# (Baby Rudin Remark 11.11) Every closed set is measurable

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!

• In any $\sigma$-field, if $A$ is a member, then $A^c$ is a member. – copper.hat Feb 18 '18 at 6:15
• 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 – Sungmin Park Feb 18 '18 at 6:18

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).