Limits of the Lebesgue measure Let $(\mathbb{R}^d, \mathfrak{M}, m)$ be Lebesgue measure space and $A \subset \mathfrak{M}$ with $m(A) < \infty$.
Suppose $f : \mathbb{R}^d \rightarrow \mathbb{R}$ as $f(x)=m(A \cap(x+A))$ with $x+A=\{x+a: a \in A\}$
I want to show $\lim _{|x| \rightarrow 0} f(x)=m(A)$
At first, I substitue f(x) and get $\lim _{|x| \rightarrow 0} f(x) = \lim _{|x| \rightarrow 0} m(A \cap(x+A)) = m(A)$.
If I change the order of Lebesgue measrue and limit, I can prove the result.
In my opinion, I can use monotone convergence theroem of measure by constructing increasing sequence of subsets of $A$, but I find it difficult to construct such sequence of the sets.
Is my approach is correct? or any other approach should be considered?
 A: Monotone convergence theorem is not helpful.
Actually, $f$ is a continuous function on $\mathbb R^{n}$. Proof of this requires an approximation result.
Theorem
If $f$ is a bounded intgerable  function then $g(x)=\int f(y)f(x-y)dy$ defines a continuous function.
[ More genatrlly we can consider $g(x)=\int f(y)h(x-y)dy$ where $f$ and $h$ are bounded integrable functions].
Once this theorem is proved we can take $f=\chi_A$ to finish the proof.
To prove this theroem we use that fact given any $\epsilon >0$ we can  find a continuous function $\phi$ with compact support such that $\int |f-\phi| <\epsilon$.  Note that $|\int f(y)f(x-y)dy -\int f(y)\phi(x-y)dy |\leq M \epsilon$ for all $x$ where $M$ is a bound for $|f|$. If we show that $\int f(y)\phi(x-y)dy $ is a continuous function of $x$ it follows (by letting $\epsilon \to 0$) that $g$ is continuous because uniform limit of continuous functions is continuous.
But  uniform continuity of $\int f(y)\phi(x-y)dy $  follows easily from uniform continuity of $\phi$.
