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My problem is this statement

$\mathfrak{M}$ contains all Borel sets in $\mathbb{R}^k$; more precisely, $E\in \mathfrak{M}$ if and only if there are sets $A$ and $B\subset \mathbb{R}^k$ such that $A\subset E\subset B$, $A$ is an $F_\sigma$, $B$ is a $G_\delta$, and $m(B-A)=0$. Also, m is regular.

Why is it that

$\mathfrak{M}$ contains all Borel sets in $\mathbb{R}^k$

is equivalent to

$E\in \mathfrak{M}$ if and only if there are sets $A$ and $B\subset \mathbb{R}^k$ such that $A\subset E\subset B$, $A$ is an $F_\sigma$, $B$ is a $G_\delta$, and $m(B-A)=0.$

user160110
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  • Why do you think they're equivalent? – zhw. Dec 20 '16 at 22:25
  • They are not equivalent. The former is included in the latter. – Henricus V. Dec 20 '16 at 22:29
  • @zhw. I assumed it was implied when he said "more precisely". In my mind the second statement is saying that $E\in \mathfrak{M}$ if it is borel almost everywhere (which is the strongest statement you can make about a set when given its measure). But I don't really see how both statements are exactly the same. – user160110 Dec 20 '16 at 22:31
  • @HenryW. Do you mean to say that there are sets which aren't borel yet still measurable? – user160110 Dec 20 '16 at 22:33
  • @user160110 Indeed. Such examples are pathological and can be found by considering the Cantor set. See this – Henricus V. Dec 20 '16 at 22:34
  • I would agree that "more precisely" is strange word usage there. That first clause is perfectly precise; it needs no further explanation. Rather than "more precisely", Rudin could have said "More is true: ... " to better effect. – zhw. Dec 20 '16 at 23:10

1 Answers1

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The two statements you mention are not equivalent.

Let $\mathfrak{M}$ be a $\sigma-$algebra containing the Borel sets, then equivalent are the following two statements:

$\mathfrak{M}$ is the minimum complete $\sigma-$algebra which contains all Borel sets in $\mathbb{R}^k$

and

$E\in \mathfrak{M}$ if and only if there are sets $A$ and $B\subset \mathbb{R}^k$ such that $A\subset E\subset B$, $A$ is an $F_\sigma$, $B$ is a $G_\delta$, and $m(B-A)=0.$

Yiorgos S. Smyrlis
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  • Ahh so first statement is the borel $\sigma$-algebra and the second statement is the completed borel $\sigma$-algebra. – user160110 Dec 20 '16 at 22:38