Lebesgue Measure of the Cartesian Product

If $E$ is Lebesgue measurable in $\mathbb{R}^n$ and $I=[a,b]$ how do I show that $E\times I$ is measurable in $\mathbb{R}^{n+1}$?

Jonas:

I'm using $\mu^*(E)=\inf \{ \sum \mathrm{Vol}(I_k) \mid E\subseteq \cup I_k\}$ and for every $\epsilon \gt 0$ there exists an open set $G$ containing $E$ such that $\mu^*(G-E)\lt\epsilon$ ($\mu^*$ is the outer measure).

I tried using the first definition since I think it would be easier, but I don't know how to make it fit together.

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1 Answer

An equivalent criterion for measurability of a set $E$ is the existence of a $G_\delta$ set $G$ containing $E$ such that $\mu^*(G\setminus E)=0$. (If you haven't already seen this, you can prove it.) You can use this along with the fact that $(G_1\times G_2)\setminus(E_1\times E_2)=((G_1\setminus E_1)\times G_2)\cup(G_1\times(G_2\setminus E_2))$ to show that if $E_1$ is measurable in $\mathbb{R^n}$ and $E_2$ is measurable in $\mathbb{R^m}$, then $E_1\times E_2$ is measurable in $\mathbb{R^{n+m}}$.

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