Decomposing real line as a union of a nullset and a set of first category $\Bbb R$ can be written of the form $A\cup B$ such that $A$ is of measure zero and $B$ is of the first category!
can anybody prove this?
I guess $A$ must be an $G_{\delta}$ set which is dense in $\Bbb R$ and obviously $B=\Bbb R-A$.

Enumerate the rational numbers as a sequence $\{ r_n;\; n\in\mathbb N\}$. For each $n\in\mathbb N$ and all $j\in\mathbb N$, set $$I_{n,j}:=\left] r_n-\frac{1}{j}\, 2^{-n}, r_n+\frac{1}j\, 2^{-n}\right[\, .$$ 
Then define 
$$O_j:=\bigcup_{n\in\mathbb N} I_{n,j}\, , $$
and 
$$ A:=\bigcap_{j\in\mathbb N} O_j\, .$$
Each $O_j$ is an open set containing all the $r_n$, so $A$ is a $G_\delta$ set containing all the $r_n$ and hence a dense $G_\delta$ set. Moreover, denoting by $\mu$ the Lebesgue measure on $\mathbb R$, we have
$$\mu(A)\leq\mu(O_j)\leq\sum_{n=1}^\infty \mu(I_{n,j})=\sum_{n=1}^\infty \frac{2}j\, 2^{-n}=\frac{2}j $$
for all $j\in\mathbb N$, so that $\mu(A)=0$.
