1
$\begingroup$

Let $m$ be Lebesgue measure on $\mathbb{R}^d$ and let $\nu$ be a finite Borel measure on $\mathbb{R}^d$. Suppose there is a constant $C<\infty$ such that $\nu(E)\leq C\sqrt{m(E)}$ for any Borel set $E$. Prove that $F(x)=\lim\limits_{r\to 0^+}\frac{\nu(B_r(x))}{m(B_r(x))}$ exists for $m$-a.e. $x\in\mathbb{R}^d$ and that $m(\{x:F(x)>t\})\leq\frac{C^2}{t^2}$ for each $t>0$. $B_r(x)=\{y:\lvert y-x\rvert <r\}$ denotes a ball.

$\endgroup$
  • $\begingroup$ What progress have you made and what parts are you stuck on? $\endgroup$ – spaceisdarkgreen Jan 3 '17 at 5:25
2
$\begingroup$

Since $\nu(E)\leq C\sqrt{m(E)}$, we have $\nu \ll m$, so there is a measurable function $f:\mathbb R^d\to [0,\infty)$ such that $\nu(E)=\int_Efdx$. i.e. $f$ is the R-N derivative $\frac{d\nu}{dm}$.

Now, by a standard non-trivial result (see for example Cohn's proof in his $Measure\ Theory;\ $ ch. 6.2), $f(x)=\lim\limits_{r\to 0^+}\frac{\nu(B_r(x))}{m(B_r(x))}$.

Finally with $E=\left \{ x:f(x)>t \right \}$,we have $\nu(E)=\int_Efdx\ge t\cdot m(E)$, from which it follows that $t\cdot m(E)\le C\sqrt{m(E)}\Rightarrow m(E)\le \frac{C^{2}}{t^{2}}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.