Decomposing real line as a union of a nullset and a set of first category

Question

$\Bbb R$ 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 $B=\Bbb R-A$.

Answer 1

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 $\mu(A)=0$.

Answer 2

Here is another solution along the same lines:

Let $\{r_n:n\in\mathbb{N}\}$ be an enumeration of the rational numbers.

Let $I_{nj}$ denote the open interval centered at $r_n$ of length $2^{-n-j}$. Each set $U_j=\bigcup_nI_{nj}$ is open and contained the rationals; hence $U_j$ is an open dense set in $\mathbb{R}$. By Baire's theorem $B=\bigcap_j U_j$ is dense in $\mathbb{R}$ and $A=\mathbb{R}\setminus B=\bigcup_j(\mathbb{R}-U_j)$ is a set of first category: $\emptyset=\mathbb{R}\setminus \overline{U_j}=\operatorname{Int}(\mathbb{R}\setminus U_j)$ for each $j$.

Denoting by $\lambda$ the Lebesgue measure on $\mathbb{R}$ we have

$$\lambda(B)\leq \lambda(U_j)\leq \sum_n\lambda(I_{nj})=2^{-j}\qquad\forall j\in\mathbb{N}$$

This implies that $\lambda(B)=0$.

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Another, perhaps more interesting, decomposition is presented in Oxtoby, J. C., Measure and Category, 2nd edition, Springer-Verlag, 1980 which uses a little of analytic number theory (See the Wikipedia article on Liouville numbers).

First some generalities:

• Recall that $x\in\mathbb{R}$ is an algebraic number if there is a polynomial $p(x)$ with coefficients in $\mathbb{Z}$ such that $p(x)=0$.
• If $x$ is algebraic, the smallest degree of all polynomials $p$ for which $p(x)=0$ is called the order of $x$. It follows that rational numbers are algebraic and that $x$ is an algebraic number of order $1$ iff $x$ is rational.
• Since the collection of polynomials with coefficients in $\mathbb{Z}$ is countable, then the collection of algebraic numbers is countable.
• Numbers that are not albegraic are called transcendental. It follows that the collection of transcendental numbers in uncountable.

Definition: A Liouville number $x$ is an irrational number such that for for any $n\in\mathbb{N}$, there are $p,q\in\mathbb{Z}$ such that $q>1$, and $$\Big|z-\frac{p}{q}\Big|<\frac{1}{q^n}$$

We have now all the ingredients to construing the decomposition. Let $E$ denote the set of Liouville numbers. Then $$ E=\big(\mathbb{R}\setminus\mathbb{Q}\big)\cap\bigcap_{n\in\mathbb{N}}\left(\bigcup^\infty_{q=2}\bigcup_{p\in\mathbb{Z}}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)\right) $$

For each $n\in\mathbb{N}$, the set $U_n=\bigcup^\infty_{q=2}\bigcup_{p\in\mathbb{Z}}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)$ is an open set that containes the rational numbers, thus $U_n$ is open and dense in $\mathbb{R}$, and by Baire's theorem, $\bigcap_nU_n$ is dense in $\mathbb{R}$. It follows that $\bigcup_n\mathbb{R}\setminus U_n$ is of first category and so, $$\mathbb{R}\setminus E=\mathbb{Q}\cup\bigcup_n\mathbb{R}\setminus U_n$$ is of first category.

For each integer $q\geq 2$, let $U_{q,n}=\bigcup_{p\in\mathbb{Z}}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)$. Since $E\subset U_n$, for any $m\in\mathbb{N}$, $$ \begin{align} E\cap(-m,m)&\subset U_n\cap(-m,m)=\bigcup_{q\geq2}U_{q,n}\cap(-m,m)\\ &\subset\bigcup_{q\geq2}\bigcup_{|p|\leq mq}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big) \end{align} $$ Consequently, for any $n>2$ $$ \begin{align} \lambda(E\cap(-m,m))&\leq \sum_{q\geq2}\sum^{mq}_{p=-mq}\frac{2}{q^n}=\sum_{q\geq2}(2mq+1)\frac{2}{q^n}\\ &\leq (4m+1)\sum_{q\geq2}\frac{1}{q^{n-1}}\leq (4m+1)\int^\infty_1\frac{dx}{x^{n-1}}=\frac{4m+1}{n-2} \end{align} $$ whence we conclude that $\lambda(E\cap(-m,m))=0$ and so, $\lambda(E)=0$.

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What makes this construction interesting is that it shows that $E$ is of second category (so it is a rather fat set in the topological set) and yet it has measure $0$. This is to be compared with the following well known result in analytic number theory:

The following is a well known result in analytic number theory:

Theorem: Every Liouville number is transcendental.

Thus, mostly all transcendental numbers are not of the Liouville type!