I was asked to prove or give a counterexample of the statement.

Assume $P$ is a subset of $[0,1]$ with Lebesgue measure equal to $1$. Then $P$ is not a countable union of nowhere dense sets.

I find that there are nowhere dense sets with positive measure. So I am not able to derive this from subadditivity.

  • 2
    $\begingroup$ This is a classic problem. Otherwise phrased: $[0,1]$ is the union of a meager set and a set of measure zero. I learned this many years ago in Oxtoby's nice little book, Measure and Category. $\endgroup$
    – GEdgar
    Commented Sep 15, 2020 at 20:51

1 Answer 1


Here is a counterexample:

For any $n$, there is a fat cantor set $C_n$ of measure $> 1 - 1/n$. Each $C_n$ is nowhere dense, since it is closed and has empty interior.

The union $C_1 \cup C_2 \cup \ldots $ has measure one, and is a countable union of nowhere dense sets.


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