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We call a measurable set $E\subset\mathbb R^N$ porous if every ball $B_r(x)$ contains a smaller ball $B_{cr}(y)$ for some $c\in(0,1)$ such that $$ B_{cr}(y)\subset B_r(x)\setminus E. $$

So I've read in my book that all porous sets have Lebesgue-measure zero. Why does this hold?

Add: I do know $\limsup_{r\to0}\frac{|E\cap B_r(z)|}{|B_r|}<1$ for some $z\in E$.

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I assume that $c$ is fixed (otherwise, a fat Cantor set should provide a counter-example). You can prove it quickly with the Lebesgue density theorem. Assume that $E$ has non-zero measure. By this theorem, Lebesgue-almost every point in $E$ is a density point, so there exists $x \in E$ such that :

$$\lim_{r \to 0} \frac{|E \cap B_r (x)|}{|B_r (x)|} = 1.$$

But then, for all $r > 0$, there exists $y$ such that $B_{cr} (y) \subset B_r (x) \cap E^c$. Hence, for all $r > 0$,

$$\frac{|E \cap B_r (x)|}{|B_r (x)|} = 1-\frac{|E^c \cap B_r (x)|}{|B_r (x)|} \leq 1-\frac{|B_{cr} (y)|}{|B_r (x)|} = 1-c^N,$$

which provides a contradiction.

Edit: the point is that $\limsup_{r \to 0} \frac{|E \cap B_r (z)|}{|B_r (z)|} < 1$ not merely for one, but for every $z \in E$; from there we can conclude with the theorem.

D. Thomine
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    Two observations: (1) The Wikipedia version of the Lebesgue density theorem only shows the result if we assume $E$ is measurable. However, there is an outer Lebesgue density result (basically the density result holds a.e. for points in the set, but not necessarily a.e. for points outside the set) that can be used if measurability of $E$ isn't assumed. (2) We get measure zero even if we just had $\liminf_{r \to 0} \frac{|E \cap B_r (z)|}{|B_r (z)|} < 1$ for every $z \in E,$ and upper porous sets (defined and used before the OP's notion, in fact) leads to this weaker density property. – Dave L. Renfro Jul 15 '15 at 19:33
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    Regarding (1) in my previous comment, I just remembered that the notion of "porous set" that the OP is working with has the property that its (topological) closure is also a porous set. (I seem to recall that you have to use a constant $c'$ such that $c' < c,$ although the value of $c - c'$ can always be made arbitrarily small.) Thus, by replacing the original set with its closure, we can then apply the measurable set version of the Lebesgue density theorem. This "fix" isn't available for upper porous sets, however. – Dave L. Renfro Jul 15 '15 at 19:48
  • @DaveL.Renfro I am sorry to bother you with a comment to a 4 years old question but I am really interested in the outer Lebesgue density theorem that you mentioned here. May I ask where can I find a reference or a proof to such a theorem? Coincidentally, a couple of days ago I just came across your name a few times when reading about porosity so I figured that you are an expert in this field. – BigbearZzz Sep 02 '19 at 01:47
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    @BigbearZzz: Lebesgue stated/proved the density theorem for measurable sets (essentially stated in 1903, 1905; fully proved in 1910), and the extension to arbitrary subsets of $\mathbb R$ was independently proved by Sierpinski (June 1917) and Blumberg (December 1918). In 1923 Sierpinski published an elementary proof that also applies to ${\mathbb R}^n$ in general (Lebesgue's 1910 proof was also in this setting). (continued) – Dave L. Renfro Sep 02 '19 at 08:37
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    Some books where the arbitrary version can be found are: Kannan/Krueger (pp. 40-41), Foran (pp. 209-210), Jeffery (pp.114-116), Saks (pp. 128-131), Hahn/Rosenthal (pp. 281-286; a relatively complete bibliography on Lebesgue density, up to 1944, is given on pp. 285-286), Sz.-Nagy (pp. 116-119). – Dave L. Renfro Sep 02 '19 at 09:21
  • @DaveL.Renfro Thank you so much! The answer is more detailed than I could ever hope for. – BigbearZzz Sep 02 '19 at 09:36
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    Of possible additional relevance: Lebesgue density theorem in the line *AND* Lebesgue points of density and similar notions *AND* A short proof of Lebesgue's density theorem by Faure (2002) *AND* For every null set $E$ there is ... and this mathoverflow followup. – Dave L. Renfro Sep 02 '19 at 09:41
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Suppose $|E| > 0$. Given $\epsilon>0$, there exists a sequence of balls $B_n$ such that $E \subseteq \bigcup_n B_n$ and $$\sum_n |B_n| \le (1+\epsilon) |E| .\tag1$$ (This follows from the definition of outer-measure.)

Now, since $E$ is porous, there exists open balls $C_n \subset B_n \setminus E$ with $|C_n| = c^N |B_n|$. Then $E \subset \bigcup_n (B_n\setminus C_n)$, and hence $$|E| \le \sum_n |B_n \setminus C_n| = (1-c^N) \sum_n |B_n | .\tag2 $$

Putting (1) and (2) together we get $$ (1+\epsilon) (1 - c^N) > 1 .$$ Since $c$ is fixed, and $\epsilon>0$ can be picked arbitrarily, we obtain a contradiction.

Stephen Montgomery-Smith
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