# Is there a measure on $[0,1]$ that is 0 on meagre sets and 1 on co-meagre sets

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.

## 2 Answers

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.

No. For any finite Borel measure $$\mu$$ on $$[0,1]$$, there is a comeager Borel set of $$\mu$$-measure zero.

First note that $$\mu$$ has at most countably many atoms, so it will be possible to find a countable dense set $$D \subset [0,1]$$ containing no atoms, i.e. $$\mu(D) = 0$$. Now any finite Borel measure on a metric space is outer regular, so for any $$n$$ there is an open set $$U_n$$ containing $$D$$ and with $$\mu(U_n) < 1/n$$. Setting $$G = \bigcap_n U_n$$, we see that $$G$$ is a dense $$G_\delta$$ (hence comeager) and $$\mu(G) = 0$$.

Relevant to your second question, the answers in the question linked above also mention that any finite Borel measure on a Polish space is Radon.