Showing that a precursor to the packing measure (on $\mathbb{R}^n$) is not a measure

Question

I am trying to prove the highlighted sentence. What countable dense sets should I consider? and how am I trying to prove this is not a measure?

I am using the usual definition of a measure (and do not mean outer measure when I say measure).

Answer 1

We should assume $s\le n$. If $s>n$, then $\mathscr P_0^s$ is in fact a measure: the zero measure.

Any countable dense subset will do. The important points is:

1. A one-point set $\{p\}$ satisfies $\mathscr P_0^s(\{p\})=0$, since in the definition there is just one ball of radius approaching $0$.
2. A dense subset $D$ of unit cube $Q=[0,1]^n$ has $\mathscr P_0^s(D)>0 $ (possibly infinite). Indeed, pick an integer $k$ so that $ 2\delta<1/k\le 4\delta$ and consider the partition of $Q$ into $k^n$ equal cubes of edgelength $1/k$. Each of these small cubes contains a ball of radius $\delta$ with center in $D$; just pick a point of $D$ that is sufficiently close to the center of the small cube. Thus, $\mathscr P_0^s(D)\ge k^n \delta^s\ge (1/4)^s$.