TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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Closed Euclidean balls are Jordan measurable and their volumes satisfy the slicing recursion

Statement

For every integer $n \geq 1$ and every $r \geq 0$ , put ${\overline{B}}_{n} (0, r) : = {x \in R^{n} : ∥ x ∥_{2} \leq r}$ , extending the positive-radius notation of Euclidean spheres and closed balls as subspaces of $R^{n}$ to $r = 0$ . This closed Euclidean ball is Jordan measurable; write its content as $V_{n} (r)$ . One has $V_{1} (r) = 2 r$ . For $n \geq 2$ and $r \geq 0$ , $V_{n} (r) = V_{n - 1} (1) \int_{- r}^{r} (r^{2} - t^{2})^{(n - 1) / 2} d t$ .

Facts & Assumptions

Given: Positive integer dimension $n$ , radius $r \geq 0$ , and the closed Euclidean balls defined in the Statement.

[F1]

A solid between continuous graphs over a compact Jordan base is compact and Jordan measurable (A solid between continuous graphs over a compact Jordan base is Jordan measurable and integrates by vertical sections).

[F2]

If a linear map $T$ has matrix $A$ and $E$ is a bounded Jordan set, then $cont (T (E)) = ∣ \det A ∣ cont (E)$ (A linear endomorphism of $R^{n}$ sends bounded Jordan sets to bounded Jordan sets and scales their content by the absolute determinant).

[F3]

If a bounded Jordan set has Jordan-measurable sections outside a content-zero parameter set, then its completed sectional-content function, with empty sections assigned content $0$ , is integrable and its integral is the set's content (Cavalieri: Jordan content is the integral of sectional contents, and equal sections give equal content).

[F4]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[F5]

The inverse of a continuous injective real function on an interval is continuous on its image (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

[F6]

Every nonnegative real has a unique nonnegative square root (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[F7]

A Euclidean set is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F8]

The Euclidean distance is $d_{2} (u, v) = ∥ u - v ∥_{2}$ and satisfies the metric triangle inequality ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

Proof

technique · induction

1.1

For $n = 1$ , the ball is the closed bounded interval $[- r, r]$ , hence compact by [F7], and it is Jordan measurable with content $2 r$ , including $r = 0$ .

givenF7base

2.1

Assume closed balls in dimension $n - 1$ are compact and Jordan measurable. The triangle inequality in [F8], applied in both orders, gives $∣ ∥ u ∥_{2} - ∥ v ∥_{2} ∣ \leq ∥ u - v ∥_{2}$ , so the Euclidean norm is continuous. On $[0, r]$ , the estimate $∣ (r^{2} - s^{2}) - (r^{2} - t^{2}) ∣ = ∣ s - t ∣ ∣ s + t ∣ \leq 2 r ∣ s - t ∣$ makes $s \mapsto r^{2} - s^{2}$ continuous; [F5] and [F6] make its nonnegative square root continuous, and [F4] makes the resulting composite continuous on the ball. Thus the $n$ -ball is the solid between two continuous square-root graphs over ${\overline{B}}_{n - 1} (0, r)$ , and [F1] makes it compact and Jordan measurable.

step 1.1ihF1F4F5F6F8algebra

3.1

For $∣ t ∣ \leq r$ , the section at last coordinate $t$ is the $(n - 1)$ -ball of radius $ρ (t) : = \sqrt{r^{2} - t^{2}}$ . It is the image of the unit $(n - 1)$ -ball under scalar multiplication by $ρ (t)$ , whose determinant has absolute value $ρ (t)^{n - 1}$ ; [F2] gives section content $V_{n - 1} (1) (r^{2} - t^{2})^{(n - 1) / 2}$ . At $t = \pm r$ this is zero.

step 2.1F2algebra

4.1

By [F3], integration of the section contents in step 3.1 gives the displayed recursion. Thus the induction proves Jordan measurability in every positive dimension and the recursion for every $n \geq 2$ .

step 3.1F3discharge-induction∎

Depends on

Used by

The closed form for the volume of the unit n-ball Corollary
The volume of a radius-r closed n-ball is π^n/2rⁿ/Γ(n/2+1) Corollary
Slicing gives the unit-ball volumes through dimension five Example

Dependency tree · two levels

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Sources

University of Toronto MAT237Y1, The Gamma Function and the Beta Function, §2.4 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis, §5C (standard reference, not scraped)