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Euclidean balls have positive finite Lebesgue measure

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $n \geq 1$ , let $x \in R^{n}$ , and let $r > 0$ . Then the Euclidean ball $B (x, r)$ is Lebesgue measurable and satisfies $0 < λ (B (x, r)) < \infty .$

Facts & Assumptions

Given: The Axiom of Countable Choice, an integer $n \geq 1$ , a point $x \in R^{n}$ , and a real radius $r > 0$ .

[L1]

The Euclidean ball is $B (x, r) = {y \in R^{n} : ∥ y - x ∥_{2} < r}$ . (Open ball, closed ball and sphere in a metric space, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it)

[L2]

Every box between its open and closed forms is Lebesgue measurable with its usual volume. (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)

[L3]

Open balls in any metric space are open; the Euclidean distance is a metric on $R^{n}$ for $n \geq 1$ . (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, clause 1; $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it)

[L4]

Under Countable Choice, every Borel subset of $R^{n}$ , in particular every open subset, is Lebesgue measurable. (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable)

[L5]

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

Proof

technique · direct

1.1L1algebra

If $∥ y - x ∥_{\infty} < r / \sqrt{n}$ , then [L1, algebra] $∥ y - x ∥_{2} \leq \sqrt{n} ∥ y - x ∥_{\infty} < r,$ so the open cube $x + (- r / \sqrt{n}, r / \sqrt{n})^{n}$ is contained in $B (x, r)$ . On the other hand, $∥ y - x ∥_{2} < r$ implies $∣ y_{i} - x_{i} ∣ < r$ for every coordinate, so $B (x, r) \subseteq x + (- r, r)^{n}$ .

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] $(2 r / \sqrt{n})^{n} > 0 and (2 r)^{n} < \infty .$ Since $B (x, r)$ lies between them, monotonicity [L5] gives $0 < (2 r / \sqrt{n})^{n} \leq λ (B (x, r)) \leq (2 r)^{n} < \infty .$

3.1step 2.1∎

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

Depends on

Dependency tree · 2 levels

Sources

Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Lemma 3.16