) = \int_{a}^{b}f(x,y)\,dx$$ continuous?  
 A: I am not entirely convinced of your argument, Math1000.  For one thing, those $\delta$ you initially chose are really "$\delta_n$", as each $f^{y_n}(x)$ is ${\it individually}$ uniformly continuous, and this gets in the way of your subsequent compactness argument.
A correct argument would have to employ continuity of the function ${\it as}$ ${\it a}$ ${\it function}$ ${\it in}$ ${\it two}$ ${\it variables}$, not solely continuous as $x$ and $y$-sections.  Consider the following:
By uniform continuity of $[a,b]$ we may find $\delta$ such that $| x_1-x_2| < \delta$ implies $| f(x_1,y_0)-f(x_2,y_0)| < \epsilon/3$, and so we layer $[a,b] \times y_0$ with a finite $\delta$-net $\{x_1, \dots ,x_n \}$.  Then we may find $\delta_i^{'} < \delta$ such that, for each $(x_i,y_0)$, $\gamma\big(s,(x_i,y_0)\big) < \delta_i^{'}$ implies $|f(s
) - f(x_i,y_0)|<\epsilon/3$, $s \in [a,b] \times Y$. Here $\gamma$ is a natural product metric arising from the individual spaces, say $$\gamma(s,t) = \gamma\big((s_1,t_1),(s_2,t_2)\big) := \text{max}\{|s_1-t_1|, d(s_2,t_2)\}.$$ Now define $\delta^{'} := \min_{i}\delta_i \le \delta$ and refine the existing $\delta$-net to obtain a new $\delta^{'}$-net $P':=\{x_1, \dots ,x_{n^{'}}\}$, $n' \ge n$.  Ensure that $(x,y) \in B_{\delta^{'}}\big((x_i,y_0)\big) := B$ for some $i$, $1 \le i \le n'$, so long as $d(y,y_0) < \delta^{'}$.  (This is because of the way the $\gamma$ metric is defined.)  Now we may safely apply the $\epsilon/3$ argument to get the following inequality: $$|f(x,y)-f(x,y_0)| \le |f(x,y) - f(x_i,y)| + |f(x_i,y)-f(x_i,y_0)| + |f(x_i,y_0) - f(x,y_0)|.$$  The first two quantities on the right-hand side are each $<\frac{\epsilon}{3}$ because $(x,y)$, $(x_i,y)$, and $(x_i,y_0)$ all lie in the same ball $B$.
  The last term is $<\frac{\epsilon}{3}$ by uniform continuity, upon noting that $$|x_i - x| \le \gamma\big((x_i,y_0),(x,y_0)\big) \le \delta^{'} < \delta.$$ 
A: Since $f$ is continuous, the $x$- and $y$-sections of $f$ are continuous, that is, the maps $f_x:Y\to\mathbb R$ and $f^y:[a,b]\to\mathbb R$ defined by
$$f_x(y)=f^y(x) = f(x,y). $$
Moreover, the $y$-sections are uniformly continuous as $[a,b]$ is compact. Suppose $y_n\stackrel{n\to\infty}\longrightarrow y$ in $Y$. Then for any $x\in[a,b]$, $$f^{y_n}(x)=f_x(y_n)\stackrel{n\to\infty}\longrightarrow f_x(y) = f^y(x).$$ Set $$h(x) = |f^{y_n}(x)-f^y(x)| $$ for $x\in[a,b]$. Let $x,x'\in[a,b]$, then
$$|f^{y_n}(x)-f^y(x)|\leqslant |f^{y_n}(x)-f^{y_n}(x')|+|f^{y_n}(x')-f^y(x')|+|f^y(x')-f^y(x)|, $$
so that 
\begin{align}
h(x)-h(x')&=|f^{y_n}(x)-f^y(x)|-|f^{y_n}(x')-f^y(x')|\\
&\leqslant |f^{y_n}(x)-f^{y_n}(x')|+|f^y(x')-f^y(x)|,
\end{align}
and by symmetry,
$$|h(x)-h(x')| \leqslant |f^{y_n}(x)-f^{y_n}(x')|+|f^y(x)-f^y(x')|. $$
Let $\varepsilon>0$ and choose $\delta>0$ such that $|x-x'|<\delta$ implies
$$\min\{|f^{y_n}(x)-f^{y_n}(x')|, |f^y(x)-f^y(x')|\}<\frac\varepsilon2.$$
Then for $|x-x'|<\delta$ we have
$$|h(x)-h(x')|\leqslant \frac\varepsilon2+\frac\varepsilon2=\varepsilon, $$
so that $h$ is continuous. Since $[a,b]$ is compact, it follows that there exists $x_n\in[a,b]$ such that $$x_n=\underset{x\in[a,b]}{\arg\max}|f^{y_n}(x)-f^{y}(x)|.$$ By compactness there exists a convergent subsequence $x_{n_k}$ with limit $x\in[a,b]$. It follows that
$$\lim_{k\to\infty}|f^{y_{n_k}}(x_{n_k})-f^y(x_{n_k})| = \lim_{k\to\infty}|f_{x_{n_k}}(y_{n_k})-f_{x_{n_k}}(y)|=0. $$
This implies that $$\limsup_{n\to\infty}|f^{y_n}(x)-f^y(x)|=0$$ and hence $$\lim_{n\to\infty}|f^{y_n}(x)-f^y(x)|=0.$$ 
It follows that
\begin{align}
|F(y_n)-F(y)| &= \lim_{n\to\infty}\left|\int_a^b(f(x,y_n) - f(x,y))\ \mathsf dx\right|\\
&= \left|\int_a^b(f^{y_n}(x) - f^y(x))\ \mathsf dx\right|\\
&\leqslant \int_a^b|f^{y_n}(x)-f^y(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow0,
\end{align}
from which we conclude.
