I am aware that applying the topological definition of continuity is very useful versus other methods like epsilon-delta, e.g. as in $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$.
I am a bit uneasy when functions of measurable sets are involved though. The most common example I've seen is:
Let $A \subset \mathbb{R}^d$ be Lebesgue measurable and define the measure to be $m(E)$. Let $f = m(A \cap B(r, 0))$, where $B$ is an open ball. If $r \in (0, \infty)$ then $f$ is continuous. Could we prove this using the topological definition?
My idea: We have
$$f^{-1}(A \cap B(r, 0)) = f^{-1}(A) \cap f^{-1}(B(r, 0)).$$
The inverse image of an open set is open, so $f^{-1}(B(r, 0))$ open. But what about $f^{-1}(A)$?