I see the following theorem in Lihe Wang's  [A geometric approach to the Calderon--Zygmund estimates](http://www.math.uiowa.edu/~lwang/cccalderon.pdf)



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> (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 $|D|\geq\frac{1}{20^n\varepsilon}|C|$.
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The first line of the proof reads:



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> 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 $2 > r > r\_x$.
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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!



On the proof of  Modified Vitali Lemma.
The function $f\_x \colon r \mapsto \frac{|C\cap B\_r(x)|}{|B\_r(x)|}$ is clearly continuous, so define
$$r\_x = \sup\{r < 2 ~:~ f\_x(r) = \epsilon\}.$$


