Is the integral over a component of a doubly continuous function continuous?

Question

Given a continuous function $$f:(0,1)\times(0,1)\to\Bbb R$$ so that $t\mapsto f(t,x)$ is integrable for all $x$, does it follow that $$x\mapsto\int_0^1 f(t,x)\,dt$$ is continuous in $x$? I would think not necessarily, in part because we are considering open intervals which blocks trivial proofs using uniform continuity. But I cannot think of a counter example.

Answer 1

Let $f(t,x)=\frac{1}{|x|}|t|^\frac{1}{|x|}$ for $x\neq 0$ and zero otherwise. It is clearly continuous for $x\neq 0$. To show that it is continuous at a point $(t_0,0)$ with $|t_0|=1-r<1$, just note that if $t$ is close enough to $t_0$, then $|t|<1-r/2$ so that $$|f(t,x)-f(t_0,0)|\leq |f(t,x)|+|f(t_0,0)|<|f(1-r/2,x)|+|f(1-r/2,0)|\to 0 $$ as $x\to 0$.

When $x\neq 0$, the integral is $$\int_0^1f(t,x) dt=\frac{1}{|x|} \frac{1}{\frac{1}{|x|}+1}=\frac{1}{|x|} \frac{1}{\frac{1}{|x|}+1}=\frac{1}{1+|x|}$$ and when $x=0$ the integral is $0$, so it is not continuous at $x=0$.

EDIT: I consider the function as $f:(0,1)\times (-1,1)\to \mathbb{R}$.