A brief introduction to
Lebesgue-Stieltjes integral
Shiu-Tang Li
Finished: July 13, 2013
Last updated: November 2, 2013
Definition 1. Let A : [0, a] → R be a RC function of finite variation.
We define an finitely additive set function µ on ([0, a], F ) by µ([0, t]) := At
for any 0 ≤ t ≤ a, where F is the collection of any finite union of sets of
form {0} or (c, d].
Theorem 2. (Caratheodory Extension Theorem) There exists a
unique signed measure µ∗ on ([0, a], B([0, a])) so that µ∗ (E) = µ(E) for all
E ∈ F , where µ is defined in Definition 1.
Proof. Decompose A = A1 − A2 , where A1 , A2 are both RC increasing functions. Write µ = µ1 − µ2 , where µ1 ([0, t]) = A1t and µ2 ([0, t]) = A2t , and it’s
easily seen that µ1 , µ2 are two finitely additive set functions on ([0, a], F ). By
[Theorem 1.2, Varadhan], for any Dn ∈ F , Dn ↓ ∅, we have µ1 (Dn ) ↓ 0 and
µ2 (Dn ) ↓ 0. Therefore, by [Theorem 1.3, Varadhan], there exists two positive
measures (µ1 )∗ and (µ2 )∗ on ([0, a], B([0, a])), so that (µ1 )∗ (E) = µ1 (E) and
(µ2 )∗ (E) = µ2 (E) for all E ∈ F . We then define µ∗ := (µ1 )∗ − (µ2 )∗ , and
the proof of existence is complete.
The uniqueness of µ∗ follows from the monotone class theorem.
Definition 3. The measure µ∗ constructed in Theorem 2 is called the
Lebesgue-Stieltjes measure associated with A.
Lebesgue-Stieltjes integral
Since the Lebesgue-Stieltjes measure
R is a measure defined on ([0, a], B([0, a])),
any f : [0, a] → R is a candidate for [0,a] f (s) dAs to make sense.
1
We first considerR the case that A is increasing on [0, a]. It is a routine job to define [0,a] f (s) dAs for f simple, f bounded Borel measurable, and then f positive Borel measurable. For arbitraryR Borel measurable function f , we decompose it into f + − f − , and define [0,a] f (s) dAs :=
R
R
+
f
(s)
dA
−
f − (s) dAs if at least one of these two integrals is finite.
s
[0,a]
[0,a]
2
1
2
For A (RC) of finite variation on [0, a], we write A = A1 −A
R , where A , A
are both RC increasing functions on [0, a]. The integral [0,a] f (s) dAs :=
R
R
1
−
f
(s)
dA
f (s) dA2s , where we require at least one of these two ins
[0,a]
[0,a]
tegrals to be finite.
Note that we may also define the Lebesgue-Stieltjes measure on ([a, b], B([a, b])).
It follows the same construction procedure.
Decomposition of dAs
By [Section 1.3, Chung], when A is RC increasing on [0, a], we may decompose it into a convex combination of three different increasing functions:
a RC discrete increasing function, a singular continuous increasing function
(not identically zero but with zero derivatives a.e.), and an absolutely continuous increasing function.
This implies dAs can be decomposed into three parts. Usually, it is the
most difficult to compute the Lebesgue-Stieltjes integral when A is singular
continuous. On the other hand, when A is absolutely continuous, we have
the following result:
Theorem 4. Let A be absolutely continuous,
and letR f be a bounded
R
Borel measurable function on [0, a]. Then [0,a] f (s) dAs = [0,a] f (s)A0s ds.
Proof. Monotone class theorem.
Total variation |dAs | of dAs
Let us recall the definition of signed measure, and its decomposition theorems. Our main reference is [Chap 2.10, James Yeh].
Definition 5. [Definition 2.10.1, James Yeh] Given a measurable
space (X, F ). A set function λ on F is called a signed measure on F if it
satisfies the following conditions:
2
(1) λ(E) ∈ (−∞, ∞] for every E ∈ F or λ(E) ∈ [−∞, ∞) for every
E ∈ F.
(2) λ(∅) = 0.
sequence {En : n ∈ N} in F ,
P∞(3) countable additivity: for
P∞every disjoint S
∞
λ(E
)
exists
in
R
and
λ(E
)
=
λ(
n
n
n=1
n=0
n=1 En ).
