306 CHAPTER 13. THE GENERALIZED RIEMANN INTEGRAL

Now define a gauge function

δ (x)≡{

min(δ 1,δ 3) on [a,c]min(δ 2,δ 3) on [c,b]

Then letting P1 be a δ fine division on [a,c] and P2 be a δ fine division on [c,b] , it followsthat P = P1 ∪P2 is a δ 3 fine division on [a,b] and all the above inequalities hold. Thusnoting that S (P, f ) = S (P1, f )+S (P2, f ) ,∣∣∣∣∫I

f dF−(∫ c

af dF +

∫ b

cf dF

)∣∣∣∣≤ ∣∣∣∣∫If dF− (S (P1, f )+S (P2, f ))

∣∣∣∣+

∣∣∣∣S (P1, f )+S (P2, f )−(∫ c

af dF +

∫ b

cf dF

)∣∣∣∣≤

∣∣∣∣∫If dF−S (P, f )

∣∣∣∣+ ∣∣∣∣S (P1, f )−∫ c

af dF

∣∣∣∣+ ∣∣∣∣S (P2, f )−∫ b

cf dF

∣∣∣∣< ε/3+ ε/3+ ε/3 = ε.

Since ε is arbitrary, the conclusion of the corollary follows.The following lemma, sometimes called Henstock’s lemma is of great significance.

When you have a tagged division of [a,b] denoted as {(Ii, ti)}qi=1 ≡ P with

Ii ⊆ (ti−δ (ti) , ti +δ (ti))

so the division is δ fine, it would make perfect sense to write ∑i∈I f (ti)∆Fi for I ⊆{1,2, ...,q}. If you know that f ∈ R∗ [a,b] and that∣∣∣∣∣ q

∑i=1

f (ti)∆Fi−∫ b

af dF

∣∣∣∣∣≤ ε

whenever P is δ fine, then it follows from Corollary 13.1.13 and induction that∣∣∣∣∣ q

∑i=1

f (ti)∆Fi−∫ b

af dF

∣∣∣∣∣=∣∣∣∣∣ q

∑i=1

f (ti)∆Fi−q

∑i=1

∫Ii

f dF

∣∣∣∣∣≤ ε

Henstock’s lemma says that you can replace the sum over all i ∈ {1,2, ...,q} with the sumover i ∈I ⊆{1,2, ...,q} and preserve the same inequality. That is, you can take the sumover any subset of the tags. The reason such a remarkable result holds is that δ is a function,not a single number. Thus you can change it some places and not others.

Lemma 13.1.14 Suppose that f ∈ R∗ [a,b] and that whenever Q is a δ fine division ofI = [a,b], ∣∣∣∣S (Q, f )−

∫I

f dF∣∣∣∣< ε.

Then if P = {(Ii, ti)}ni=1 is any δ fine division of I, meaning that

Ii ⊆ (ti−δ (ti) , ti +δ (ti)) ,

and P′ ={(

Ii j , ti j

)}rj=1

is a subset of P, then∣∣∣∣∣ r

∑j=1

f(ti j

)∆Fi−

r

∑j=1

∫Ii j

f dF

∣∣∣∣∣≤ ε.