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Finite coordinate boxes and their faces are compact Jordan sets

Statement

Let $m \geq 1$ be an integer. For a canonical coordinate vector $x : {0, \dots, m - 1} \to R$ , write $x_{j} : = x (j - 1)$ for $1 \leq j \leq m$ . Accordingly, $\prod_{j = 1}^{m} I_{j}$ denotes ${x \in R^{m} : x (j - 1) \in I_{j} for 1 \leq j \leq m}$ ; this is only a relabelling of the coordinates in the cited suppliers. Let $Q = \prod_{j = 1}^{m} [a_{j}, b_{j}] \subseteq R^{m}, a_{j} < b_{j} .$ Then $Q$ is compact and Jordan measurable, and its boundary has Jordan content zero. For any product $R = \prod_{j = 1}^{m} I_{j}$ where each $I_{j}$ is one of $[a_{j}, b_{j}]$ , $(a_{j}, b_{j})$ , $[a_{j}, b_{j})$ or $(a_{j}, b_{j}]$ , the set $R$ is a bounded Jordan set with closure contained in $Q$ . Every continuous $f : Q \to R$ is Riemann integrable over $R$ . When $m \geq 2$ , each closed coordinate face of $Q$ is compact and Jordan measurable in its own $(m - 1)$ -dimensional coordinate space; the same endpoint-variant and integrability conclusions hold for products in that face.

Facts & Assumptions

Given: A positive integer $m$ and finite real endpoints $a_{j} < b_{j}$ for $j = 1, \dots, m$ .

[F1]

An axis-parallel rectangle has volume equal to the product of its side lengths (Axis-parallel rectangles in $R^{m}$ and their volume).

[F2]

Jordan outer content is the infimum of total volumes of finite axis-parallel rectangle covers; Jordan measurability is equality of inner and outer content (Jordan inner and outer content and Jordan measurable bounded sets in $R^{m}$ ). Content zero implies nullity in the covering sense used in Lebesgue's criterion, and content zero passes to subsets (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[F3]

A closed axis-parallel box in Euclidean $R^{m}$ is compact (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 clause 1).

[F4]

A bounded set is Jordan measurable exactly when its boundary has Jordan content zero (A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero).

[F5]

A continuous real function on a compact Jordan set is Riemann integrable (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

[F6]

A bounded function on a rectangle whose discontinuity set is null in the covering sense is Riemann integrable (Lebesgue's criterion in $R^{m}$ : a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null).

[F7]

A bounded function on a bounded Jordan set is Riemann integrable when its zero extension to a containing rectangle is Riemann integrable (The Riemann integral of a bounded function over a bounded Jordan measurable set).

Proof

technique · direct finite-cube covering

1.1givenF3algebra

The box $Q$ is closed and bounded because it is a finite product of closed bounded intervals; [F3] therefore makes $Q$ compact. Its boundary is the union of the $2 m$ coordinate faces $F_{r, -} : = {x \in Q : x_{r} = a_{r}}$ and $F_{r, +} : = {x \in Q : x_{r} = b_{r}}$ , since a point of $Q$ is interior exactly when all its coordinates lie strictly between the endpoints.

2.1F1F2step 1.1algebra

Put $C : = \max (1, \max_{1 \leq r \leq m} \prod_{j \neq r} (b_{j} - a_{j} + 1))$ , with empty product $1$ . Given $ε > 0$ , choose $0 < δ < \min (1, ε / (2 m C))$ . For each face $x_{r} = c$ , where $c = a_{r}$ or $b_{r}$ , cover every other interval by the $N_{j} : = ⌈ (b_{j} - a_{j}) / δ ⌉$ intervals $[a_{j} + k δ, a_{j} + (k + 1) δ]$ , $0 \leq k < N_{j}$ , and use $[c, c + δ]$ in coordinate $r$ . Their products are finitely many closed cubes of side $δ$ covering that face. By [F1], their total volume is $δ^{m} \prod_{j \neq r} N_{j} \leq δ \prod_{j \neq r} (b_{j} - a_{j} + δ) \leq δ C .$ For $m = 1$ this is one interval of length $δ$ per endpoint. Joining the covers of all $2 m$ faces gives total cube volume at most $2 m δ C < ε$ . Hence $\partial Q$ has content zero in the finite-cube sense of [F2], as well as zero Jordan outer content.

3.1F2F4step 1.1step 2.1

The box is bounded and has boundary of content zero by step 2.1, so [F4] makes $Q$ Jordan measurable. For any stated endpoint variant $R$ , its boundary is contained in the same coordinate faces from step 1.1, hence has content zero by step 2.1; since $R$ is bounded, [F4] makes it Jordan measurable as well.

4.1F2F5F6F7step 3.1

Let $f : Q \to R$ be continuous. By [F5], $f$ is Riemann integrable on $Q$ and therefore bounded there. For any endpoint variant $R$ , extend $f ∣_{R}$ by zero to $Q$ . At every point outside $\partial R$ this extension is either $f$ on a neighborhood in $R$ or zero on a neighborhood in $Q ∖ \overline{R}$ ; thus its discontinuities lie in $\partial R$ . The boundary has content zero by step 3.1, hence is null by [F2]. Then [F6] makes this zero extension integrable on $Q$ , and [F7] gives the Riemann integrability of $f ∣_{R}$ over $R$ .

5.1F3step 1.1step 2.1step 3.1step 4.1∎

For $m \geq 2$ , a closed coordinate face of $Q$ is a closed rectangle in $R^{m - 1}$ , so it is compact by [F3]. Applying steps 1.1–4.1 in dimension $m - 1$ proves its Jordan measurability and the stated integrability of every endpoint variant on that face.

Remarks

This lemma proves only finite-dimensional box and face regularity. It asserts no integrability or convergence result for unbounded sets or exhaustions.

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis II, §10.5.1 'Volume and Jordan measurable sets', Proposition 10.5.1