If λ is a signed measure on F , the triple (X, F , λ) is called a signed measure
space.
Theorem 6. (Hahn Decomposition of Signed Measure Spaces)
[Theorem 2.10.14, James Yeh] For an arbitrary signed measure space
(X, F , λ), a Hahn decomposition exists and is unique up to null sets of λ,
that is, there exist a positive set P and a negative set N for λ such that
P ∩ N = ∅ and P ∪ N = X, and moreover if P 0 and N 0 are another such
pair, then P 4P 0 and N 4N 0 are null sets for λ.
Theorem 7. (Jordan Decomposition of Signed Measures) [Theorem 2.10.21, James Yeh] Given a signed measure space (X, F , λ). A
Jordan decomposition for (X, F , λ) exists and is unique, that is, there exists
a unique pair {µ, ν} of positive measures on (X, F ), at least one of which
is finite, such that µ ⊥ ν and λ = µ − ν. Moreover with an arbitrary Hahn
decomposition {P, N } of (X, F , λ), if we define two set functions µ and ν
on F by setting µ(E) = λ(E ∩ P ) and ν(E) = −λ(E ∩ N ) for all E ∈ F ,
then {µ, ν} is a Jordan decomposition for (X, F , λ).
Definition 8. [Definition 2.10.22, James Yeh] Given a signed measure space (X, F , λ). Let µ and ν be the unique positive measures on F ,
at least one of which is finite, such that µ ⊥ ν and λ = µ − ν . Let us
call µ and ν the positive and negative parts of λ and write λ+ for µ and λ−
for ν. The total variation of X is a positive measure |λ| on F defined by
|λ|(E) = λ+ (E) + λ− (E) for E ∈ F .
Since dA on ([0, a], B([0, a])) is a signed measure, we may find P, N ∈
B([0, a]), P ∪ N = [0, a], P ∩ N = ∅, so that dA = µ − ν, where µ(E) =
dA(E ∩P ) ≥ 0 and ν(E) = −dA(E ∩N ) ≤ 0 for all E ∈ B([0, a]). Therefore,
we may define the total variation |dA| of dA by |dA| := µ + ν.
It’s easily seen that
R
[0,a]
f · (1P − 1N ) dA =
R
[0,a]
f · |dA| for f ∈ B([0, a]).
Next, we show the connection between the variation of signed measures
and the variation of a BV function. (Thanks for the assistance by Wei-Da
3
Chen.)
Theorem 9. Let V : [0, a] → R, V (t) be the variation of A on [0, t].
We also assume that A0 = 0. We claim that dV = |dA|, as a measure on
([0, a], B([0, a])).
Proof. |dA| ≤ dV : we first notice that |dA(s, t]| = |A(t) − A(s)| ≤ V [s, t] =
V (t) − V (s) = dV (s, t] for all intervals (s, t] ⊂ [0, a]. By the monotone
class theorem, |dA(B)| ≤ dV (B) for any B ∈ B([0, a]). Now, let {P, N } be
the Hahn decomposition of ([0, a], B([0, a]), A). For any E ∈ B([0, a]), we
have |dA|(B) = dA(B ∩P )−dA(B ∩N ) ≤ dV (B ∩P )+dV (B ∩N ) = dV (B).
P
dV ≤ |dA|:Psince dV (s, t] = V (t) − V (s) = V [s, t] = sup4 |Ati −
Ati−1 | ≤ sup4 |dA|(ti − ti−1 ] = |dA|(s, t]. The rest of the proof follows
from our old friend, the monotone class theorem.
Properties of Lebesgue-Stieltjes integral
Throughout this section we write
Rt
0
f (s) dAs :=
R
(0,t]
f (s) dAs .
Ra
Theorem 10. (Right continuity) Let 0 |f (s)| |dAs | < ∞, where f ∈
Rt
B([0, a]). Then g(t) := 0 f (s) dAs is RC on (0, a].
Proof. Dominated convergence theorem.
Ra
Theorem 11.
Let 0 |f (s)| |dAs | < ∞, where f ∈ B([0, a]). Then
Rt
g(t) := 0 f (s) dAs is of BV on [0, a].
Proof. g(t) is the difference of two increasing functions on [0, a].
