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Finite Jordan covers bound upper integrals, while interior-disjoint Jordan subfamilies bound lower integrals

Statement

Let $n \geq 1$ , let $E \subseteq R^{n}$ be a bounded Jordan set, and let $h : E \to [0, \infty)$ be bounded and Riemann integrable.

If Jordan sets $E_{1}, \dots, E_{N}$ cover $E$ , and each $M_{i} \geq 0$ satisfies $M_{i} \geq \sup_{E \cap E_{i}} h$ whenever that intersection is nonempty, then $\int_{E} h \leq \sum_{i = 1}^{N} M_{i} cont (E_{i}) .$
If Jordan sets $F_{1}, \dots, F_{N}$ lie in $E$ and have pairwise disjoint interiors, and each real $m_{i}$ satisfies $m_{i} \leq \inf_{F_{i}} h$ whenever $F_{i}$ is nonempty, then $\sum_{i = 1}^{N} m_{i} cont (F_{i}) \leq \int_{E} h .$

For an arbitrary bounded integrable real $h$ and a nonempty Jordan set $F \subseteq E$ , $∣ \int_{F} h ∣ \leq \sup_{F} ∣ h ∣ cont (F) .$

Facts & Assumptions

Given: The Jordan sets and bounded integrable function in the statement.

[L1]

On a bounding rectangle, the Riemann integral is linear, monotone, and bounded by the integral of the absolute value (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ); zero extension transfers these properties to Jordan-set integrals.

[L2]

The indicator of a bounded Jordan set is Riemann integrable with integral equal to its content (A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content), and a bounded set is Jordan measurable exactly when its boundary has content zero (A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero).

Proof

technique · direct

1.1

Extend all functions by zero to one bounding rectangle. The indicators in [L2] are integrable, and the nonnegativity of every $M_{i}$ makes the following inequality valid both on and off $E$ . Monotonicity and finite linearity from [L1] give the cover bound, because each indicator integral is the content of its Jordan set: $h \leq \sum_{i = 1}^{N} M_{i} 1_{E_{i}} .$

L1L2given

1.2

The restriction of $h$ to each $F_{i}$ is integrable: away from grid cells meeting $\partial F_{i}$ its Darboux gap is inherited from $h$ , while [L2] makes the total volume of boundary cells arbitrarily small. Hence $\int_{F_{i}} h \geq m_{i} cont (F_{i})$ by [L1]. Pairwise interior-disjoint Jordan sets intersect only on their content-zero boundaries, so the sum of their zero-extended restrictions equals the restriction to their union outside a content-zero set. The same boundary-cell argument and linearity [L1] therefore add these integrals without overcounting; their union lies in $E$ , and $h \geq 0$ , giving the lower bound.

L1L2

2.1

The boundary-cell argument in step 1.2 also makes the restriction of a signed integrable $h$ to $F$ integrable. A nonempty $F$ makes $\sup_{F} ∣ h ∣$ a real number, since $h$ is bounded. On $F$ , the inequalities $- ∣ h ∣ \leq h \leq ∣ h ∣ \leq \sup_{F} ∣ h ∣$ and [L1], together with the indicator identity in [L2], give the last estimate.

L1L2step 1.2∎

Depends on

Used by

Dependency tree · two levels

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Sources

A. Leibman, Multidimensional Real Analysis, Lemma 5.5.1 (standard reference, not scraped)