 Solution

For $c>0$, Brownian scaling and the bijection $t\mapsto c^{2}t$ of $[0,\infty )$ give $Z\stackrel{d}{=}{c}^{-1}Z$, or $\boxed{cZ\stackrel{d}{=}Z}$. The supremum is a measurable extended random variable because continuity allows the supremum over rational times, and $Z\ge 0$.

For $z>0$, the function $F(z)=\mathbb{P}(Z\le z)$ must satisfy $F(z)=F(z/c)$ for every $c>0$, so it is constant on $(0,\infty )$. Letting $z\downarrow 0$ and $z\uparrow \infty$ shows $\mathbb{P}(Z=0)=\mathbb{P}(Z<\infty )$, hence $\boxed{Z\in \{0,+\infty \}\text{ almost surely}}$. This scaling argument alone does not decide the mass at zero; part (d) does.