Theorem
12. (Associativity)
Let f, g be as above. Let Rh ∈ B([0, a])
Ra
Ra
a
so
R a that 0 |h(s)| |dgs | < ∞ or 0 |h(s)f (s)| |dAs | < ∞. Then 0 h(s) dgs =
h(s)f (s) dAs .
0
Proof. First, it is easily seen that the identity holds for h(s) = 1(a,b] (s). By
the monotone class theorem, the identity holds for all bounded Borel measurable h’s.
Let {P, N } be the Hahn decomposition of dA. This implies {P 0 , N 0 } is
the Hahn decomposition of dg, where P 0 = ({f ≥ 0} ∩ P ) ∪ ({f < 0} ∩ N ),
N 0 = ({f ≥ 0} ∩ N ) ∪ ({f < 0} ∩ P ). The decomposition of dg follows from
taking h = 1A∩P 0 and h = 1A∩N 0 .
4
Ra
Ra
For arbitrary h ∈ B([0, a]) so that 0 |h(s)| |dgs | < ∞ or 0 |h(s)f (s)| |dAs | <
∞, we write h = h+ − h− = h+ 1P 0 + h+ 1P 0 − h− 1N 0 − h− 1N 0 . We then approximate each component with bounded Borel measurable functions, using
the monotone convergence theorem.
R a Remarks 13. (1) We say f satisfies property (*) if f ∈ B([0, a]) and
|f (s)| |dAs | < ∞. When f is continuous on [0, a], f of BV on [0, a], or
0
f bounded Borel measurable, then f satisfies property (*). (2) If A is of BV
on [0, a], then A− is of BV on [0, a], thanks to the Jordan decomposition for
BV functions.
Theorem 14. (Integration by parts) [Revuz. Yor] LetRA, B be two
t
functions of finite variation. Then for any t > 0, At Bt = A0 B0 + 0 As− dBs +
Rt
Rt
Rt
P
Bs dAs = A0 B0 + 0 As− dBs + 0 Bs− dAs + 0<s≤t (As −As− )·(Bs −Bs− ).
0
Proof. Each term is equal to dA ⊗ dB[0, t]2 .
Theorem 15. [Revuz. Yor] If F is a C 1 -function and A is of finite
variation, then F (A) is of finite variation and
F (At ) = F (A0 ) +
As− ) .
Rt
0
F 0 (As− )dAs +
F (As ) − F (As− ) − F 0 (As− )(As −
P
0<s≤t
Proof. 1. The first assertion
follows from
P
|Ati − Ati−1 | ≤ M · 4 |Ati − Ati−1 |.
P
4
|F (Ati )−F (Ati−1 )| ≤
P
4
|F 0 (ξi )|·
2. It’s easily seen that the identity holds for F ≡ c, and if F1 , F2 both
satisfy the identity, so does F1 + cF2 .
Step 1. I’d like to show the identity holds for F (x) = xn , using induction.
First we perform integration by parts formula on (At )n−1 and At , and we have
Z t
Z t
n−1
n
n
As− d(As )n−1
(At ) =(A0 ) +
(As− )
dAs +
0
0
X
n−1
+
(As − As− ) · ((As )
− (As− )n−1 ).
(1)
0<s≤t
By induction hypothesis, we have
Z t
n−1
n−1
(At )
=(A0 )
+
(n − 1)(As− )n−2 dAs
0
X
n−1
+
(As )
− (As− )n−1 − (n − 1)(As− )n−2 (As − As− ).
0<s≤t
5
(2)
Apply monotone class theorem to (2) above, for all bounded Borel measurable f we have
Z t
Z t
n−1
f (s)(As− )n−2 dAs
f (s) d(As )
= (n − 1)
0
0
X
n−1
+
f (s)(As )
− f (s)(As− )n−1 − (n − 1)f (s)(As− )n−2 (As − As− ). (3)
0<s≤t
Let f (s) = As− in (3) and substitute (3) back to (1), we have
Z t
Z t
n
n
n−1
(At ) =(A0 ) +
(As− )
dAs + (n − 1) (As− )n−1 dAs
0
0
X
n−1
n
+
(As− )(As )
− (As− ) − (n − 1)(As− )n−1 (As − As− )
0<s≤t
+
X
(As − As− ) · ((As )n−1 − (As− )n−1 )
0<s≤t
n
Z
=(A0 ) +
t
n(As− )n−1 dAs +
0
X
(As )n − (As− )n − n(As− )n−1 (As − As− ),
0<s≤t
which proves the claim made in Step 1.
