Showing that $m(E_1\times E_2)=m(E_1)m(E_2)$

Question

Please find the sentence underlined with red. Can anyone tell me how to use the corollary to attain the result? Is $$\int_{\mathbb{R}^{d_2}} m((E_1\times E_2)^y)\mathrm{d}y$$ helpful? I'm stuck with this problem. Thank you.

Answer 1

If $E=E_1\times E_2$ is measurable, by Corollary 3.3, we have \begin{align*} m(E)=\int_{\mathbb{R}^{d_2}} m(E^y)dy =\int_{\mathbb{R}^{d_2}} m(E_1)\chi_{E_2}dy=m(E_1)\int_{\mathbb{R}^{d_2}}\chi_{E_2}dy=m(E_1)m(E_2) \end{align*}

because $$E^y=\begin{cases} E_1&,y\in E_2 \\ \emptyset&,y\notin E_2. \end{cases}$$