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!
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2$\begingroup$ Come to think of it, it is actually a duplicate of this question, I believe. $\endgroup$– Pierre PCCommented Nov 28, 2021 at 10:49
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$\begingroup$ The question has already been answered satisfactorily, but it's worth pointing out that the existence of a Borel measure on $X$ such that meager sets have measure zero imposes topological restrictions on $X$ which are not satisfied by the usual topology on the real line (but are, e.g., by the density topology). Chapter 22 of Oxtoby's book Measure and Category has more on this. I'll try to post an excerpt as an answer later. $\endgroup$– Gro-TsenCommented Nov 28, 2021 at 14:33
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$\begingroup$ Does this answer your question? Is there a measure on $[0,1]$ that is 0 on meagre sets and 1 on co-meagre sets $\endgroup$– Eric PetersonCommented Nov 30, 2021 at 14:23
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2 Answers
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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.
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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.
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1$\begingroup$ I just want to mention that starting from any probability space $(X,\mu)$ there is a canonical way to generate an associated category measure space $(\hat X, \hat \mu)$, by taking $\hat X$ to be the Gelfand dual of $L^\infty(X,\mu)$ (viewed as a $C^*$-algebra), and defining $\hat \mu$ via the Riesz representation theorem; alternatively, one can apply the Loomis-Sikorski theorem to the measure algebra of $X$. See Section 7 and Remark 9.1 of this paper of myself and Asgar Jamneshan: arxiv.org/abs/2010.00681 . We call $(\hat X,\hat \mu)$ the canonical model of $(X,\mu)$. $\endgroup$ Commented Nov 29, 2021 at 18:46