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I was trying to prove this statement:

Prove that every measurable set $E \subset \mathbb{R}$ is the union of a Borel set and a set of measure 0.

And I found its prove here Prove that Lebesgue measurable set is the union of a Borel measurable set and a set of Lebesgue measure zero

But the proof in the link is unclear for me (seems to be not organized) and it uses the first part in the question in the link, So I am wondering is there a nice and easier proof than this? I am studying measure from Royden "Real Analysis" and I have read until theorem 12 on section 2.4. Any help in an organized simple proof will be greatly appreciated.

EDIT:

I have taken this theorem:

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So what can I do after using (iii) in this theorem as a given?

cmk
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Intuition
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  • Is the (2) in the subject a typo or copy-paste error? If not, what does it mean? –  Sep 09 '19 at 23:04
  • I think u just have to work with the definition of being the inf, make the intersection of the open sets , and then use the fact that its lebesgue measure to get the other set that has measure zero. – Someone Sep 09 '19 at 23:05
  • What is Royden's definition of Lebesgue measurability? Is he using the Caratheodory condition applied to Lebesgue outer measure? –  Sep 09 '19 at 23:05
  • No, it is not a typo, (2) because there is another question by the same title on this site@Bungo – Intuition Sep 09 '19 at 23:07
  • Although since u want the union i think we will have to work with the closed sets from within, i.e, use the fact that its complement is also lebesgue measure and is approximated by open sets. – Someone Sep 09 '19 at 23:09
  • The definition in Royden is: a set $E$ is said to be measurable provided for any set $A$: $$m^{}(A) = m^{}(A \cap E) + m^{*}(A \cap E^{c})$$.@Bungo – Intuition Sep 09 '19 at 23:10
  • Rudin has a very clear proof of this in his Real and Complex Analysis, 2d edition. – Matematleta Sep 10 '19 at 00:06

1 Answers1

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Hint: For any measurable set $A$ of finite measure, the measure of $A$ is the supremum of the measures of closed sets contained in $A$. Use this to show that for any measurable $A$ (maybe of infinite measure) and $n$, there is a closed $F\subseteq A$ such that $A\setminus F$ has measure smaller than $1/n$.

tomasz
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  • and this will be the set of measure zero that the measurable set contains? – Intuition Sep 09 '19 at 23:13
  • @hopefully: No, since it will have positive measure. – tomasz Sep 09 '19 at 23:14
  • I have edited my question please check it .... so your hint will be the Borel set? what about the set of measure zero? – Intuition Sep 09 '19 at 23:29
  • @hopefully: If you take the theorem you cited for granted, then this is immediate by (iv). – tomasz Sep 10 '19 at 01:02
  • And why $F_{\sigma}$ is considered as a Borel set? – Intuition Sep 10 '19 at 01:06
  • @hopefully Borel sets are a $\sigma$-algebra containing the open sets, hence closed sets are Borel, and countable unions ($F_\sigma$) of closed sets are Borel. –  Sep 10 '19 at 01:10
  • got it Thank you very much ...... so the solution is much easier than what is stated in the link above – Intuition Sep 10 '19 at 01:16
  • @hopefully: I did not read it carefully, but I believe that, presentation aside, the main difference is that the answer you linked does not take the theorem you cite for granted, but instead, contains its proof. – tomasz Sep 10 '19 at 10:04