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The Saks–Henstock lemma for fine partial tagged partitions

Statement

If f is Henstock–Kurzweil integrable on [a,b], then for every ε>0 there is a gauge δ such that every δ-fine partial tagged partition {([ui,vi],ξi)}i=1m satisfies

∑i=1m∣f(ξi)(vi−ui)−∫uivif∣<ε.

The assertion includes the empty partial partition.

Fine partial tagged partitions have uniformly small sums of local integration errors.

Facts & Assumptions

Given: An HK-integrable f and a fine partial tagged partition for a sufficiently accurate gauge.

[L1]

Every gauge on each complementary compact interval admits a fine tagged partition (Cousin's lemma: every gauge on a compact interval admits a fine tagged partition).

[L2]

Henstock–Kurzweil integrals restrict to subintervals and add over adjacent intervals (Henstock–Kurzweil integrability on subintervals and additivity over adjacent intervals).

[L3]

HK integrability means that one gauge makes every fine tagged sum lie within a prescribed error of the integral value (The Henstock–Kurzweil integral on a compact interval).

Proof

technique · direct
1.1givenL1L2L3

The empty family has error 0. Otherwise fix a whole-interval gauge whose full-partition error is below ε/4. After a partial partition fine for that fixed gauge is given, [L2] makes f integrable on each of its finitely many complementary compact intervals. For any prescribed complement error, [L3] supplies a local accuracy gauge there; [L1] supplies a partition fine for the minimum of that local gauge and the already fixed whole-interval gauge. Thus the resulting completions are both arbitrarily accurate and fine for the original gauge.

2.1step 1.1L1L2algebra∎

For the cells with nonnegative local error, complete their complement with fine partitions whose total local error is below ε/4. Additivity [L2] identifies the resulting full-partition error with the selected positive errors plus those complement errors, so the positive total is below ε/2. Repeating the construction for the negative cells bounds the absolute value of their total by ε/2; adding the two bounds gives the displayed strict estimate.

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Sources