Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A\subset\mathbb{R}^n$ be a bounded semi-algebraic subset with Lebesgue measure $\mu^n(A)=0$, $\mu^{n-1}(A)>0$ on some $(n-1)$-dimensional Hyperplane $H$ ("bounded" added ss, Bernd). Let $B(\epsilon)$ be an open $\epsilon$-Ball around zero.

Obviously, for any $\epsilon>0$, $\mu^n(A+B(\epsilon))>\mu^n(B(\epsilon))>0$

It seems just as obvious that also the following holds:

$\forall \delta>0, \exists \epsilon>0$ with $\mu^n(A+B(\epsilon))<\delta \ \ \ \ \ $ (1)

Since it is easy to find $A$ not being semi-algebraic for which the statement doesn't hold, I have problems in capturing the "niceness" of A in a proof.

Questions:

  • is (1) correct?
  • if so, what is a necessary / sufficient condition of A for (1) to hold?

Your help is very much appreciated!

share|improve this question
    
I assume that you mean that $\mu^{n-1}(A)$ is finite, otherwise it is of course false. –  Goldstern Jan 16 '12 at 12:41
    
Thanks Goldstern and Jakob, you're absolutely right. $A$ is bounded and $\mu^{n−1}(A)$ is finite. Still there are non-semi-algebraic sets for which (1) does not hold. What I can see for now, in my application $A$ is closed. However it would be even better if there is a more general solution working for any $A$. Thanks again for your efforts! –  Bernd Jan 16 '12 at 14:34
    
Hmm... For closed bounded sets one has $A=\cap_n A+B(1/n)$, hence $\mu^n(A)=\lim \mu^n(A+B(1/n))$ and we are done. I guess that for semialgebraic $A$ its boundary has always measure 0, so we may replace $A$ to its closure. –  Fedor Petrov Jan 16 '12 at 19:50
add comment

2 Answers

Would make this a comment if I had the points for it.

(1) is always false if A is not bounded, even if $\mu^{n-1}(A)$ is finite. For example:

$$A = \lbrace(x,y,z) \in \mathbb{R}^3|z=0,y=0\rbrace.$$

Obviously, $\mu^{n-1}(A)=0$.

But $A$ is the $x$ axis, so for any $\varepsilon>0$, $A+B(\varepsilon)$ is a cylinder with radius $\varepsilon$ around the $x$ axis, so you're not even going to get a finite n-measure no matter how small you choose $\varepsilon$.

(1) should hold for any bounded open set, though, for example.

Edit:

Actually, for your situation (measurable $A$ contained in a hyperplane, no matter whether it's semialgebraic), I'm pretty sure (1) holds iff $A$ is bounded. If it isn't bounded, $A+B(\varepsilon)$ contains a sequence of disjoint $\varepsilon$-balls; if it's bounded, $A+B(\varepsilon)$ is contained in a cylinder of bounded radius (as $\varepsilon \rightarrow 0$) and height $2\varepsilon$.

share|improve this answer
add comment

My guess is that compactness (thus almost being a compact set) helps. This is an idea for a proof.

Partition $R^n$ like a Voronoi diagram ($x$ is in the Voronoi cell of $a$ if $a$ is the closest point of $A$ to $x$). Because of compactness, almost every point belongs to a Voronoi cell.

The Voronoi cell of an individual point should be well-behaved because for any $a \in A$ and $V_a$ its Voronoi cell we have $$A+B(\epsilon) \cap V_a = \big (a+B(\epsilon) \big) \cap V_a$$

Partition $A$ according to the dimension of the Voronoi cell of its individual points. Let $A_i$ be the set of $a \in A$ with $V_a$ $i$ dimensional. Now $$\mu^n(A+B(\epsilon)\big) = \sum_i \mu^n(A_i + B(\epsilon) \big) \le \sum_i \mu^{n-i}(A_i)\mu^i(B_i(\epsilon) \big)$$ where $\mu^i (B_i(\epsilon) \big)$ is the measure of an $i$-ball and $\mu^0$ is interpreted to be the counting measure. Now it is sufficient to prove that $\mu^n(A_i + B(\epsilon) \big)$ is finite for all $0 \le i \le n$.

This is easy to picture for polytopes: this is the condition that it has finitely many vertices, its edges are finite and (no more than) $1$-dimensional, its faces are $2$-dimensional, etc.

I believe this can be proved in general using (for example) induction on the number of (in)equalities defining your semi-algebraic set.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.