Prove that if $f(x,t)$ is continuous in $D=\{(x,t):x\in[a,b]\land t\in[c,d]\}$ then $F(x)=\int_c^d f(x,t)\mathrm dt$ is continuous in $[a,b]$
This is about Riemann integration. I dont know how to show this formally. I understand that if we took $x=x_0$ as a constant then the function
$$F(y)=\int_c^y f(x_0,t)\mathrm dt$$
is continuous in $[c,d]$. I can show that with the $\epsilon-\delta$ definition of continuity, when we fix a variable as a constant then the one-dimensional result is continuous too.
And if I can show that the general case
$$F(x,y)=\int_c^y f(x,t)\mathrm dt$$
is continuous in $D$ then Im done. Then
$$F(x,y)=\int_c^y f(x,t)\mathrm dt=\lim_{n\to\infty}U(f,P_{n,t})=\lim_{n\to\infty}\frac{y-c}{n}\sum_{k=1}^n M_k(x)$$
where $M_k(x)=\sup\{f(x,t):t\in[t_{k-1},t_k]\land x\in[a,b]\}$. Then if I prove that $M_k(x)$ is continuous for every $k$ and every partition $P_{n,t}$ and converges uniformly to some function $M(x)$ then $M(x)$ is continuous and so $F(x,y)$ is continuous.
But I get stuck here. Can you help me with some hint? Maybe Im overcomplicating the proof, there is a more easy path to do it? Thank you in advance.
As $f$ is continuous on $D$ (which is compact), it is so uniformly. Let $\epsilon > 0$. Then, there is $\delta > 0$ such that if $\|(x,y) - (x',y')\| < \delta$, then $|f(x,y) - f(x',y')| < \epsilon/(d-c)$. Suppose that $x,y \in [a,b]$ are such that $|x - y| < \delta$. Then, for any $t \in [c,d]$, $\|(x,t) - (y,t) \|= |x - y| < \delta$, so for all $t \in [c,d]$, $|f(x,t) - f(y,t)| < \epsilon/(d-c)$. Hence,
$$|F(x) - F(y)| = \left| \int_c^d f(x,t) dt - \int_c^d f(y,t)dt \right| = \left| \int_c^d (f(x,t) - f(y,t))dt \right| \\ \le \int_c^d |f(x,t) - f(y,t)|dt < \epsilon$$
This shows that $F$ is uniformly continuous.