Alphabeta MathHow statement and proof provenance work
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The Henstock–Kurzweil integral has at most one value
Statement
A function on a compact interval has at most one Henstock–Kurzweil integral value.
Facts & Assumptions
Given: Alleged integral values and for the same function on .
Every gauge on a compact interval admits a fine tagged partition (Cousin's lemma: every gauge on a compact interval admits a fine tagged partition).
Proof
Suppose, for contradiction, that ; choose gauges controlling errors below , take their pointwise minimum, and use [L1] to obtain one tagged partition fine for both.
The triangle inequality gives , a contradiction, so .
Depends on
Used by
Dependency tree · two levels
6 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
- Alessandro Fonda, The Kurzweil-Henstock Integral for Undergraduates, Ch. 1 (standard reference, not scraped)
- Andrew Bruckner, Judith Bruckner and Brian Thomson, Real Analysis, Sections 1.2 and 1.21 (standard reference, not scraped)