1
$\begingroup$

Assume the closure of a set of points $\mathcal{P}$ has $0$ Lebesgue measure. For any interval $[a,b] \subset \mathbb{R}$ and any $\delta > 0$, does there exist a finite number of open intervals whose union has measure $\delta$ that covers $\mathcal{P} \cap [a,b]$? This intuitively makes sense if we consider the rationals, which has $0$ Lebesgue measure but whose closure ($\mathbb{R}$) has non-zero Lebesgue measure. So it looks like any set whose closure has $0$ Lebesgue measure must have cardinality less than the rationals, but I don't know if this helps in showing the statement.

$\endgroup$
  • $\begingroup$ Sorry, I mean the closure of the set of points has 0 lebesgue measure, not the closure of the interval $\endgroup$ – reinin Jun 9 at 17:59
1
$\begingroup$

Assume the closure of a set of points $\mathcal{P}$ has $0$ Lebesgue measure. For any interval $[a,b] \subset \mathbb{R}$ and any $\delta > 0$, does there exist a finite number of open intervals whose union has measure $\delta$ that covers $\mathcal{P} \cap [a,b]$?

Yes. If $\overline{P}$ has measure zero, then $\overline{P} \cap [a,b]$ is a compact set of measure zero. Since it has measure zero, you can cover it with a countable number of open intervals whose total measure is less than $\delta$. Since it is compact, this open cover has a finite subcover, whose total measure will be even smaller. Since it covers $\overline{P} \cap [a,b]$, it necessarily also covers $P \cap [a,b]$.

So it looks like any set whose closure has $0$ Lebesgue measure must have cardinality less than the rationals

That is false; the standard counterexample is the Cantor set, which is closed, has measure zero, and has cardinality equal to that of $\mathbb{R}$.

$\endgroup$
  • $\begingroup$ Thank you for the informative response $\endgroup$ – reinin Jun 9 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.