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Could any one tell me how to solve this one?

Let $K$ be a compact subset of $\mathbb{R}^n$, and $$A:=\{x\in\mathbb{R}^n:d(x,K)=1\}.$$ Show that $A$ has Lebesgue measure $0$.

Thank you!

Davide Giraudo
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Myshkin
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  • Assuming that $d(x,K)$ is induced by the Euclidean metric, I suspect one could argue that, locally, $A$ is a smooth $n-1$ manifold, i.e., locally the graph of some smooth function. This graph has measure zero. – Jas Ter Mar 12 '13 at 14:50
  • @SimenK. Well, I don't think you will get smooth function, but merely a Lipschitz one. Which is still sufficient for your line of reasoning. – Thomas Mar 12 '13 at 15:12
  • You'd still need to argue carefully, because there will be points where the graph is not locally the graph of a function. E.g., take $K={(-1,0),(1,0)}$ and look around the origin. – Harald Hanche-Olsen Mar 12 '13 at 15:15
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    I don't have an answer, but here is an argument that could lead to one: The set of points which do not have a unique nearest point in $K$ is called the ridge of the distance function. The ridge is known to have Hausdorff dimension at most $n-1$, while $A$ is expected to be the graph of a Lipshitz function near points which are not on the ridge. Put these together, and a proof might emerge. – Harald Hanche-Olsen Mar 12 '13 at 16:55
  • Minor correction to my comment: I am looking at a paper by Nirenberg, and find that the ridge is actually the closure of the set I mentioned above. Which is even better, I think. – Harald Hanche-Olsen Mar 12 '13 at 17:11
  • Umm, it appears that the Hausdorff dimension argument for the ridge requires some smoothness of the boundary. Argh. – Harald Hanche-Olsen Mar 12 '13 at 17:43
  • I just asked a related question whose solution might solve this one as well. – Harald Hanche-Olsen Mar 12 '13 at 18:02
  • Maybe one can try something like this: We can suppose that $d(x,K)$ is an function from a ball containing $K$ to $\mathbb{R}$. Now, we aproximate $d$ by an $C^1$ function $f$ in the Lipschitz norm. We can suppose that $1$ is an regular value of $f$ and then the set $f^{-1}(1)$ has zero measure. Now this set is close to $A$ in some way that I couldn't not precise yet... – Tomás Mar 12 '13 at 18:27
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    I'm afraid this is way beyond my abilities, so I did the case $n=1$... Writing the open set $\mathbb{R}\setminus K=\bigcup (a_j,b_j)$ we see that $A$ is countable. So it has Lebesgue measure $0$. – Julien Mar 12 '13 at 23:30
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    Out of curiosity, where did this problem come from? – Nate Eldredge Mar 13 '13 at 02:13
  • @Nate: in hindsight after posting my answer, it could conceivably be a textbook exercise. :-) – Willie Wong Mar 13 '13 at 10:07

4 Answers4

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Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$.

Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$.

This implies that for all $\epsilon < 1/2$, we have that $$ \mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0)) $$ where $\mu$ is the Lebesgue measure. (The factor $1/2^n$ is very loose: Since a sphere tangent to $x_0$ is contained in $B$, locally an orthant centered at $x_0$ is contained in $B$.)

Hence we have that for every $x_0\in A$, $$ \limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} < 1 $$

By the Lebesgue differentiation theorem, the set of $x_0\in A$, for $A$ measurable, such that the above condition holds must be measure zero. Hence $A$ has measure zero.

Willie Wong
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    This is very nice! – Nate Eldredge Mar 13 '13 at 13:00
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    And sooo obvious – in retrospect. Tiny nitpick, though: The limit should be a limsup. – Harald Hanche-Olsen Mar 13 '13 at 15:34
  • @HaraldHanche-Olsen: you are right of course. Fixed. – Willie Wong Mar 13 '13 at 16:23
  • Any idea if $K$ is any subset ? I will be impressed if it doesn't hold. – user10676 Mar 13 '13 at 18:00
  • @user10676: use that $d(x,K) = d(x,\bar{K})$. Then cut-off at radius $R$. By Heine-Borel we now have a compact set. Take a countable increasing sequences of $R$ and use sigma sub-additivity. – Willie Wong Mar 14 '13 at 09:03
  • Actually, on second thought, Heine-Borel is unnecessary, since we can restrict to the ball $B_3(x)$ for $x \in A$. So closed is sufficient. – Willie Wong Mar 14 '13 at 09:07
  • I have two questions: why is $ \mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0)) $? And why is $ \limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} $ ? – Ri-Li Dec 04 '20 at 17:33
  • @Ri-Li: $A$ and $B$ are disjoint, so $\mu(B_\epsilon(x_0) \cap A) \leq \mu (B_\epsilon(x_0) \cap B^{C}) = \mu(B_\epsilon(x_0)) - \mu(B_\epsilon(x_0) \cap B)$. – Willie Wong Dec 04 '20 at 17:40
  • For your first question: it is just a very rough estimate of the volume contained in $B_1(y_0) \cap B_\epsilon(x_0)$. If you draw a picture you will see it is obvious. – Willie Wong Dec 04 '20 at 17:46
  • How does the volume of $B_1(y_0) \cap B_\varepsilon(x_0)$ relate to $\mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0))$? How is the $\frac{1}{2^n}$ related to the orthant? By orthant do you mean cube? – INQUISITOR Dec 07 '20 at 08:02
  • @user439126: orthant means the higher-dimensional analogue of a quadrant. You can see the geometry by first looking at the plane case. If you take $B_1(y_0)\cap B_\epsilon(x_0)$, then you can draw a (rotated) set of axes centered at $x_0$ such that of the four quadrants of $B_{\epsilon}(x_0)$, one is entirely contained in $B_1(y_0)\cap B_\epsilon(x_0)$. Now generalize to higher dimensions. – Willie Wong Dec 07 '20 at 14:12
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I was going to make this a comment but it occurred to me there might be sufficient interest that perhaps I should not bury it in a comment.

