Lebesgue measure and continuity of a function. 
Let $A,B$ be Lebesgue-measurable sets on $\mathbb{R}$ and $m$ be Lebesgue measure and $m(A)<\infty.$

Prove that $f(x):=m((A+x)\cap B)$ is continuous.
$\Big( A+x=\{ a+x | a\in A\}\Big)$.
I can use the following Fact.
Fact
If $E$ is Lebesgue measurable on $\mathbb{R}$ and $m(E)<\infty$, then $\lim_{x\to 0} m((E+x)-E)=0.$
I tried but couldn't prove.
Set $\forall a \in \mathbb{R}.$
I'll prove $|f(x)-f(a)|\to 0 \quad (x\to a)$.
If I could prove $|f(x)-f(a)|<m((A+x-a)-A)$, let $x\to a$ and I could prove that $f(x)$ is continuous, using the Fact.
But I cannot prove
$|f(x)-f(a)|<m((A+x-a)-A)$.
I would like you to give me some ideas.
 A: $\newcommand\e\varepsilon$There might be a smarter way to go about this, but here is a naive approach. Let $\e>0$. Then there exists $V$ open with $V\supset A$ and $m(V\setminus A)<\e/4$. Note that this implies that $m(V)<\infty$. Now
$$
m((V+x)\cap B)-m((A+x)\cap B)=m((V\setminus A)+x)\cap B)\leq m(V\setminus A). 
$$
Then
\begin{align}
|f(y)-f(x)|
&\leq \e/2+m((V+y)\cap B)-m((V+x)\cap B).
\end{align}
Now since $V$ is open, we may write it as a disjoint union of open intervals, $V=\displaystyle\bigcup_n(a_n,b_n)$. The finite measure of $V$ gives $\sum_n(b_m-a_n)<\infty$. We have $$m(E)-m(F)=m(E\setminus F)-m(F\setminus E).$$ Then, for $x<y$,
\begin{align}
m((a_n,b_n)+y)\cap B)-m((a_n,b_n)+x)\cap B)&\leq m((a_n+x,a_n+y))+m((b_n+x,b_n+y))\\[0.3cm]
&\leq2(y-x). 
\end{align}
So, we choose $n_0$ such that $\sum_{n>n_0}(b_n-a_n)<\e/4$. Then if $|y-x|<\e/(2n_0)$,
\begin{align}
|f(y)-f(x)|&\leq \frac\e2+\sum_nm((a_n,b_n)+x)\cap B)\\[0.3cm]
&\leq\frac\e2+\frac\e4+\sum_{n>n_0}m((a_n,b_n)+x)\cap B)\\[0.3cm]
&\leq\frac\e2+\frac\e4+\frac{2n_0\e|y-x|}{2n_0}=\e.
\end{align}
