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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?

alryosha
  • 563

1 Answers1

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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 $\phi$.

Kavi Rama Murthy
  • 311,013