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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.


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

Your help is very much appreciated!

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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

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.


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 $2\varepsilon$.

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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.

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