Baire's theorem from a point of view of measure theory

Question

According to Baire's theorem, for each countable collection of open dense subsets of $[0,1]$, their intersection $A$ is dense. Are we able to say something about the Lebegue's measure of $A$? Must it be positive? Of full measure?

Thank you for help.

Answer 1

Let $q_1,q_2,\ldots$ be an enumeration of the rationals. Let $I_n^m$ be an open interval centered at $q_n$ with length at most $1/m 1/2^n$. Then $\bigcup_n I_n^m$ is an open dense set with Lebesgue measure at most $1/m$. The intersection of these open dense sets has Lebesgue measure zero.

Answer 2

No, not really. For $\varepsilon\in[0,1)$ let $C_\varepsilon$ be a fat Cantor set whose measure is $\varepsilon$. We know that such sets exist for every $\varepsilon$ in the given range.

The fat Cantor sets are nowhere dense, and closed, so their complement is open and dense. Pick $\varepsilon_n$ to be a sequence which approaches $1$, and let $O_n$ be the complement of $C_{\varepsilon_n}$, note that the measures of $O_n$ approach zero. Then $\bigcap O_n$ is the intersection of dense open sets, but its measure is zero.