does $f(x,y)$ continuous imply $\int_{a}^{b} f(x,y) \,dx $ is continuous with respect to y? let $f(x,y)$ be a continuous function define $ \int_{a}^{b} f(x,y) \,dx $ as a function of y $g(y)$. Is $g(y)$ continuous ?
I tried to link this by Fundamental Theorem of calculus by setting $F(x,y)=  \int_{a}^{x} f(t,y) \,dt $.
However I could not show it is continuous or not with respect to y. Any help would be appreciated!
Edit:
Is this valid ? $\lim_{y \to y_0} \int_{a}^{b} f(x,y) \,dx = \int_{a}^{b}\lim_{y \to y_0}f(x,y)=\int_{a}^{b}f(x,y_0)=g(y_o)$ .
 A: Set $F(y)=\int^a_bf(x, y)\,dx$, where $f$ is continuous on a finite box $[a,b]\times[c,d]$. For $y_0\in[c,d]$ you wish to see how the difference
$$ F(y)-f(y_0)=\int^b_af(x,y)-f(x,y_0)\,dx$$
behave when $y$ is close to $y_0$. The assumption of continuity of $f$ is key here.
I ignore if you are aware of the notion of uniform continuity. Here is more or less how the story goes (I am not a great story teller though):

For a continuous function on say, a box $C=[a,b]\times[c,d]$, uniform continuity >implies that for any $\varepsilon>0$, there is $\delta>0$ such that for any part >of points $(x,y)$ and $(x',y')\in C$,
$$ \begin{align}\|(x,y) - (x',y')\|<\delta\quad\text{implies}\quad|f(x,y)-f(x',y')|<\varepsilon\tag{1}\label{one}\end{align}$$
This is stronger that mere continuity in the sense that the $\delta>0$ depends only  $\varepsilon$ and not on any particular point in $C$.

This can then be used in your favor to attack the problem. For
$$
|F(y)-F(y_0)|\leq\int^b_a|f(x,y)-f(x,y_0)|\,dy
$$
So, given $\varepsilon>0$, there is $\delta>0$ such that \eqref{one} holds,  what can you say of $|F(y)-F(y_0)|$?
