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For every positive $ε$ there is a dense open subset of $(0, 1)$ of Lebesgue measure below $ε$

Example

Assume the Axiom of Countable Choice. For every real $ε > 0$ there is an open set $U \subseteq (0, 1)$ such that $U$ is dense in $(0, 1)$ and $λ_{1} (U) < ε$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a real $ε > 0$ .

[L1]

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).

[F1]

Let $μ$ be a measure and let $(E_{k})_{k \in N}$ be measurable. Then $μ (⋃_{k} E_{k}) \leq \sum_{k} μ (E_{k})$ (Finite and countable subadditivity of measures).

[F2]

The rationals are countably infinite ( $Q$ is countably infinite).

[F3]

If $∣ r ∣ < 1$ then the series $\sum r^{k}$ converges (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Verification

technique · direct

1.1

Since the rationals in $(0, 1)$ are countably infinite, fix an enumeration $(q_{k})_{k \in N}$ of $Q \cap (0, 1)$ .

F2choose

2.1

Put $U : = ⋃_{k \in N} ((q_{k} - ε 2^{- k - 3}, q_{k} + ε 2^{- k - 3}) \cap (0, 1))$ . This set is open, and [L1], [F1] and [F3] give $λ_{1} (U) \leq \sum_{k = 0}^{\infty} ε 2^{- k - 2} = ε / 2 < ε$ .

step 1.1L1F1F3algebra

3.1

Every rational point of $(0, 1)$ lies in $U$ , so $U$ is dense in $(0, 1)$ .

step 1.1step 2.1∎

Depends on

Used by

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For every positive εε there is a dense open subset of (0,1)(0,1) of Lebesgue measure below εε