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Does there exist a partition of real numbers (with standard topology; Lebesgue measure) into two measurable sets $A$ and $B$, satisfying the following properties:

  1. $A$, $B$ are both dense in the real numbers.
  2. $A$, $B$ both have positive measure.

Edited: as pointed out by Henry, actually I am looking for A B such that for any open interval I, the intersection of I and A, the intersection of B and I both have positive measure. Sorry for the confusion.

Tongou Yang
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  • Let $A$ be all reals in [0,1] and all rational numbers outside of [0,1] and $B$ be $A's$ complement in real. – cr001 Oct 01 '15 at 07:10
  • @cr001 $B$ would not be dense in the reals - the nbhd $(0,1)$ does not intersect $B$. Thomas do $A$ and $B$ need to be measurable, or just of positive outer measure? – Titus Oct 01 '15 at 07:12
  • I am sorry for the typo. It should be "Let $A$ be all irrationals in [0,1] and all rational numbers outside of [0,1] " – cr001 Oct 01 '15 at 07:15
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    It might be more interesting to look for the intersections of $A,B$ with any open interval having positive measure. – Henry Oct 01 '15 at 07:48
  • I would not quite say that this question is a duplicate of the question about a Borel set, but an answer to that question is definitely an answer to this one. So Thomas Yang should look there. – Rory Daulton Oct 01 '15 at 20:11

2 Answers2

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A=Positive rational numbers and negative irrational numbers.

B=Negative rational numbers, positive irrational numbers and 0.

Stig Hemmer
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There is a rather famous construction of such a set. It helps to know the following facts:

  1. if $O$ is a nonempty open set then there exists a nowhere dense closed set $C \subset O$ with the property that $0 < m(C)$, and
  2. if $C_1,\ldots,C_n$ is a collection of closed nowhere dense sets and $O$ is a nonempty open set, then $O \setminus (C_1 \cup \cdots \cup C_n)$ is nonempty and open.

In particular, if $O$ is open we can find two disjoint nowhere dense closed sets $C_1,C_2 \subset O$ with the property that $0 < m(C_1)$ and $0 < m(C_2)$: just select $C_2 \subset O \setminus C_1$.

Let $\{I_n\}$ be a sequence of open intervals that forms a basis for the topology of the line.

Step 1: Select two disjoint nowhere dense closed sets $K_1,K_2 \subset I_1$ such that $0 < m(K_1)$ and $0 < m(K_2)$.

Step 2: Select two disjoint nowhere dense closed sets $K_3,K_4 \subset I_2 \setminus (K_1 \cup K_2)$ such that $0 < m(K_3)$ and $0 < m(K_4)$.

Step 3: Select two disjoint nowhere dense closed sets $K_5,K_6 \subset I_3 \setminus (K_1 \cup K_2 \cup K_3 \cup K_4)$ such that $0 < m(K_5)$ and $0 < m(K_6)$.

Step 4: Proceed inductively to obtain a sequence $\{K_j\}$ of pairwise disjoint nowhere dense closed sets with the property that $K_{2j-1},K_{2j} \subset I_j$. Define $$E = \bigcup_j K_{2j-1},\quad F = \bigcup_j K_{2j}.$$

That ends the construction. If $O \subset \mathbb R$ is open, there exists an interval $I_j \subset O$ so that $$ E \cap O \supset E \cap I_j \supset K_{2j-1}$$ and $$ E^c \cap O \supset F \cap O \supset F \cap I_{j} \supset K_{2j}.$$ Thus $$m(E \cap O) \ge m(K_{2j-1}) > 0 \quad \text{and} \quad m(E^c \cap O) \ge m(K_{2j}) > 0.$$

We conclude that both $E$ and $E^c$ occupy a set of positive measure in every open subset of the line.

Umberto P.
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