312 CHAPTER 13. THE GENERALIZED RIEMANN INTEGRAL

Suppose∫ 3√

20 f dF exists, the ordinary Stieltjes integral, and let Pn consist of points{

k 3√

2n ,k = 0,1, · · · ,n

}. Thus all division points are irrational. If this integral exists, then

we would need to have a number I such that

|S (Pn, f )− I|< ε

for all n large enough. However, we can choose the tags in such a way that S (Pn, f ) = 0and another way such that S (Pn, f ) = 2. Therefore, this Riemann Stieltjes integral does notexist.

However, consider the following piecewise linear continuous functions. fm is piece-wise linear, equal to 0 on (−∞,1], 1 on [1+1/m,2−1/m] , and then 0 on [2,∞). Then∫ 3√

20 fmdF does exist and it is clear that the integrals are increasing in m. Using the Proce-

dure 9.7.1, it follows that the Riemann Stieltjes integral∫ 3√

20 fmdF = 0. By the monotone

convergence theorem, since limm→∞ fm (x) = X(1,2) (x) , it follows that in terms of the gen-eralized Riemann integral, ∫ 3

√2

0X(1,2) (x)dF

is defined and equals 0.Why do we consider gauges instead of the norm of the partition? The next example will

illustrate this question. Compare with the above in which the Riemann Stieltjes integraldoes not exist because of the freedom to pick tags on either side of the jump. A suitablegauge function can prevent this.

Example 13.3.3 Consider the same example. This time, let

δ (x) ={

min( 1

2 |x−1| , 12 |x−2|

)if x /∈ {1,2}

1/2 if x ∈ {1,2}

This gauge function forces 1,2 to be tags because if you have a tagged interval whichcontains either 1 or 2, then if t is the tag, and if it is not either 1 or 2, then the intervalis not contained in (t−δ (t) , t +δ (t)). Also 1,2 are both interior points of the intervalcontaining them. Consequently if the division is δ fine, then the sum equals 0. From thedefinition,

∫ 3√

20 fmdF = 0.

As pointed out in Theorem 13.1.7, if the Riemann Stieltjes integral exists, then so doesthe generalized Riemann Stieltjes integral and they are the same. Therefore, if you haveF (x) = x, and I is an interval contained in [a,b], then

∫ ba XI (x)dF is the length of the

interval. In general, we have the following result about the indicator function of intervals.

Proposition 13.3.4 Suppose I is an interval and F (x) = x. Then XI∩[a,b] ∈ R∗ [a,b].

Proof: First suppose I = (p,q) . Then consider the sequence of continuous functionswhich are increasing in n and converge pointwise to X(p,q) (x) .

graph of fn

limn→∞ fn(x) = X(p,q)(x)

p qp+ 1n q− 1

n