Let $U$ be the function defined by $$\begin{cases} U(x) = 1 & \text{if } x \geqslant 0 \\ U(x) = 0 & \text{if } x < 0 \end{cases}$$ Justify that $U$ defines a distribution on $\mathcal{D}$.
Let $U$ be the function defined by
$$\begin{cases} U(x) = 1 & \text{if } x \geqslant 0 \\ U(x) = 0 & \text{if } x < 0 \end{cases}$$
Justify that $U$ defines a distribution on $\mathcal{D}$.