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A product grid bounds the Darboux sums of the lower and upper section-integral functions

Statement

Let $A \subseteq R^{p}$ and $B \subseteq R^{q}$ be nondegenerate closed rectangles, let $f : A \times B \to R$ be bounded, and let $P$ and $R$ be grids of $A$ and $B$ . With $ℓ_{B}, u_{B} : A \to R$ the lower and upper $B$ -section-integral functions of Sections, lower and upper section integrals, and iterated Riemann integrals on product rectangles and Jordan sets, $L (f, P \times R) \leq L (ℓ_{B}, P) \leq U (ℓ_{B}, P) \leq U (u_{B}, P) \leq U (f, P \times R) .$ The symmetric chain holds after exchanging $A$ and $B$ . No individual section is assumed integrable.

Facts & Assumptions

Given: Rectangles $A, B$ , a bounded $f : A \times B \to R$ , grids $P, R$ , and the section envelopes $ℓ_{B}, u_{B}$ .

[L1]

For a bounded function on a product rectangle, the lower and upper section integrals are defined for every parameter even when the section is not Riemann integrable (Sections, lower and upper section integrals, and iterated Riemann integrals on product rectangles and Jordan sets).

[L2]

A grid cell of a product rectangle is the product of the corresponding cells of the two factor grids, and a sum over cells is the associated finite iterated sum (Grid partitions of a rectangle in $R^{m}$ , their cells, refinements and mesh).

[L3]

Lower and upper Darboux sums are the finite sums of the cell infima and suprema weighted by cell volume (Lower and upper Darboux sums over a grid partition in $R^{m}$ ); finite sums may be regrouped and preserve inequalities term by term (Laws of finite sums and finite products).

Proof

technique · direct

1.1

For a cell $I$ of $P$ and a cell $J$ of $R$ , put $m_{I J} : = \inf_{I \times J} f$ and $M_{I J} : = \sup_{I \times J} f$ . Regrouping the finite sums over the product grid gives $L (f, P \times R) = \sum_{I} vol (I) \sum_{J} m_{I J} vol (J)$ and the analogous formula with $M_{I J}$ for the upper sum.

L2L3given

2.1

If $x \in I$ , then $\inf_{J} f_{x} \geq m_{I J}$ and $\sup_{J} f_{x} \leq M_{I J}$ for every $J$ . Hence $ℓ_{B} (x) \geq \sum_{J} m_{I J} vol (J)$ and $u_{B} (x) \leq \sum_{J} M_{I J} vol (J)$ . Taking the infimum or supremum over $x \in I$ preserves these inequalities.

L1L3step 1.1algebra

3.1

Multiply the cellwise inequalities by $vol (I)$ and sum over $I$ . Together with $ℓ_{B} \leq u_{B}$ , this gives the displayed chain. Exchanging the coordinate blocks gives the symmetric chain.

L1L3step 2.1algebra∎

Depends on

Used by

Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections Theorem

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, §10.2, Exercises 10.2.2-10.2.3 (standard reference, not scraped)
A. Leibman, Multidimensional Real Analysis, §5.4 (standard reference, not scraped)