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!