If we have $$F(t):=\int_V f(t,x)dx$$ where $V$ is some measurable subset of $\mathbb R$ and $x\mapsto f(t,x)$ is a measurable function. Moreover let $F$ be defined for all $t\in U$ a open subset of $\mathbb R$.
Then if $t\mapsto f(t,x)$ is continous for almost all $x$ and we can find an integrable function $g(x)$ such that $|f(t,x)|\leq g(x)$ for all $t\in U$ it follows that $F(t)$ is continous.
I wanted to ask, since continuity is a local property, is it not sufficent to find a bounding function $g(x)$ for every $t$ on some interval $(t+\epsilon,t-\epsilon)$ for an arbitrary small $\epsilon$ ? The idea came up because the proof of the first statement uses dominated convergence to show that $\lim_{s\to t} F(s)=\int_V \lim_{s\to t}f(s,x)dx=\int_Vf(t,x)dx$ and hence this would work for every $t$ on a arbitrary neighborhood of $t$.
I appreciate any replies.
Thanks in advance.