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I'm reading Principles of Mathematical Analysis Chapter 11 written by Walter Rudin. I came up to Remark 11.11 (a) where it says "If $A$ is open in $R^p$, then $A \in \mathfrak{M}(\mu)$, since it is possible to construct a countable base whose members are open intervals".

But the next statement is confuses me. It says, "By taking complements, it follows that every closed set is in $\mathfrak{M}(\mu)$"

I have searched for similar questions here at Math Stack Exchange and most of the answers make use of Borel Set or $\mathscr{B}$, a collection of all Borel Set, or $\sigma$-algebra, which is closed under "complement", but they are yet to be defined in Remark 11.11.

My idea is as follows. Let $A$ be a set in $R^p$ which is open and is a union of countable open interval. $$A = \bigcup_{n=1}^{\infty}I_n$$ then, $$A^c = \bigcap_{n=1}^{\infty}{I_n}^c$$ $\mathfrak{M}(\mu)$ is a $\sigma$-ring which means that it is closed under countable union and therefore countable intersection. Therefore if ${I_n}^c \in \mathfrak{M}(\mu)$, then the proof is complete.

I tried to think of unit intervals $U_{(n_1,n_2,\dots,n_p)}$, which I defined, $$U_{(n_1,n_2,\dots,n_p)} = \{ x \in R^p | n_i \le x_i \lt n_i+1 \text{ for } i=1,2,\dots,p \}, \text{ $n_1,n_2,\dots,n_p$ are integers.}$$ Every unit intervals are pairwise disjoint, and collection of all unit intervals in $R^p$ is countable. For some unit interval $U^*$ which intersects $I_n$, $(U^* - I_n)$ is an elementary set which can be broken down into a union of disjoint finite intervals. Therefore it seems that ${I_n}^c$ can be experessed as a countable union of disjoint intervals. $U_{(n_1,n_2,\dots,n_p)}$ belongs to $\mathscr{E}$(a family of all elementary subsets in $R^p$). It is possible to construct a sequence where every member of the sequence is the interval itself, therefore $U_{(n_1,n_2,\dots,n_p)} \in \mathfrak{M}_F(\mu)$. Since ${I_n}^c$ can be expressed as a countable union of unit intervals in $\mathfrak{M}_F(\mu)$, ${I_n}^c \in \mathfrak{M}(\mu)$ and proof is complete.

But I am uncertain whether this solution is correct. Any corrections or guidance into the right direction is appreciated. Thanks!

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  • $\begingroup$ In any $\sigma$-field, if $A $ is a member, then $A^c$ is a member. $\endgroup$ – copper.hat Feb 18 '18 at 6:15
  • $\begingroup$ Yes, I read that in other answers, but the book hasn't yet introduced the definition and property of $\sigma$-field so I was curious of other answers. Thanks anyway! @copper.hat $\endgroup$ – Sungmin Park Feb 18 '18 at 6:18
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Your argument is perfectly fine, but intersecting with unit intervals is unnecessary. You can actually tile ${\mathbb R}^p$ with a countable number of translates of the original interval $I$ (after possibly adding/deleting faces on a few of them), so upon deleting $I$ from ${\mathbb R}^p$ you immediately see that $I^c$ is the countable union of intervals.

An example to illustrate what I'm saying: take $I=[a,b]$, and put $l=b-a$. Then $${\mathbb R} = \dotsm\cup[a-2l,a-l)\cup [a-l,a)\cup[a,b]\cup(b,b+l)\cup[b+l, b+2l)\cup\dotsm$$ The idea is you edit the faces of the intervals adjacent to $I$ to make everything fit together nicely, and you take all the other translates to be half-open. The same argument works in higher dimensions (think about why).

Another way you could have verified Rudin's remark is by noting every closed set in ${\mathbb R}^p$ is a countable intersection of open sets (see here if you're unfamiliar with this fact: Closed set as a countable intersection of open sets).

As an aside, Chapter 11 of Baby Rudin is easily the worst part of the book. If you're not being forced to use it, there are much better introductions to measure theory (Wheeden and Zygmund, Royden, Stein and Shakarchi, Terry Tao's notes, etc.)

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Let A be a closed set.
Note that R^p is an open set, so it is measurable.
Also, since A is closed, its complement is an open set, so it is measurable.
Then since M(u) is a ring, A=R^p-A^c is measurable.

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