 Solution

For $r\ge 0$, Lebesgue measure of the ball is $v_d{r}^d$, so the Poisson random measure gives $N_r\sim \mathrm{Poisson}(v_d{r}^d)$ andBy monotonicity of the balls, the set of radii with no points is an interval starting at zero. Thus $\{R\ge r\}=\{N_r=0\}$: if all smaller balls are empty then their union, the open ball of radius $r$, is empty; if that ball is empty then its radius belongs to the defining set. We obtainThe continuous tail gives $\mathbb{P}(R=0)=0$ and $\mathbb{P}(R=\infty )=0$. Differentiating the cumulative distribution function gives the probability density functionThe substitution $u=v_d{r}^d$ proves that this density integrates to one; equivalently $v_d{R}^d$ has the unit-rate exponential distribution. This is the first-neighbour case of the Kth-nearest-neighbour distance in a homogeneous Poisson point process.