I would like to prove that:
\begin{equation} H_{(a,b)}=H_a \otimes H_b \end{equation} So far I have:
\begin{equation} \langle H_a(x) \otimes H_b(y), \phi\rangle=\langle H_a(x),\langle H_b(y),\phi(x,y)\rangle\rangle =\\ \langle H_a(x), \psi(x)\rangle= \int_{a}^{\infty}\psi(x)dx=\int_{a}^{\infty}\int_{b}^{\infty}\phi(x,y)dxdy \end{equation} where $\phi \in C_0^{\infty}(\mathbb{R^2})$, with $H_a(x)$ I mean that the Heaviside distribution acts on the $x$ variable and that $\psi(x)=\langle H_b(y), \phi(x,y)\rangle=\int_{b}^{\infty}\phi(x,y)dy, \psi\in C_0^{\infty}(\mathbb{R})$.
First of all, I am not sure if that is correct.
Moreover, I am not sure if I can identify the double integral as a consecutive integral. I know that I should be able to, but which are the conditions that have to be met, in order for this to hold?
Thank you!