13
$\begingroup$

Does there exist $K \subseteq \mathbb{R} \backslash \mathbb{Q}$ such that $K$ is compact, and has Lebesgue measure greater than $0$? As I have been trying to think of examples, I suspect that any subset of $\mathbb{R} \backslash \mathbb{Q}$ that is closed can be at most countable, since the closure of an uncountable subset of irrationals should contain some rationals. And, the Lebesgue measure of a countable set is $0$. If there are any examples of such a set, I would be very interested to know how it is constructed.

$\endgroup$
2

1 Answer 1

18
$\begingroup$

The answer is yes. Count the rationals in $[0,1]$ as $r_1,r_2,\ldots$, let $I_k$ be an open interval containing $r_k$ of length $3^{-k}$, and let $K=[0,1]\setminus\cup_k I_k$.

This question is somewhat related to the question Perfect set without rationals, but there measure did not come up. For example, the Cantor set-like construction given there by JDH could be made to have positive measure.

$\endgroup$
12
  • 2
    $\begingroup$ You could also appeal to inner regularity of Lebesgue measure. $\endgroup$
    – t.b.
    Commented Feb 9, 2011 at 1:26
  • $\begingroup$ @Theo: Good point, but wouldn't the most straightforward way to see that be to cover the complement with an approximating open set and then take complements? Either way, thanks, that is a nice way to think of it which hadn't occurred to me. $\endgroup$ Commented Feb 9, 2011 at 1:33
  • $\begingroup$ Sorry I didn't read the question properly, where an explicit construction was asked. You're of course right, all proofs of inner regularity I can think of at the moment essentially boil down to this. $\endgroup$
    – t.b.
    Commented Feb 9, 2011 at 1:33
  • $\begingroup$ Come to think of it, "most straightforward" depends on how Lebesgue measure is constructed. The way I first learned it is in terms of defining outer measure using covering sequences of intervals and then imposing the Carathéodory criterion (or equivalent) for measurability. However, in Rudin's Real and complex analysis, Lebesgue measure comes from an application of the Riesz representation theorem, and its regularity follows by an application of a general theorem on the locally compact & $\sigma$-compact case. $\endgroup$ Commented Feb 9, 2011 at 1:49
  • $\begingroup$ @Theo, inner regularity of Lebesgue measure helped a lot in solving my problem. Thanks! $\endgroup$
    – PFHayes
    Commented Feb 9, 2011 at 2:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .