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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.

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    $\begingroup$ You can't just take $\mathbf{O}=E$? $\endgroup$ – Randall Jun 27 '18 at 16:00
  • $\begingroup$ Depends how the author defined "$\subset$". If $E$ has to be a real subset from $O$ there is some more work to do. $\endgroup$ – Gono Jun 27 '18 at 16:09
  • $\begingroup$ So m*(O\O)=m*(∅)=0? $\endgroup$ – T J Jun 27 '18 at 16:09
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    $\begingroup$ @Gono . Stein better mean that $\subset $ doe not mean proper subset, else he is saying that $E$ is not measurable when $E=\Bbb R^n.$ $\endgroup$ – DanielWainfleet Jun 28 '18 at 19:43
  • $\begingroup$ There are many equivalent ways to define "Lebesgue-measurable set". It is helpful to be familiar with several of them. There also seems to be a modern teaching style that omits the mention of "inner measure", which I find perplexing. $\endgroup$ – DanielWainfleet Jun 28 '18 at 19:47
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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 $0=m^*(\phi)=m^*(O \backslash E).$

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  • $\begingroup$ I strongly dispute that this is "modern usage". It's certainly a usage that exists, but it is far from dominant. $\endgroup$ – Eric Wofsey Jun 29 '18 at 4:16

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