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