Let $f(x,y):[0,1]^2 \rightarrow R$ be a continuos function, and $F(x)$ defined as
$$F(x)= \int^{1}_{0} f(x,y)1_{(y\leq x)} dy.$$ where $1_{(y\leq x)}= 1$ if $y\leq x$ and $1_{(y\leq x)}=0$ if $y> x$ (indicator function).
Clearly, we are working under the conditions $x\in[0,1]$ and $y\in[0,1].$ Can we say that $F(x)$ is continuous?
I would be particularly interested in the case where $f(x,y)=f_1(x-y) f_2(y)$ were $f_1$ and $f_2$ are continuous functions. Any help would be much appreciated.
One way to see this is via the dominated convergence theorem. Indeed, suppose that $x_n\rightarrow x$, which yields:
\begin{equation} \begin{split} \lim_{n\rightarrow \infty }F(x_n)&=\lim_{n\rightarrow \infty}\int_0^1f(x_n,y)1_{(y\leq x_n)}dy\\ &=\int_0^1\lim_{n\rightarrow \infty}f(x_n,y)1_{(y\leq x_n)}dy\\ &=\int_0^1 f(x,y)1_{(y\leq x)}dy\\ &=F(x), \end{split} \end{equation} where the dominated convergence theorem application is valid because everything is bounded.