PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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Riemann area between continuous graphs equals Jordan content

Statement

Let $a < b$ , let $α, β : [a, b] \to R$ be continuous with $α \leq β$ , and set

$K : = {(x, y) : a \leq x \leq b, α (x) \leq y \leq β (x)} .$

Then $K$ is compact and Jordan measurable. Its Jordan content equals its Riemann area between continuous graphs (Riemann area between two continuous graphs and the disc as a vertically simple region):

$cont (K) = \int_{a}^{b} (β (x) - α (x)) d x .$

Facts & Assumptions

Given: Reals $a < b$ and continuous functions $α \leq β$ on $[a, b]$ , with $K$ as in the Statement.

[L1]

For this $K$ , the region-between-graphs theorem states: $K$ is compact and Jordan measurable (A region between two continuous graphs is Jordan measurable, and a continuous integrand extending to its closure integrates by vertical sections).

[L2]

For a continuous $H : K \to R$ , that theorem gives $\int_{K} H = \int_{a}^{b} (\int_{α (x)}^{β (x)} H (x, y) d y) d x$ (A region between two continuous graphs is Jordan measurable, and a continuous integrand extending to its closure integrates by vertical sections).

[L3]

If $E$ is Jordan measurable, then the integral of its indicator over a bounding rectangle equals $cont (E)$ (A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content).

[L4]

The Riemann area between continuous graphs $α \leq β$ on $[a, b]$ is $\int_{a}^{b} (β (x) - α (x)) d x$ (Riemann area between two continuous graphs and the disc as a vertically simple region).

Proof

technique · direct

1.1

Apply [L2] to the constant function $H = 1$ on the compact Jordan set supplied by [L1]; by [L3], the left side is $cont (K)$ , while the right side is $\int_{a}^{b} (\int_{α (x)}^{β (x)} 1 d y) d x$ .

L1L2L3

2.1

The inner integral is $β (x) - α (x)$ , including a zero contribution when the two graphs coincide, so step 1.1 is exactly [L4] and proves the formula.

step 1.1L4∎

Depends on

Used by

Dependency tree · two levels

23 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

W. F. Trench, Introduction to Real Analysis, Theorem 7.2.6 (standard reference, not scraped)
M. E. Taylor, Introduction to Analysis in Several Variables, Theorem 3.1.9 (standard reference, not scraped)