# Take a set of points in a closed interval whose closure has 0 lebesgue measure. Can we cover them with finitely many intervals of small measure?

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.

• Sorry, I mean the closure of the set of points has 0 lebesgue measure, not the closure of the interval – reinin Jun 9 '19 at 17:59

## 1 Answer

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 $$\mathbb{R}$$.

• Thank you for the informative response – reinin Jun 9 '19 at 18:06