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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.

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  • $\begingroup$ Sorry, I mean the closure of the set of points has 0 lebesgue measure, not the closure of the interval $\endgroup$
    – reinin
    Commented Jun 9, 2019 at 17:59

1 Answer 1

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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 $\mathbb{R}$.

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  • $\begingroup$ Thank you for the informative response $\endgroup$
    – reinin
    Commented Jun 9, 2019 at 18:06

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