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Assuming Countable Choice, the integral function of a Riemann-integrable function is Lipschitz and differentiable almost everywhere, with derivative equal to the integrand there
Statement
Assume the Axiom of Countable Choice. Let , let be Riemann integrable, and let
Then is Lipschitz on . If is the set of discontinuities of , then has measure zero and
Thus the set of points where fails to exist or differs from is contained in a set of measure zero. At and , the displayed derivative is the relative one-sided derivative on .
Facts & Assumptions
Given: Reals , an integrable , its integral function , and Countable Choice.
An integrable is bounded, and if then its integral function satisfies (The integral function of a bounded integrable is Lipschitz, hence uniformly continuous).
Assuming Countable Choice, a bounded function is Riemann integrable if and only if its discontinuity set has measure zero (Lebesgue's criterion for Riemann integrability: a bounded on is Riemann integrable if and only if its set of discontinuities has measure zero).
If is integrable and continuous at , then its integral function is differentiable at and , with relative endpoint derivatives (The first fundamental theorem: if is integrable on and continuous at , then ; in particular a continuous has as a primitive).
Proof
Integrability supplies a bound , so is Lipschitz by [L1].
The discontinuity set has measure zero by the forward implication of [L2], whose stated proof is the sole use of Countable Choice.
Every is a continuity point of , so [L3] gives there, including the relative derivative at an endpoint outside .
Hence every point at which the derivative conclusion fails lies in , and that containing set has measure zero.
Depends on
- The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous
- Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero
- The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$, then $F'(c) = f(c)$; in particular a continuous $f$ has $F$ as a primitive
Used by
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Sources
- J. Lebl, Basic Analysis I & II, Section 5.3 (standard reference, not scraped)