I'm trying to solve such task:
Let $\mu$ be a Radon measure on $\mathbb{R}^2$ and let $K \subset \mathbb{R}^2$ be a compact set, such that: $$ \forall a\in K, \epsilon>0 ~~~\exists r\in(0,\epsilon)~~:~~\mu(B(a, r)) \leq r^2$$ where $B(a, r) = \lbrace x : d(a, x) \leq r \rbrace$ is closed ball.
Show that $\mu(K) \leq \frac{1}{\pi}\lambda_2(K)$, where $\lambda_2$ is Lebegue measure.
I'll want to use Besicovitch covering theorem. If I understand it correctly, this theorem goes as follows:
Let $\mu$ be a Radon measure on $\mathbb{R}^n$, let $A$ be a Borel set in $\mathbb{R}^n$ and $\mathscr{B}$ be a collection of closed non-degenerate balls in $\mathbb{R}^n$ such tat every point in A is the center of balls in $\mathscr{B}$ with arbitrary small diameter.
Then there exists disjoint collection $\mathscr{L} \subset \mathscr{B}$ such that $\mu(A\smallsetminus \bigcup\mathscr{L})=0$.
Some helpful facts I noticed:
$K$ is bounded, as $K$ is compact in $\mathbb{R}^2$.
As $\mu$ is Radon measure, for every positive $\epsilon$, I can choose bounded open set $U \supset K$ in $\mathbb{R}^2$, such that $\mu(U \smallsetminus K) < \epsilon$.
$\lambda_2(B(a,r)) = \pi r^2$, so $\mu(B(a, r)) \leq \frac{1}{\pi}\lambda_2(B(a,r))$ indeed.
$\lambda_2$ is Radon measure as well.
(Result of 3) As $\mu$, as well as $\lambda_2$, is countable additive, if there is family of disjoint balls $\mathscr{L}$ we have: $$ \mu(\bigcup\mathscr{L}) = \sum_{B\in \mathscr{L}}\mu(B) \leq \sum_{B\in \mathscr{L}}\frac{1}{\pi}\lambda_2(B) = \frac{1}{\pi} \lambda_2(\bigcup\mathscr{L}) $$
My attempt: Let choose $\epsilon > 0$ and $U$ as described above in (2).
Now my idea is to choose such family of balls $\mathscr{B}$ that every point in $K$ is center of non-degenerated balls with arbitrary small diameter and $K \subset \bigcup\mathscr{B} \subset U$.
Edit: such family that $B(a, r) \in \mathscr{B} \Rightarrow \mu(B(a, r)) \leq r^2$.
Based on Besicovitch covering theorem there are families of disjoint balls $\mathscr{L}_1$, $\mathscr{L}_2 \subset \mathscr{B}$ such that $\mu(K\smallsetminus\bigcup\mathscr{L}_1) < \epsilon$ and $\lambda_2(K\smallsetminus\bigcup\mathscr{L}_2) < \epsilon$. (I can cause $K \subset U$.)
Moreover, as $\bigcup \mathscr{L}_1, \bigcup\mathscr{L}_2 \subset \bigcup\mathscr{B} \subset U$ also true are inequalities $\mu(\mathscr{L}_1 \smallsetminus K) < \epsilon$ and $\lambda_2(\bigcup\mathscr{L}_2\smallsetminus K) < \epsilon$.
Ending attempt: If I can somehow show that I can choose $\mathscr{L}_1 = \mathscr{L} = \mathscr{L}_2$, then based on (5): $$ \mu(K) < \epsilon + \mu(\bigcup \mathscr{L}) \leq \epsilon + \frac{1}{\pi}\lambda_2(\bigcup\mathscr{L}) < 2\epsilon + \frac{1}{\pi}\lambda_2(K) $$
And as $\epsilon$ may be arbitrary small I obtained $\mu(K) \leq \frac{1}{\pi}\lambda_2(K)$.
However, I don't know if I can choose $\mathscr{L}_1 = \mathscr{L}_2$. Could you show me formal proof of that or other formal ending of that proof?
Also, if I made mistake above or you think something is not obviously correct, please let me know.
Actually, according to Besikovitch covering lemma, there are finite number of disjoint families $\mathscr L_i$, $\ i=1,\ldots, b_2$ such that $K\subset \cup_i \cup \mathscr{L}_i$. Covering lemmas for balls are not very suitable for obtaining sharp estimate (because essentially, in most cases balls overlap too much.) What can be expected most would be that there exists $b_2>1$ such that for every compact set $K'\subset K$, $$ \mu(K')\le \frac{b_2}\pi\lambda_2(K'). $$ (Constant $b_2$ in Besikovitch covering lemma cannot be improved to $1$.) But at least we can see that $\mu$ is absolutely continuous with respect to the Lebesgue measure on $K$. Based on this fact, we take 'infinitesimal' approach. By Radon-Nikodym theorem, there is $f\in L^1(K,\lambda_2)$ such that $$ d\mu = fd\lambda_2 . $$ And by Lebesgue's differentiation theorem, for almost every $a\in K$, we have $$ f(a) = \lim_{r\downarrow 0} \frac1{\lambda_2(B(a,r))}\int_{B(a,r)}f\ d\lambda_2=\lim_{r\downarrow 0} \frac{\mu(B(a,r))}{\lambda_2(B(a,r))}. $$ By the assumption, for each $a\in K$, there is a sequence $r_n\downarrow 0$ such that $$ \lim_{n\to\infty} \frac{\mu(B(a,r_n))}{\lambda_2(B(a,r_n))}\le \frac1{\pi}. $$ So we have $f\le \frac1{\pi}$ a.e. on $K$ and it follows $$ \mu(K)= \int_K f(x)\ \mathrm d\lambda_2 \le \frac1{\pi}\lambda_2(K). $$