Show that $f(x)=\mu((A+x) \cap B)$ is continues, where $A,B \subset \Bbb R$ are measurable sets with a positive and finite measure

Let $A,B \subset \Bbb R$ be measurable sets with a positive and finite measure, and let $\mu$ be the Lebesgue measure. Define $f(x)=\mu((A+x) \cap B)$. Show that $f$ is continues.

I'm not sure how to approach this. Any clues?

• Is $\mu$ Lebesgue measure? (The assertion will fail if $\mu$ puts positive weight on certain singletons, or even if $\mu$ is not absolutely continuous.) – John Dawkins Sep 23 '17 at 16:08
• Thanks @JohnDawkins. Yes, it is. – mibarg Sep 23 '17 at 16:21

Therefore, there exists $A_1$, $B_1$ finite unions of intervals so that $$\mu(A\Delta A_1) < \epsilon \\ \mu(B \Delta B_1) < \epsilon$$
Now, denote consider the function $$f_1(x)= \mu((A_1+ x) \cap B_1)$$
Show that: $$|f_1(x)- f(x)| < 2 \epsilon$$ $$|f_1(x) - f_1(y)| \le m\, n|x-y|$$ if $A_1$ consists of $m$ intervals, and $B_1$ of $n$intervals (this last one is a poor estimate, but sufficient).
We conclude: if $|x-y| < \frac{\epsilon}{mn}$, then $|f(x) - f(y)|< 3 \epsilon$. Thus $f$ is (uniformly) continuous.