Prove that there is no finite Borel measure $\mu$ such that set of$\mu$ negligible sets equal the set of meager set Suppose that $([0,1],B([0,1]),\mu)$ is a measure space, here $B([0,1])$ is the set of all Borel sets on [0,1], let $N_{\mu}$ be the set of all subset S of $[0,1]$ such that S is $\mu$-negligible, let $M$ be the set of all meager sets contained in $[0,1]$, I want to show that there is no finite Borel measure $\mu$ on $[0,1]$ such that $N_{\mu}=M$, how to show this? can anyone help me? thank you in advance
 A: First step: If there is a point $x\in [0,1]$ such that $\mu(\{x\})>0$, then we have a meager set (namely $\{x\}$) of positive measure.
Second step: Suppose that for all $x\in [0,1]$, $\mu(\{x\})=0$. Then, since $\mu$ is finite, for any sequence of open intervals $\{U_n\}_n$ such that $\{x\}= \bigcap_n U_n$, we have $\lim_n \mu(U_n) =0$.
So we can mimic the excellent answer by Bjørn Kjos-Hanssen in MO for the case of Lebesgue measure (see here):
Let $p_i$ be a list of the rational numbers. Let $U_{i,n}$ be an open interval centered on $p_i$ such that $\mu(U_{i,n})<2^{-i}/n$. Then $V_n=\cup_i U_{i,n}$ is an open cover of the rationals, of measure at most $\sum_i 2^{-i}/n=2/n$. Then $\cap_n V_n$ is a co-meager set of measure zero. By Baire Category Theorem, $\cap_n V_n$ is not meager.
So yes, there is a measure zero set that is not meager, and so no, not every measure zero set is meager.
Remark: For the second step above, Kavi Rama Murphy'suggestion is also fine:  mimicing the construction of fat Cantor set using $\mu$, instead of the Lebesgue measure, will produce a set $C$ closed and nowhere dense (so meager) such that $\mu(C)>0$.
