# How to prove every open set is Lebesgue measurable?

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.

• You can't just take $\mathbf{O}=E$? – Randall Jun 27 '18 at 16:00
• Depends how the author defined "$\subset$". If $E$ has to be a real subset from $O$ there is some more work to do. – Gono Jun 27 '18 at 16:09
• @Gono . Stein better mean that $\subset$ doe not mean proper subset, else he is saying that $E$ is not measurable when $E=\Bbb R^n.$ – DanielWainfleet Jun 28 '18 at 19:43
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 $0=m^*(\phi)=m^*(O \backslash E).$