$\liminf_{r\to 0} f(r)/r^n=0$ implies $\mu=0$ if $\mu$ is a positive Borel measure on $\Bbb R^n$, $f(r)=\sup _{x\in \Bbb R^n} \mu(B(x,r))<\infty$

Question

Suppose $\mu$ is a positive Borel measure on $\Bbb R^n$. For $r>0$, define $f(r)=\sup _{x\in \Bbb R^n} \mu(B(x,r))$, and suppose $f(r)<\infty $ for all $r$. I am trying to show that if $\liminf_{r\to 0} f(r)/r^n=0$ then we must have $\mu=0$. By assumption, $\mu(K)$ is finite for any compact set $K$, and this implies that $\mu$ is regular (Riesz representation theorem). Assuming $\mu \neq 0$, I tried to show$\liminf_{r\to 0} f(r)/r^n>0$, but I got stuck. Any hints?

Answer 1

Here is a proof using a covering argument:

Fix $R>0$. The open ball $B(0,R)$ can be covered by $(\lfloor \frac{2R}{r} \rfloor+1)^n$ cubes of side length $r$. Each of the cubes, in turn, is contained in a ball of radius $\sqrt{n} r$. Thus, in conclusion, the ball $B(0,R)$ can be covered by $(\lfloor \frac{2R}{r} \rfloor+1)^n$ balls of radius $\sqrt{n} r$. (This is pretty rough upper bound but it will do the job.) By the subadditivity of $\mu$, we get

$$\mu(B(0,R)) \leq \left( \frac{2R}{r} +1 \right)^n f(\sqrt{n} r). \tag{1}$$

For $r<2R$, this implies

$$\mu(B(0,R)) \leq \left( \frac{2R}{r} +\frac{2R}{r} \right)^n f(\sqrt{n} r) = 4^n R^n \sqrt{n}^n \frac{ f(\sqrt{n} r)}{(\sqrt{n} r)^n}$$

As $r \in (0,2R)$ is arbitrary, we find that

$$\mu(B(0,R)) \leq 4^n R^n \sqrt{n}^n \underbrace{\liminf_{r \to 0} \frac{ f(\sqrt{n} r)}{(\sqrt{n} r)^n}}_{=0 \,\text{by assumption}} =0.$$

Hence,

$$\mu(\mathbb{R}^d) = \sup_{R>0} \mu(B(0,R)) =0.$$