13.3. COMPUTING GENERALIZED INTEGRALS 313

Then fn ∈ R∗ [a,b] and by the monotone convergence theorem, X(p,q)∩[a,b] ∈ R∗ [a,b].Next suppose I = [p,q]. In any case, [p,q]∩ [a,b] is a closed interval which is con-

tained in [a,b]. There is nothing to show if this intersection is /0 because then X[p,q]∩[a,b]is 0. Consider the complement, [p,q]C ∩ [a,b]. This equals either the intersection of oneor two open intervals with [a,b]. As shown above, each of these open intervals is theincreasing limit of continuous functions. Hence, by the monotone convergence theorem,X

[p,q]C∩[a,b] ∈ R∗ [a,b] and so X[p,q]∩[a,b] = 1− X[p,q]C∩[a,b] ∈ R∗ [a,b] because the sum of

generalized Riemann integrable functions is generalized Riemann integrable. Next supposeI = (p,q]. Then I ∩ [a,b] = (p,b],(p,q], or [a,q] . The first and last case were just consid-ered, the first by expressing the indicator function of the interval as the increasing limit ofcontinuous functions. It is only the second one which maybe is not clear. However, thecomplement of this set is [a, p]∪ (q,b]≡U and X(p,q] = 1−XU where XU ∈ R∗ [a,b] andso, even in this case, X(p,q] ∈ R∗ [a,b]. In case I = [p,q), the situation is exactly similar.

Suppose the following conditions.

1. F is continuous and increasing on (a,b) but may have jumps at a and b.

2. Let F̂ (a)≡ F (a+) , F̂ (b)≡ F (b−)

3. Let f ∈ R([a,b] , F̂

)Is f ∈ R∗ [a,b]? What is

∫ ba f dF?

Let ηn > 0 be such that if ∥P∥< ηn, then∣∣∣∣S( f ,P, F̂)−∫ b

af dF̂

∣∣∣∣< εn, εn→ 0

Now let δ n (x) ≡ min( 1

2 |x−a| , 12 |x−b| ,ηn,

12 |b−a| if x /∈ {a,b}

),δ (a) = δ (b) = εn.

Then let P = {(Ii, ti)}mni=1 be a δ fine division of [a,b]. If a is in some Ii, then the tag

for Ii must be a since if it is t ̸= a, (t−δ (t) , t +δ (t)) cannot contain a. Also, the inter-val containing a is not just a so in fact, a is the left end point of this interval. Similarconsiderations apply to b. Hence P is of the form

a = xn0 < xn

1 · · ·< xnmn = b

and the tags for [a,x1] , [xmn−1,b] are a,b. Thus S ( f ,P,F) is of the form

(F (xn1)−F (a)) f (a)+

mn−1

∑k=2

f (tk)(F (xk)−F (xk−1))+ f (b)(F (b)−F

(xn

mn−1))

Since F = F̂ on (a,b) , we can add in some terms on the top and bottom of the sum and seethat this equals

(F (xn1)−F (a)) f (a)+

mn

∑k=1

f (tk)(F̂ (xk)− F̂ (xk−1)

)+ f (b)

(F (b)−F

(xn

mn−1))

−(

f (b)(F (b−)−F

(xn

mn−1))

+ f (a)(F (xn1)−F (a+))

)