$$\theta(t)=\begin{cases} 1&, t>0 \\ 0&, t \leq 0 \end{cases}$$
Let $\vec r(x,y,z)$ be a position vector. In a solution to an exercise, the following is stated: $$\theta(vt-|\vec r|)(\nabla\cdot \frac{\hat r}{r^2})=4\pi\theta(vt-|\vec r|)\delta^3(\vec r)\overset{?}{=}4\pi\theta (t)\delta^3(\vec r)$$ Can somebody explain the equality?
I would like to offer a good idea, why $\theta(vt-|\vec r|)=\theta (t)$, but I have no good idea:
I know that $f(\vec r)\delta^3(\vec r)=f(0)\delta^3(\vec r)$ but $\theta$ depends on $vt-|\vec r|$ not $\vec r$.
PS: $$\theta(vt-|\vec r|)=\begin{cases}1&, vt-|\vec r|>0 \\ 0&, vt-|\vec r| \leq 0 \end{cases}$$
EDIT: The book is "Introduction to Electrodynamics" (David J. Griffiths ), problem 7.34, Publisher: Pearson