1
$\begingroup$

I think the following should be true but can’t seem to figure out a proof:

Let $M$ be a connected compact Riemannian manifold (without boundary) with the volume measure, normalized so that the total measure is 1.

Let $A$ be a Borel measurable subset such that $0 < \mathrm{vol}(A) < 1$. For $r > 0$, let $$ B(A, r) := \bigcup_{x \in A} B(x, r) $$ be the $r$-neighborhood of $A$, where $B(x, r) := \{y \in M \mid d(x, y) < r\}$.

Show that for all $r > 0$, $\mathrm{vol}(B(A,r)) > \mathrm{vol}(A)$.

If A is closed (hence compact) this is clear. But I am having trouble proving it in general. I have thought of using the fact that $A$ can be approximated from the inside by compact sets but it is not clear to me how to complete the proof. Any suggestions are appreciated. Thanks.

$\endgroup$
3
  • 1
    $\begingroup$ I guess you can do the following: You always have $B(A,r) \supset \overline{A}$, so if $\mu(\overline{A})>\mu(A)$, the claim is clear. Now assume $\mu(\overline{A})=\mu(A)$. Note that $B(A,r) = B(\overline{A},r)$. Since you claim that you can prove the claim for closed $A$, you then get $\mu(B(A,r)) =\mu(B(\overline{A},r))>\mu(\overline{A})\geq\mu(A)$. $\endgroup$
    – PhoemueX
    Dec 7 '19 at 12:51
  • $\begingroup$ Oh yes of course. You're absolutely right. Of course $\overline{A} = \bigcap_{r > 0}B(A, r)$, so one can reduce to the problem to the closed case by taking the closure. Thanks a lot!! $\endgroup$
    – B Chung
    Dec 7 '19 at 20:07
  • $\begingroup$ For the record the proof when $A$ is closed is the following: Since $\mathrm{vol}(A) \in (0, 1)$, $A$ cannot be the empty set or $M$. Since $M$ is connected and the complement $A^c$ is open and nonempty, $A$ is not open (and nonempty). Therefore there exists $x \in A$ such that for all $r > 0$, $B(x, r) \cap A^c \neq \emptyset$, thus as an open set it contains some open ball $B(y, r')$ for some $y \in A^c$ and $r' > 0$. Hence the disjoint union $B(y, r') \sqcup A \subset B(A, r)$. Note that $\mathrm{vol}(B(y, r')) > 0$. Therefore $\mathrm{vol}(B(A, r)) > \mathrm{vol}(A)$. $\endgroup$
    – B Chung
    Dec 7 '19 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.