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Euclidean balls have positive finite Lebesgue measure

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)).

Let n≥1, let x∈Rn, and let r>0. Then the Euclidean ball B(x,r) is Lebesgue measurable and satisfies 0<λ(B(x,r))<∞.

Facts & Assumptions

Given: The Axiom of Countable Choice, an integer n≥1, a point x∈Rn, and a real radius r>0.

[L4]

Under Countable Choice, every Borel subset of Rn, in particular every open subset, is Lebesgue measurable. (Assuming countable choice, every Borel subset of Rn is Lebesgue measurable)

[L5]

Under Countable Choice, λn is a measure on the Lebesgue sigma-algebra, and measures are monotone on measurable inclusions. (Assuming countable choice, L(Rn) is a sigma-algebra containing every elementary set and λn is a complete measure extending elementary volume, clause 1; Measures are monotone)

Proof

technique · direct
1.1L1algebra

If ∥y−x∥∞<r/n, then [L1, algebra] ∥y−x∥2≤n ∥y−x∥∞<r, so the open cube x+(−r/n,r/n)n is contained in B(x,r). On the other hand, ∥y−x∥2<r implies ∣yi−xi∣<r for every coordinate, so B(x,r)⊆x+(−r,r)n.

2.1step 1.1L2L3L4L5algebra

The ball B(x,r) is open in the Euclidean metric by [L3], hence Borel and Lebesgue measurable by [L4] under the stated Countable Choice. The two cubes from step 1.1 are measurable by [L2], with measures [step 1.1, L2, L3, L4, algebra] (2r/n)n>0and(2r)n<∞. Since B(x,r) lies between them, monotonicity [L5] gives 0<(2r/n)n≤λ(B(x,r))≤(2r)n<∞.

3.1step 2.1∎

Therefore B(x,r) is Lebesgue measurable of positive finite measure.

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