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Theorem : There exists a measurable subset of $\Bbb{R}$ without the property of Baire and a non measurable set with the property of Baire.

Lemma $ 1$: $ \Bbb{R}$ can be decomposed into disjoint union of two sets, one is null and other is meager.

Lemma $ 2$ : Any set with positive outer measure contains a non measurable subset.

Lemma $3$: Any set of second category contains a set that lacks the property of Baire.

Lemma $4$ : Any subset of a null set is (null) measurable.

Lemma $5$ : Any subset of a first category set (is first category) has the property of Baire.

Proof :

By Lemma $1$ , $\Bbb{R}=N\cup M$ where $N$ null and $M$ meager.

Clearly $N$ is of second category otherwise contradict $\Bbb{R}$ is Baire space. Hence by Lemma $3$ , there exists $A\subset N$ that lacks the property of Baire. But by Lemma $4$ , $A$ is measurable. Hence $A$ is measurable set without the property of Baire.

Clearly $M$ has positive outer measure (in fact positive measure, as $M=\Bbb{R}\setminus M$ measurable) , by Lemma $ 2$ , there exists $B\subset M$ such that $B$ is non measurable.But by Lemma $5$ , $B$ has the property of Baire. Hence $B$ is a non measurable set that lacks the property of Baire.

Any comment, suggestion,criticism.

Am I need any further improvement?

Sourav Ghosh
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1 Answers1

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A different approach: Let $b$ be a Bernstein set. That is, $b\subset\Bbb R$ and $b\cap c\ne\emptyset\ne (\Bbb R\setminus b)\cap c$ for every uncountable closed $c\subset\Bbb R.$

Note that $b$ is dense in $\Bbb R.$

Any closed subset of $\Bbb R$ that is a subset of $b$ or of its complement is a countable set so $b$ and its complement each have inner Lebesgue measure $0,$ so $b$ is not measurable.

Suppose by contradiction that $F$ is a countable non-empty family of dense open subsets of the space $b$ with $\cap F=\emptyset.$ For each $f\in F$ let $f'$ be an open subset of $\Bbb R$ with $f'\cap b=f.$ Now $f$ is dense in $b,$ and $b$ is dense in $\Bbb R,$ so $f$ is dense in $\Bbb R.$ So $f'$ is dense in $\Bbb R.$

By the Baire Category Theorem the set $d=\cap_{f\in F}f'$ is a dense $G_{\delta}$ subset of $\Bbb R.$ And $d\cap b=\cap_{f\in F}(f'\cap b)=\cap F=\emptyset.$

But there exists a closed uncountable $c$ in $\Bbb R$ with $c\subset d$, so $c\cap b$ is empty, contrary to $b$ being a Bernstein set. The existence of this $c\subset d$ comes from:

(1). A $G_{\delta}$ subset of $\Bbb R$ is a completely metrizable space.

(2).If $d$ is a non-empty completely metrizable space with no isolated points then $d$ has a subspace $c$ which is homeomorphic to the Cantor set.

DanielWainfleet
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  • Excellent approach. – Sourav Ghosh Jun 02 '22 at 17:03
  • $\cap F$ must be dense in $b$. Let $x,y\in\Bbb R$ with $x<y$. Let $\psi:(x,y)\to\Bbb R$ be a homeomorphism. The image $b^=\psi [b\cap (x,y)]$ is a Bernstein set and $F^={\psi [f]:f\in F}$ is a non-empty countable family of dense open subsets of the space $b^$. So by the argument in my A we have $\cap F^\ne\emptyset$, so $\emptyset\ne\psi^{-1}[\cap F^*]=(\cap F)\cap (x,y).$ – DanielWainfleet Jun 02 '22 at 17:45