Let $z(x):=\int_{Y} f(x,y)d\mu(y)$ for $x \in \mathbb R$ be an integral function where $\mu$ is a finite(!) Borel measure on $Y$ and $x \mapsto f(x,y)$ is continuous for every $y.$
Moreover, we know that $\int_{ Y} \left\lvert f(x,y) \right\rvert d\mu(y)$ is uniformly bounded in $x.$
My question is whether this implies that $z$ is continuous as well?
What is the obstacle: It seems that the dominated convergence theorem does not apply so easily as $f(x,y)$ is not known to be uniformly bounded in both $x$ and $y.$