I'm curious if there is a finite measure on the $\sigma$-algebra of subsets of $[0,1]$ with the Property of Baire, whose null sets are exactly the meagre sets.


I'd also be interested how "nice" such a measure can be like can it be Radon(when restricted to Borel sets) for example.



Is there a measure on $[0,1]$ that is 0 on meagre sets and 1 on co-meagre sets
The answer is no. Assume that such a measure $\mu$ exists.


First, since every singleton in $[0,1]$ is closed with empty interior, $\mu(\{x\}) = 0$ for all $x \in [0,1]$. Write $B\_{x,\epsilon}$ for the open ball around $x$ of radius $\epsilon$ with respect to the standard metric on $[0,1]$. By countable additivity, for all $x \in [0,1]$, $\mu(B\_{x,2^{-n}}) \to 0$. If we take an enumeration of the rationals $(q\_i)\_{i \in \mathbb{N}}$, for each $i$ there exists an $n\_i$ such that $\mu(B(q\_i,2^{-n\_i})) < 2^{-i}$. So $D\_1 = \bigcup\_{i=1}^\infty B(q\_i,2^{-n\_i})$ is a dense open set with $\mu(D\_1) \leq 1$. 


By re-doing the previous construction, picking $\mu(B(q\_i,2^{-n\_i})) < 2^{-(i+k)}$, we can define dense open sets $D\_k$ with $\mu(D\_k) \leq 2^{-k}$. Now, by countable additivity $N = \bigcap\_{k=1}^\infty D\_k$ has measure zero. The set $[0,1]\setminus N$ is a union of closed sets with empty interior, *i.e.* a meagre set, so $\mu([0,1] \setminus N) = 0$ as well, so $\mu([0,1]) = 0$. 


I see that Nate Eldredge was a bit quicker than I was, so I'll add that it *is* possible to find a finitely-additive probability measure whose null sets are exactly the meagre sets -- this is most easily done using the isomorphism between the Baire property algebra modulo meagre sets and the algebra of regular open sets.


