Analysis Problem of the Day 55

Today’s problem appeared as Problem 5 on the UCLA Spring 2022 Analysis Qual:

Problem 5. Let $\mu$ be a Borel measure on $\mathbb{R}^2$ such that for fixed $r>0, \mu(B(x,r))$ is independent of $x.$

a) Prove that $\mu(B(x,r)) \leq cr^2$ for some $c>0$ for $0 < r \leq 1.$

b) Prove that $\mu$ is a constant multiple of the Lebesgue measure.

Solution: a) We write $\mu(B(x,r))=B(r)$ and note that $B$ is decreasing as $r \to 0.$ We are asked to show that the measure scales quadratically. This makes intuitive sense since the “area” of a ball in $\mathbb{R}^2$ scales quadratically with radius. Rigorously, note that a circle of radius $r$ contains a square of length $\sqrt{2} r,$ each side of which contains at least $\lfloor\frac{\sqrt{2}r}{\frac{r}{2^n}}\rfloor = \lfloor 2^n \sqrt{2}\rfloor \geq 2^n$ subdivisions of side length $\frac{r}{2^n},$ each of which corresponds to a square that can fit a circle of radius $\frac{r}{2^{n+1}}.$ By the additivity of measures, it follows that $B(r) \geq 2^{2n} B(\frac{r}{2^{n+1}}).$ We let $c=\sup_{r \in [1,2]} B(r),$ and note that for $r \in [2^{-(n+1)},2^{-n}],$ the above inequality implies $B(r) \leq 2^{-2n} c \leq c r^2.$ Thus, $B(r) \leq cr^2$ for all $0<r \leq 1.$

b) For a Borel measurable set $A$ and any Borel measure $\mu$ on $\mathbb{R}^2,$ $\mu(A)$ is the infimum of the measures of countable unions of open balls containing $A.$ Let $m$ be the Lebesgue measure on $\mathbb{R}^2.$ Since $\mu \leq C m$ on balls, it follows that $\mu(A) \leq C m(A)$ for all measurable $A$ with $m(A)<1.$ This implies that $\mu \ll m,$ so by Radon-Nikodym, there exists a measurable function $f$ s.t. $\int_A f dm = \mu(A)$ for all measurable $A.$ By the Lebesgue differentiation theorem, it follows that $f$ is a.e. the limit of its averages over shrinking balls. But $\frac{1}{m(B(x,r))}\int_{B(x,r)} f dm = \frac{\mu(B(x,r))}{m(B(x,r))},$ which is constant for fixed $0 < r \leq 1$ for all $x.$ It follows that $f$ is a.e. the limit of the same sequence, i.e. $f$ is constant a.e. It immediately follows that $\mu$ is a constant multiple of the Lebesgue measure.

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