Prove that if $\int_{a}^{+\infty}f(x,y)dx$ converges uniformly to $F(y)$ on $[b,c]$, then $F$ is continuous. Let $a,b,c \in \Bbb R, b < c $ and let $f$ be a continuous real-valued function on the set $\{ (x,y) \in E^2 : x \ge a, y \in [b,c] \}$. Let $F:[b,c] \to \Bbb R$ be another function.  Prove that if $\int_{a}^{+\infty}f(x,y)dx$ converges uniformly to $F(y)$ on $[b,c]$, then $F$ is continuous.
I saw this problem on a past version of an exam published by the class I am following (but not taking, I just want to learn some RA). I was wondering how you would prove this.
 A: Assuming "uniform convergence" is meant in the sense that $$\lim_{t\rightarrow\infty}\int_a^t f(x,y) dx = F(y)$$ uniformly, then note that, since $F$ is assumed to be finite and uniform convergence implies pointwise convergence, that $\lim_{t\rightarrow\infty}\int_a^t f(x,y) dx $ exists as pointwise limit for each $y$. Also $ \int_a^t f(x,y) dx $ is a continous function in $y$ for each fixed $t$ (why?). Continuity of $F$ then follows because the uniform limit of continous functions is continous.
A: Let $\varepsilon>0$. Since the convergence is uniform, there is a $M>a$ such that $$\sup_{y\in[b,c]}\left|\int_{a}^tf(x,y)\mathrm dx-F(y)\right|<\frac{\varepsilon}{3}$$
if $t> M$.
Let $t>M$.
$$|F(y+h)-F(y)|\leq \left|\int_a^t f(x,y)\mathrm dx-F(y)\right|+\left|\int_{a}^{t}(f(x,y+h)-f(x,y))\mathrm dx\right|+\left|\int_a^{t}f(x,y+h)\mathrm dx-F(y+h)\right|\leq \frac{2\varepsilon}{3}+\int_a^t|f(x,y+h)-f(x,y)|\mathrm dx$$
Since $f$ is continuous on the compact, $[a,t]\times [b,c]$, $f$ is  uniformly continuous on $[a,t]\times [b,c]$, and thus, there is a $\delta=\delta_t>0$ s.t. $$|f(x,y+h)-f(x,y)|<\frac{\varepsilon}{3(t-a)}$$
if $|h|<\delta$ and thus $$\int_a^t|f(x,y+h)-f(x,y)|\mathrm dx\leq \frac{\varepsilon}{3}.$$
Finally, we conclude, that $|F(y+h)-F(y)|<\varepsilon$ if $|h|<\delta$ and thus $F$ is continuous.
