Behavior of certain double integral when domain shrinks to zero

Question

$\newcommand{\BR}{\mathbb{R}}$Let $B$ be the unit ball in $\BR^d$. Let $f : B\times B \to \BR$ be a measurable function satisfying $$\int_B \left(\int_B |f(x,y)|dy\right)^2 dx < \infty.$$ Is it always true that $$\int_B \left(\int_B 1_{|x-y|<\epsilon}|f(x,y)|dy\right)^2 dx \to 0\text{ as }\epsilon \to 0?$$ Also, would it be true if the square were omitted in both expressions? If the square is omitted then the set being integrated over has measure $\to 0$ so intuitively I would expect the integral $\to 0$ as well. With the square there I'm not so sure.

Answer 1

Yes, it is true by the dominated convergence theorem.

It does not matter if you omit the square or not. What matters is the absolute value sign.