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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.

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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.

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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.

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