Prove that there is no finite Borel measure $\mu$ such that set of $\mu$-negligible sets equals the set of meager sets 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 subsets $S$ 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: It follows easily from the following result.

Lemma.
For any finite Borel measure $\mu$ on $[0,1]$ (possibly $\mu=0$) and $\varepsilon>0$, there exists $1/3<a<1/2<b<2/3$ such that $[0,a]\cup[b,1]$ has measure at least $(1-\varepsilon)\mu([0,1])$.

Indeed, you can then iterate the argument with $\varepsilon$ going to zero fast enough that the closed sets you construct always have measure at least 1/2 and the diameters of the connected components go to zero. The intersection of those is a closed set of empty interior and positive measure.
To prove the lemma, notice that there are finitely many atoms of mass at least $\varepsilon/2$, then choose $a=x-\delta$, $b=x+\delta$ for $x$ outside of this finite set and $\delta>0$ small enough.
A: The question has already been answered satisfactorily, but let me give a reference to a more general statement.  If $X$ is a topological space with a finite measure $\mu$ defined on the $\sigma$-algebra of sets having the property of Baire, and if $\mu(E)=0$ iff $E$ is meager, then $(X,\mu)$ is said to be a category measure space.  So your question, after completing the measure to the $\sigma$-algebra of sets having the property of Baire, is whether $[0,1]$ admits a measure $\mu$ such that the pair is a category measure space (a category measure for short).
The answer is negative by virtue of theorem 22.2 in John Oxtoby's book Measure and Category (2d ed. 1980, Springer GTM 2): if $X$ is a regular Baire space [recall that this means that every comeager space is dense, which is the case of $[0,1]$] which admits a category measure as defined in the previous paragraph, then every meager set is nowhere dense.  Clearly this is not the case if $[0,1]$ (as the set of rationals is meager but dense).
On the other hand, if we replace the usual topology on $[0,1]$ by the density topology, then meager sets are exactly Lebesgue null sets, so the Lebesgue measure becomes a category measure.