At the beginning of the paper below Erdős gives a short proof (that he attributes to Tibor Radó) making use of the Lebesgue density theorem that $E_r$ has Lebesgue measure zero, where $E$ is a closed set in ${\mathbb R}^n$ and $$E_r \, = \; \{ x \in {\mathbb R}^n : \; d(x,E)=r \} $$

Paul Erdős, Some remarks on the measurability of certain sets, Bulletin of the American Mathematical Society 51 #10 (October 1945), 728-731.

Later in this paper (item 6), Erdős proves the stronger result that, for $K$ compact, the Hausdorff $(n-1)$-measure of $K_r$ is finite, and hence $E_r$ has $\sigma$-finite Hausdorff $(n-1)$-measure when $E$ is closed.

More precise results can be found in a 1985 paper by Oleksiv/Pesin Zbl 573.28010 [English translation: Mathematical Notes 37 (1985), 237-242], and I'm sure there are quite a few related results in the literature. For instance, each of the sets $E_r$ is $[1]$-very porous in the sense defined in this conference talk of mine, and among other things I gave a short argument there that each such set has $\sigma$-finite packing $(n-1)$-measure.

The analysis of results related to these in infinite dimensional normed spaces has also generated a fair amount of interest. One possible entry point into this is Ludek Zajicek's 1983 paper Differentiability of the distance function and points of multi-valuedness of the metric projection in Banach space.

(Added 12 Weeks Later) Someone recently gave me a vote on this answer and, in looking at what I wrote, it occurred to me that a much better reference for the comment "one possible entry point into this" that I made in the last paragraph is the following book:

Joram Lindenstrauss, David Preiss, and Jaroslav Tišer, Fréchet Differentiability of Lipschitz Functions and Porous Sets in Banach Spaces, Annals of Mathematics Studies #179, Princeton University Press, 2012, x + 425 pages. Zbl 1241.26001 (a review)

Princeton University Press web page for the book

Amazon.com web page for the book

user642796
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Dave L. Renfro
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Disclaimer: The following approach has a flaw in the second step, because the set $A$ could be a nowhere dense set with positive measure like e.g. a fat Cantor set as Davide Giraudo pointed out in his comment. I don't see any way to work around this gap, but leave the answer here in the hope that it might still prove useful in some way.

  1. $A$ is compact. Since $A$ is clearly bounded, it only needs to be shown that it's closed. Let $x_i$ be a convergent sequence in $A$ with limit point $x$. Then $x$ is also in $A$, because for each $x_i$ you find a $y_i$ in $K$ with $d(x_i,y_i)=1$. The $y_i$ have a convergent subsequence with some limit point $y$. Because $d$ is continuous it follows that $d(x,y)=1$. Further, one can show that $d(x,y)\geq1$ for all $y\in K$ (Otherwise, say $d(x,y)<1-\epsilon$ for some $y\in K$, then $d(x_i,y)\leq d(x,x_i)+d(x,y)<\epsilon/2+1-\epsilon<1$ for some $x_i$). Thus $x\in A$.

  2. If $A$ is closed and Lebesgue measurable with $\lambda(A)>0$ it must contain at least one non-empty open set of $\mathbb R^n$, namely the interior of $A$. This in turn would contain a closed ball, call it $\bar B_\delta(x)$ for some $x\in A$ and $\delta>0$.

  3. $K$ cannot contain any point of the open ball $B_{1+\delta}(x)$, because otherwise there would be a point in $\bar B_\delta(x)$ with distance to $K$ smaller than 1. But, then $x$ would have distance larger then $1$ to $K$, i.e. $x\not\in A$, contrary to the assumption that $\bar B_\delta(x)\subset A$.

A possible remedy: If we define $B=\bigcup_{y\in K}\bar B_1(y)$, then $A=\partial B$, since the boundary of $B$ contains exactly the points which have distance $1$ to $K$. Since $B$ is a relatively "smooth" set (in the sense that it cannot have infinitely small and dense holes like a Cantor set), this may carry over to its boundary and salvage step 2. by ruling out "weird sets" like the fat Cantor.

Elmar Zander
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  • Is statement 2. obvious? I had similar thoughts for a method of proof but couldn't justify that step. That is, why must the interior be non-empty? – Dan Rust Mar 12 '13 at 22:18
  • @DavideGiraudo You're right. I didn't think about that. As the proof builds on that, there's probably no way to salvage it. I just want to think about it a bit more, before I remove it. – Elmar Zander Mar 12 '13 at 22:21
  • I wouldn't remove the answer, it may still be useful to the OP. Just add a disclaimer that the gaps may be un-fillable. – Dan Rust Mar 12 '13 at 22:23
  • The proposition (due to Vitali, I think?) that you may have been thinking of in step 2 is that if $A$ has positive Lebesgue measure then $A-A$ contains an open neighborhood of zero. I don't know if this is useful for this question. – Yoni Rozenshein Mar 12 '13 at 22:58
  • @YoniRozenshein Not that I was aware of. Maybe you can give me a pointer to this proposition? – Elmar Zander Mar 12 '13 at 23:53
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    I miscredited Vitali. Here it is: http://en.wikipedia.org/wiki/Steinhaus_theorem – Yoni Rozenshein Mar 13 '13 at 00:09