porous sets: why measure zero? 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 $z\in E$.

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.