10.10. ONE DIMENSIONAL LEBESGUE STIELTJES INTEGRAL 295

Proof: It suffices to consider the case that f is nonnegative. Otherwise, one simplyconsiders the positive and negative parts of the real and imaginary parts of the function.Thus f is a bounded function and there is a decreasing sequence of upper step functions,denoted as {un} and an increasing sequence of lower step functions denoted as {ln} suchthat ∫ b

alndt ≤

∫ b

af dt ≤

∫ b

aundt,

∣∣∣∣∫ b

aundt−

∫ b

alndt∣∣∣∣< 2−n

Since f must be bounded, it can be assumed that |un (t)| , |ln (t)|< M for some constant M.Let g(t) = limn→∞ un (t) and h(t) = limn→∞ ln (t) . Then from the dominated convergencetheorem (Why?) one obtains∫ b

af dt = lim

n→∞

∫ b

alndt = lim

n→∞

∫ b

alndm =

∫ b

ahdm

≤∫ b

agdm≤ lim

n→∞

∫ b

aundm = lim

n→∞

∫ b

aundt =

∫ b

af dt

Also, from the construction, h(t)≤ f (t)≤ g(t). From the above,∫ b

a |g(t)−h(t)|dm= 0. Itfollows that g is measurable (why?) and f (t) = g(t) for m a.e. t. (why?) By completenessof the measure, it follows that f is Lebesgue measurable and

∫ ba f dm=

∫ ba gdm =

∫ ba hdm=∫ b

a f dt. (why?) ■If you have seen the Darboux Stieltjes integral, defined like the Riemann integral in

terms of upper and lower sums, the following compares the Lebesgue Stieltjes integralwith this one also. For f a continuous function, how does the Lebesgue Stieltjes integralcompare with the Darboux Stieltjes integral? To answer this question, here is a technicallemma.

Lemma 10.10.3 Let D be a countable subset of R and suppose a,b /∈ D. Also supposef is a continuous function defined on [a,b] . Then there exists a sequence of functions {sn}of the form sn (x)≡ ∑

mnk=1 f

(zn

k−1

)X[zn

k−1,znk)(x) such that each zn

k /∈ D and

sup{|sn (x)− f (x)| : x ∈ [a,b]}< 1/n.

Proof: First note that D contains no intervals. To see this let D = {dk}∞

k=1 . If D has aninterval of length 2ε, let Ik be an interval centered at dk which has length ε/2k. Therefore,the sum of the lengths of these intervals is no more than ∑

∞k=1

ε

2k = ε. Thus D cannot containan interval of length 2ε. Since ε is arbitrary, D cannot contain any interval.

Since f is continuous, it follows from Theorem 3.7.4 on Page 82 that f is uniformlycontinuous. Therefore, there exists δ > 0 such that if |x− y| ≤ 3δ , then | f (x)− f (y)| <1/n. Now let {x0, · · · ,xmn} be a partition of [a,b] such that |xi− xi−1| < δ for each i. Fork = 1,2, · · · ,mn−1, let zn

k /∈ D and∣∣zn

k− xk∣∣< δ . Then∣∣zn

k− znk−1∣∣≤ |zn

k− xk|+ |xk− xk−1|+∣∣xk−1− zn

k−1∣∣< 3δ .

It follows that for each x ∈ [a,b] ,∣∣∣∑mn

k=1 f(zn

k−1

)X[zn

k−1,znk)(x)− f (x)

∣∣∣< 1/n. ■

Proposition 10.10.4 Let f be a continuous function defined on R. Also let F be anincreasing function defined on R. Then whenever c,d are not points of discontinuity of Fand [a,b] ⊇ [c,d] ,

∫ ba f X[c,d]dF =

∫f X[c,d]dµ.Here µ is the Lebesgue Stieltjes measure

defined above.