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Open subsets of $\mathbb{R}$ are countable disjoint unions of intervals. I naively suspected that the boundary of such a countable union is countable, and hence measure zero. This is not true since Fat Cantor sets are positive measure and closed, so their complement is an open set with boundary that is positive measure. Why is the boundary of a countable disjoint union of intervals not just the union of the endpoints, and hence countable? Could someone provide a counterexample? I cannot quite think of one.

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    $\begingroup$ You already found a counterexample. And the "reason why" is that the union of the endpoints may be not closed. A simpler counterexample is $\bigcup_{n\in\Bbb N}\left(\frac1{2n+1},\frac1{2n}\right).$ $\endgroup$ Commented Apr 16, 2023 at 0:59
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    $\begingroup$ Generally, the boundary of a (countable) union is not the union of their boundaries. $\endgroup$
    – Sambo
    Commented Apr 16, 2023 at 12:10
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    $\begingroup$ @AnneBauval Ah, right, 0 is a boundary point but not an endpoint of any interval therein. Thank you for this! Much easier to "see" what is happening in this example. $\endgroup$
    – Karambwan
    Commented Apr 16, 2023 at 20:30

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The boundary of a union of pairwise-disjoint open real intervals must contain all the limits of all convergent sequences of endpoints of those intervals, and there can be uncountably many such limits.

The open intervals that comprise the complement, in $[0,1]$, of the Cantor set $C$, are all the numbers in $[0,1]$ that $cannot$ be written in base $3$ without using the digit $1$. E.g. $1/3$ in base $3$ can be written $0.0\overline 2$ so $1/3$ is $not$ in the open complement. The members of $C$ are all the numbers of the form $\sum_{n=1}^{\infty}f(n)3^{-n}$ where every $f(n)\in \{0,2\}$. There are uncountably many of them. Indeed the function $\psi:C\to \Bbb R$ where $\psi(\sum_{n=1}^{\infty}f(n)3^{-n})=\sum_{n=1}^{\infty}\frac 1 2 f(n)3^{-n}$ maps $C$ onto the whole interval $[0,1].$

Now consider taking the open middle $1/3$ of $[0,1],$ then taking open middle $1/9$ths of what's left, taking open middle $1/27$ths of what's still left, and so on. This gives us an open set $S$ with $\overline S=[0,1]$ but the boundary $\partial S=[0,1]\setminus S$ has measure $\prod_{n=1}^{\infty}(1-3^{-n})>0.$ A subset of $[0,1]$ which is closed and nowhere-dense but has positive measure is called a fat Cantor set. Some authors may require in their definition, that a fat Cantor set must also be a perfect set, i.e. a non-empty compact subspace with no isolated points.

Footnote: An elementary theorem is that if $0\le x_n<1$ for every $n$ then $\prod_{n=1}^{\infty}(1-x_n)=0\iff\sum_{n=1}^{\infty}x_n=\infty.$ Therefore $\prod_{n=1}^{\infty}(1-3^{-n})>0$ because $\sum_{n=1}^{\infty}3^{-n}\ne\infty.$ A companion result is that if $y_n\ge 0$ for every $n$ then $\prod_{n=1}^{\infty}(1+y_n)=\infty\iff\sum_{n=1}^{\infty}y_n=\infty.$

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  • $\begingroup$ There seems to be a new style of teaching Lebesgue measure that teaches outer measure but never mentions inner measure, which is like teaching addition but not subtraction. I urge you to find some good text that covers inner measure . $\endgroup$ Commented Apr 16, 2023 at 11:55
  • $\begingroup$ This is but the OP's counterexample. $\endgroup$ Commented Apr 16, 2023 at 21:20
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Consider the boundary of the middle thirds of the Cantor set: $(1/3,2/3)$, $(1/9,2/9)\cup(7/9,8/9)$, etc. This boundary is the Cantor set, well-known to be uncountable, despite being the boundary of the union of countably-many pair-wise disjoint open sets.

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  • $\begingroup$ This is but the OP's counterexample. $\endgroup$ Commented Apr 16, 2023 at 21:20

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