Studying the Continuity of a function 
Let $U \subset \mathbb{R}^N$ be an open set, let $f : U \times [a, b] \to \mathbb{R}$ be a continuous function. Consider the function $$g(x):= \int_a^b f(x,y) \,dy$$
  with $x \in U$.
i) Prove that $g$ is continuous in $U$.
ii) consider the function $f : [−1, 1] \times  [−1, 1] \to \mathbb{R}$ defined by
  $$f (x, y) :=\begin{cases} \frac {|y|−|x|} {y^2}&\text{ if $|x|<|y|$}\\
0 &\text{ if $|x|\geq |y|$}
\end{cases}$$
  Let $g(y):= \int_{-1}^1 f(x,y) \,dx$ for $y \in [-1,1]$
  Study the continuity of $g$.

So I am a bit stuck on how to prove the continuity from the basics: i know how to prove that if $f$ is continuous on $[a,b]$, then $g=\int f$ is continuous on $[a,b]$ but i assume because of the different notation and dimension here, i have to prove a different way? In addition, how would i study the continuity?
 A: First note that continuity is a local property. You can exploit this property together with the assumption that $U$ is open in $\Bbb R ^n$. That is: we can simplify the exercise to show that $g$ is continuous in any compact subset of $U$ because any point on an open set of $\Bbb R ^n$ (assuming that $\Bbb R ^n$ is a normed space) have a compact neighborhood.
Using compact sets is easy because compacts sets behaves, in many ways, as if there were finite sets. In particular we will use the property that any continuous function in a compact set is uniformly continuous, that is, if $C\subset U$ is compact then $f$ is uniformly continuous in $C\times [a,b]$, because Cartesian product of compact sets is a compact set in the product topology.
First note that
$$
\begin{align*}
|g(x_0)-g(x)|&\leqslant \int_{a}^b|f(x_0,y)-f(x,y)|\,\mathrm d y\\
&\leqslant (b-a)\sup_{y\in[a,b]}|f(x_0,y)-f(x,y)|
\end{align*}\tag1
$$
Because $f$ is uniformly continuous in $C\times[a,b]$ then for any $\epsilon >0$ there is a $\delta >0$ such that if $\|(x_0,y_0)-(x,y)\|<\delta $ then $|f(x_0,y_0)-f(x,y)|<\epsilon $ for all $(x_0,y_0),(x,y)\in C\times [a,b]$, where $\|{\cdot}\|$ is any norm on $\Bbb R ^n$ (all norms are equivalent in finite dimensional vector spaces, so you can choose the norm you like more, by example the maximum norm).
Then from $\rm (1)$ we have that if $\|(x_0,y)-(x,y)\|_\infty=|x_0-x|<\delta $ then $|g(x_0)-g(x)|<(b-a)\epsilon $, for all $x_0,x\in C$, what finishes the proof.
A: Let $x \in U$ such that  $x_n \to x$
Then $x_n$ is bounded,thus exists a closed ball $B:=\overline{B(0,M)}$ such that $x_n \in B,\forall n \in \Bbb{N}$
Take $f_n(y)=f(x_n,y)$ we have that $f(x,y)$ on $[a,b]$ by continuity of $f$ on  $B \times [a,b]$
In this cartesian product we have the metric $d((x_1,y_1),(x_2,y_2))=\sqrt{||x_1-x_2||_2^2+|y_1-y_2|^2}$ where $||x_1-x_2||_2$ is the usual metric on $\Bbb{R}^N$
Since $B\times [a,b]$ is compact then $f$ is uniformly continuous with respect to this metric on $B \times [a,b]$
Let $\epsilon>0$
$\delta>0$ and $n_0 \in \Bbb{N}$ such that $||x_n-x||_2<\delta,\forall n \geq n_0$
So $d((x_n,y),(x,y))=||x_n-x||_2<\delta$ for every $y \in [a,b]$ and $\forall n \geq n_0$
and $|f(x_n,y)-g(x,y)| <\epsilon,\forall y \in [a,b]$ by uniform continuity.
Thus $\sup_{y \in [a,b]}|f(x_n,y)-f(x,y)| \leq \epsilon,  \forall n \geq n_0$
Thus $f_n(y) \to f(x,y)$ uniformply on $[a,b]$
So by using this  and the interchange of limit and riemman integral under uniform convergence,you have the desired conclusion.
For the second part,just study and calculate the integral  use the first part.
