$f:[-1,1]\to\Bbb{R}$ be continuous function, and let $g(x)=\int_{0}^{1}f(xy)dy\forall x\in[-1,1]$

Question

The whole question is
Suppose $f:[-1,1]\to\Bbb{R}$ be continuous function and define $g(x)=\int_{0}^{1}f(xy)dy\ \forall x\in[-1,1]$. Prove that $g$ is continuous on $[-1,1]$.
I have tried it in the following manner-
Choose $\varepsilon>0$
$f$ is uniformly continuous on $[-1,1]$, so $\exists\delta>0$ such that $|f(a)-f(b)|<\varepsilon$ if $|a-b|<\delta$.
Let $x_1,x_2\in[-1,1]$, my target is to find a $\delta'>0$ such that $|g(x_1)-g(x_2)|<\varepsilon$ if $|x_1-x_2|<\delta'$.
Now choose $y\in[0,1]$, $|f(x_1 y)-f(x_2 y)|<\varepsilon$ if $|x_1y - x_2 y|<\delta$ i.e. $|y||x_1 - x_2 |<\delta$.
From this I can't proceed further, how can I get a fixed $\delta'>0$ for any $y\in[0,1]$. Then I can get the required relation for $g$ just by integrating both sides with respect to $y$ on $[0,1]$.
Thanks for assistance in advance.

Answer 1

Note that if $|x_1-x_2|<\delta$, where $\delta$ is given by the uniform continuity of $f$, then $|x_1y-x_2y|<\delta$ for all $y\in [0,1]$, since $|x_1y-x_2y|=|x_1-x_2| |y|\leq |x_1-x_2|.$