How to prove every open set is Lebesgue measurable?

Question

I am currently using Stein's book to self study measure theory and now I'm stuck on proving this property of Lebesgue measure below.

Property: Every open set in $\mathbb R^d$ is measurable

The book just says this immediately follows from the definition of Lebesgue measure (a subset $E$ of $\mathbb R^d$ is Lebesgue measurable, is for any $\varepsilon>0$ there exists an open set $O$ with $E\subset O$ and $m^*\left(O\smallsetminus E\right)<\varepsilon$), but I'm not sure how the property is derived from the definition.

Thanks in advance.

Answer 1

Modern usage is that $E\subset E.$ The author must be using this, and not requiring that $E\subsetneqq O.$ Because in the case $E=\Bbb R^n$ there is no $O$ such that $E\subsetneqq O\subset \Bbb R^n$, but if $E=\Bbb R^n$ then $E $ $ is $ measurable. So we can let $O=E,$ and it should not be hard to prove that $0=m^*(\phi)=m^*(O \backslash E).$

Answer 2

a subset $E$ of $\mathbb{R}^{d}$ is Lebesgue measurable, if for any $\varepsilon>0$ there exists an open set $O$ with $E\subset O$ and $m^{∗}(O\setminus E)<\varepsilon.$

So $E$ is Lebesgue measurable if its outer measure differs from an open set's by less than $\varepsilon.$ Well, if $E$ is open, then it's equal to its interior, which I'll denote by $E^\circ$. Then we can write $E\subset E^\circ$ and all that is left is show that $m^*(E^\circ\setminus E)<\varepsilon $, which is already done, since they're the same set and therefore $m^*(E^\circ\setminus E)=0<\varepsilon$.