Answer by WHLin for Is this weird function with argument in the integrand continuous? (Fundamental Theorem of Calculus)

I think in general the statement is "false". Let $I = [0,2]$, $f: I^2 \rightarrow \mathbb{R}$, and $t_0 = 0$. Define

$$ f(x,t) \equiv 0, \;\; \text{if } x \in [0,1) \cup (1,2] $$$$ f(x,t) \equiv 1, \;\; \text{if }x =1 $$

Define $F(x) \equiv \int_{t_0}^x f(x,t)dt$.Then we have$F(x) = 0$ for $x<1$ or $x>1$, but $F(x) = 1$ for $x=1$. Hence, in this case $F(x)$ is not continuous at $x=1$. In general, for $F(x)$ to be continuous we would need some kinds of continuity along the $x$-component of $f(x,t)$.