The proof of $\text{lim sup}\frac{\lambda (B)}{m (B)} =0$ if $\lambda$ is singular to $m$

58 Views Asked by At

How to prove the following:

$\lambda$ is a measure on $R^d$ which is singular w.r.t lebesgue measure $m$. Then $\text{lim sup}_{r\rightarrow 0}\frac{\lambda (B(r,x))}{m (B(r,x))} =0$ $m$-almost everywhere.

Note: $B(r,x)$ is a ball centered at $x$ with radius $r$.

Could anyone provide me with the first step or a hint?

Does it relate to the shrinking rate?