Measure on the reduced Borel field of subsets of the Cantor space

Question

Let $X=2^\omega$ be the Cantor space, $B(X)$ the Borel field of $X$, and $M(X)$ the $\sigma$-ideal of meager subsets of $X$. Is it possible to define a measure on $B(X)/M(X)$?

Answer 1

No. The proof is essentially the same as the proof that there are meager sets of positive measure with respect to Lebesgue measure.

In detail, if $\mu$ is a measure on $B(X)/M(X)$, we can consider $\mu$ as a measure on $B(X)$ which vanishes on all meager sets. Now let $\{q_n:n\in\mathbb{N}\}$ be a countable dense subset of $X$. For each $n$, $\{q_n\}$ is meager and so must have measure $0$. Picking a sequence of open neighborhoods $U_m$ of $q_n$ with $\bigcap_m U_m=\{q_n\}$, we must have $\mu(U_m)\to 0$. In particular, we can pick $m$ such that $\mu(U_m)<2^{-n-1}$. Let us write $V_n$ for this chosen $U_m$.

Now let $V=\bigcup_n V_n$. Since $q_n\in V$ for all $n$, $V$ is an open dense set, and $X\setminus V$ is meager. But $\mu(V)\leq\sum_n\mu(V_n)<\sum_n 2^{-n-1}<1$, so $\mu(X\setminus V)>0$. This is a contradiction, and so no such $\mu$ can exist.