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A dense $G_{δ}$ subset of $R$ of Lebesgue measure zero containing every rational, and its meager complement of full measure

Example

Assume the Axiom of Countable Choice. Then there is a dense $G_{δ}$ set $G \subseteq R$ with $Q \subseteq G$ and $λ_{1} (G) = 0$ . Consequently $R ∖ G$ is meager and has full measure on every bounded interval.

Facts & Assumptions

Given: The Axiom of Countable Choice.

[L1]

Every subset of $R^{n}$ has a $G_{δ}$ measurable hull of the same outer measure (Every subset of $R^{n}$ has a $G_{δ}$ measurable hull of the same outer measure).

[L2]

Every at most countable subset of $R^{n}$ is Lebesgue null (Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ ).

[F1]

$A$ is a $G_{δ}$ set of $X$ when there is a sequence $(V_{n})_{n \in N}$ of open subsets of $X$ with $A = ⋂_{n \in N} V_{n}$ ( $G_{δ}$ and $F_{σ}$ subsets of a topological space, agreeing with the real-line notion).

[F2]

$A$ is nowhere dense when the interior of its closure is empty, and a set is meager when it is a countable union of nowhere dense sets (Nowhere dense, meager (first category), residual, and second category subsets of $R$ ).

[L3]

Assuming countable choice, a box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[L4]

Assuming countable choice, $λ_{n}$ is a complete measure on the sigma-algebra $L (R^{n})$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

Verification

technique · direct

1.1L1L2

The rational line $Q$ is countable, hence null by [L2]. Applying [L1] to $Q \subseteq R$ gives a $G_{δ}$ set $G \supseteq Q$ with $λ_{1} (G) = 0$ .

2.1step 1.1F1algebra

Because $Q$ is dense in $R$ and $Q \subseteq G$ , the set $G$ is dense; and [F1] records that it is $G_{δ}$ .

3.1step 1.1step 2.1F1F2L3L4algebra∎

Write $G = ⋂_{n \in N} V_{n}$ with each $V_{n}$ open, as in [F1]. Since $G$ is dense and $G \subseteq V_{n}$ , every $V_{n}$ is dense, so each $F_{n} : = R ∖ V_{n}$ is closed with empty interior; hence $R ∖ G = ⋃_{n \in N} F_{n}$ is meager by [F2]. Now let $I$ be any bounded interval in $R$ . Step 1.1 gives $λ_{1} (I \cap G) = 0$ because $I \cap G \subseteq G$ , while [L3] makes $I$ Lebesgue measurable with finite measure equal to its length; therefore [L4] applied to the disjoint union $I = (I \cap G) ⊔ (I ∖ G)$ gives $λ_{1} (I ∖ G) = λ_{1} (I)$ , so the complement has full measure on every bounded interval.

Depends on

Used by

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Dependency tree · two levels

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A dense GδGδ​ subset of RR of Lebesgue measure zero containing every rational, and its meager complement of full measure