13.2. MONOTONE CONVERGENCE THEOREM 309

Theorem 13.2.2 Let fn (x) ≥ 0 and suppose fn ∈ R∗ [a,b] ,· · · fn (x) ≤ fn+1 (x) · · ·and that f (x) = limn→∞ fn (x) for all x and that f (x) has real values. Also suppose that{∫ b

a fndF}∞

n=1is a bounded sequence. Then f ∈ R∗ [a,b] and

∫ b

af dF = lim

n→∞

∫ b

afndF (13.4)

Proof: Let ε > 0 be given. Let η be small enough that 3η +(F (b)−F (a))η < ε .

Since{∫ b

a fndF}∞

n=1is increasing and bounded, there exists I = limn→∞

∫ ba fndF . There-

fore, there exists N such that∣∣∣∫ b

a fNdF− I∣∣∣ < η so for all n ≥ N,

∣∣∣∫ ba fndF− I

∣∣∣ < η . By

assumption f (x) = lim j→∞ f j (x). Now define for j ≥ N, Fj ≡{

x :∣∣ f j (x)− f (x)

∣∣< η}.

Since f j (x) is increasing, these Fj are also increasing and since limn→∞ fn (x) = f (x) , theunion of these Fj is all of [a,b]. Let E j+1 ≡ Fj+1 \Fj, and EN ≡ FN so these E j are disjointand they partition [a,b]. On E j, E j,

∣∣ f j (x)− f (x)∣∣< η .

Next is a choice of fineness so that the sum will be close to the integral. Let δ n be suchthat when P = {(Ii, ti)}qn

i=1 is δ n fine, then∣∣∣∣∣ qn

∑i=1

fn (ti)∆Fi−∫ b

afndF

∣∣∣∣∣< η2−(n+1) (13.5)

I will choose a smaller δ in order to use just one instead of one for each n ≥ N. Let δ (x)be defined by

δ (x)≡min(δ N (x) , · · · ,δ j (x)) for x ∈ E j (13.6)

Since the union of the E j is [a,b] , this defines δ (x) > 0 on [a,b], δ (x) being positivebecause it is a minimum of finitely many positive numbers on each E j, j ≥ N. Also forany n ≥ N,δ (x) ≤ δ n (x) on En ∪En+1 ∪ ·· · because δ n is in the list so δ (x) ≤ δ n (x) onEn ∪En+1 ∪ ·· · . If x ∈ E j for some j < n, then δ (x) ≤ δ j (x) so from Henstock’s lemma,Lemma 13.1.15, ∣∣∣∣∣ ∑i∈I j

∫Ii

f jdF− ∑i∈I j

f j (ti)∆Fi

∣∣∣∣∣≤ η2−( j+1) (13.7)

Let P be δ fine. Modify δ to make δ = δ̂ n ≤ δ n on the rest of [a,b] ,other than thefinitely many intervals containing the tags from En ∪ En+1 ∪ ·· · and consider a δ̂ n finedivision of [a,b] which retains the intervals corresponding to the tags which are in En ∪En+1∪ ·· · . Then Henstock’s lemma, Lemma 13.1.15 implies the following for I j those iwhere the tag ti is in E j, j ≥ n.∣∣∣∣∣ ∞

∑j=n

∑i∈I j

fn (ti)∆Fi−∞

∑j=n

∑i∈I j

∫Ii

fndF

∣∣∣∣∣≤ η2−(n+1) ≤ η (13.8)

This is because the double sum ∑∞j=n ∑i∈I j selects intervals which have tags in En∪En+1∪

·· · on which δ ≤ δ n. Also, for δ given as above and N ≤ j ≤ n,δ ≤ δ j on [a,b] from theconstruction. Note that P does not depend on the choice of n≥ N.

I want to estimate∣∣∣S (P, fn)−

∫ ba fndF

∣∣∣ for n≥ N and this choice of a gauge function δ .The idea is to estimate in terms of two sums according to which E j contains the tags. Then