Limit of ratio between Lebesgue measure and a finite Borel measure

Question

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

Answer 1

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 $t\cdot m(E)\le C\sqrt{m(E)}\Rightarrow m(E)\le \frac{C^{2}}{t^{2}}.$