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I see the following theorem in Lihe Wang's A geometric approach to the Calderon--Zygmund estimates

(Modified Vitali) Let $0<\varepsilon<1$ and let $C\subset D\subset B_1$ be two measurable sets with $|C|<\varepsilon |B_1|$ and satisfying the following property: for every $ x\in B_1$ with $|C\cap B_r(x)|\geq \varepsilon |B_r|$, $B_r(x)\cap B_1\subset D$. Then $|D|\geq\frac{1}{20^n\varepsilon}|C|$.

The first line of the proof reads:

Since $|C|<\varepsilon |B_1|$, we see that for almost every $x \in C$, there is an $r_x < 2$ so that $|C\cap B_{r_x}(x)|=\varepsilon |B_{r_x}|$ and $|C\cap B_{r}(x)|<\varepsilon |B_{r}|$ for all $2 > r > r_x$.

Why "and $|C\cap B_{r}(x)|<\varepsilon |B_{r}|$ for all $2 > r > r_x$"?.

What I know is that $$\lim_{r\to 0}\frac{|C\cap B_r(x)|}{|B_r(x)|}=1>\varepsilon \text{ for almost every} x\in C$$ and $$\frac{|C\cap B_2(x)|}{|B_2(x)|}<\frac{|C|}{B_1}<\varepsilon.$$ Then there is an $r_x < 2$ so that $|C\cap B_{r_x}(x)|=\varepsilon |B_{r_x}|$. But why $|C\cap B_{r}(x)|<\varepsilon |B_{r}|$ for all $2 > r > r_x$ ?

So I guess $$ \frac{|C\cap B_r(x)|}{|B_r(x)|} $$ decreasing in $r$ for almost every $x\in C$ ? But it seems to be not true. Any one can help me with this proof?

Thanks!

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When you say "decreasing", you allow equality, right? And do you mean decreasing for $r$ sufficiently small in terms of $x$? Otherwise there are trivial counterexamples, such as $C$ being the unit disk. –  Greg Martin Apr 22 '13 at 6:45
    
Come to think of it, I wouldn't be surprised if a Smith-Volterra-Cantor set were a counterexample to this. en.wikipedia.org/wiki/Smith–Volterra–Cantor_set –  Greg Martin Apr 22 '13 at 6:48
    
@Greg: Thanks, and I've changed my question. –  Y.Z Apr 22 '13 at 7:29
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1 Answer

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The function $f_x \colon r \mapsto \frac{|C\cap B_r(x)|}{|B_r(x)|}$ is clearly continuous, so define $$r_x = \sup\{r < 2 ~:~ f_x(r) = \epsilon\}.$$

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