Suppose f is a continuous function and F is an increasing integrator function. How do you
find ∫_{a}^{b}f

(x)

dF? Is there some sort of easy way to do it which will handle lots of
simple cases? It turns out there is a way. It is based on Lemma 9.1.15. First of
all

F (x+ ) ≡ lim F (y),F (x− ) ≡ lim F (y)
y→x+ y→x−

For an increasing function F, the jump of the function at x equals F

(x+)

− F

(x− )

.

Procedure 9.4.1Suppose f is continuous on

[a,b]

and F is anincreasing functiondefined on

[a,b]

such that there are finitely many intervals determined by the partitiona = x_{0}< x_{1}<

⋅⋅⋅

< x_{n} = b which have the property that on

[xi,xi+1]

, the following functionis differentiable and has a continuous derivative.

(
{ F (x) on (xi,xi+1)
Gi (x) ≡ ( F (xi+) when x = xi
F (xi+1− ) when x = xi+1

Also assume F

(a)

= F

(a+)

,F

(b)

= F

(b− )

. Then

∫ b n−1∫ xj+1 n−1
f (x)dF = ∑ f (x)G ′(x )dx+ ∑ f (xi)(F (xi+ )− F (xi− ))
a j=0 xj j i=1

Here is why this procedure works. Let δ be very small and consider the partition

a = x0 < x1 − δ < x1 < x1 + δ < x2 − δ < x2 < x2 + δ <
⋅⋅⋅x − δ < x < x + δ < x − δ < x = b
n−1 n− 1 n−1 n n

where δ is also small enough that whenever

|x− y|

< δ, it follows

|f (x) − f (y)|

< ε. Then
from the properties of the integral presented above,

∫ x1−δ ∫ x2−δ ∫ b n∑−1
fdF + f dF + ⋅⋅⋅+ fdF + (f (xi)− ε)(F (xi + δ)− F (xi − δ))
a x1+δ xn−1+ δ i=1

∫ b
≤ a fdF ≤

∫ x1−δ ∫ x2−δ ∫ b n∑−1
fdF + f dF + ⋅⋅⋅+ fdF + (f (xi)+ ε)(F (xi + δ)− F (xi − δ))
a x1+δ xn−1+ δ i=1