Step 2. Fix K large so that [−K, K] contains the image of [0, t] under
A. By Weierstrass approximation theorem, we may find a sequence of poly0
nomials {pn } so that pn → F 0 uniformly
R x in [−K, K], and pn (−K) = F (−K)
for all n ∈ N. Now we let Pn (x) := −K pn (y) dy + F (−K) for all n ∈ N, and
it follows that Pn → F uniformly in [−K, K].
Step 3. Since A is of BV on [0, t], we may decompose As = Bs − Cs ,
where B, C are increasing functions on [0, t]. We have
X
X
X
|As − As− | ≤
|Bs − Bs− | +
|Cs − Cs− |
0<s≤t
0<s≤t
0<s≤t
≤ Bt − B0 + Ct − C0 < ∞.
Step 4. Let 4As := As − As− .
X
F (As ) − Pn (As ) − F (As− ) + Pn (As− ) − F 0 (As− ) − pn (As− ) · 4As
0<s≤t
≤
X
(n)
0
|F 0 (A(n)
s ) − pn (As )| · |4As | + |F (As− ) − pn (As− )| · |4As |
0<s≤t
→ 0 as n → ∞
Step 5. Approximate F using Pn . The proof is now complete.
6
Corollary 16. Settings as in Theorem 15. Let f be a bounded Borel
measurable function, then we have
Z t
Z t
f (s)F 0 (As− ) dAs
f (s) dF (As ) =
0
0
X
+
f (s)F (As ) − f (s)F (As− ) − f (s)F 0 (As− ) · 4As .
0<s≤t
Before proceeding to the next theorem, we would like to show that when
A is of bounded variation on [0, a], A0− := 0, we may define Yt := Π∞
j=1 (1 +
4Aaj ), where {a1 , a2 , · · · } is some enumeration of {0 ≤ s ≤ t : 4As 6= 0}.
Note that Y0 = 1 + A0 , and we define Y0− := 0.
Here are a few remarks.
(1) We’d like to show the limit of Πnj=1 (1 + 4Aaj ) exists as
n goes to
P
m
n
infinity.
For n > m, |Πj=1 (1 + 4Aaj ) − Πj=1 (1 + 4Aaj )| ≤ e s∈[0,a] |4As | ·
P
n
m+1≤j≤n |4Aaj |, showing that Πj=1 (1 + 4Aaj ) is Cauchy.
(2) The limit is independent of our choice of {aj }. Let {b1 , b2 , · · · } be
another enumeration of {0 ≤ s ≤ t : 4As 6= 0}. For each n ∈ N, we may
find k = k(n) ∈ N so that {b1 , b2 , · · · , bn } ⊂ {a1 , a2 , · · · , ak }. We may find n
large so that |Πnj=1 (1 + 4Abj ) − Πkj=1 (1 + 4Aaj )| is small enough.
By (2), we may write Yt = Π0≤s≤t (1 + 4As ).
(3) Now we extend the definition of Yt = Π0≤s≤t (1 + 4As ) = Πs∈I (1 +
4As ), where I = [0, a], to arbitrary finite subinterval of [0, a] or a single
point in [0, a], with the same approach of construction. The new definition
is denoted by YI . Here are some properties of YI :
(3-1) For I1 , I2 disjoint, I1 ∪ I2 = I, YI = YI1 · YI2 .
P
(3-2) For I1 , I2 disjoint, I1 ∪ I2 = I, |YI − YI1 | ≤ e
The following ones are proved using (3-2):
(3-3) Yt is of BV on [0, a].
(3-4) If In ↑ I, then YIn → YI .
7
s∈I
|4As |
·
P
s∈I2
|4As |.
(3-5) Yt− = Π0≤s<t (1 + 4As ); 4Yt = Π0≤s<t (1 + 4As ) · 4As .
P
(4) We’d like to show Yt = 0≤s≤t 4Ys . Let sn,1 < sn,2 < · · · < sn,n be a
reordering of {a1 , · · · , an }, we have
Yt = lim Πnj=1 (1 + 4Aaj )
n→∞
= lim Πnj=1 (1 + 4Asn,j )
n→∞
n
X
m−1
m
= lim (1 + 4Asn,1 ) +
Πj=1 (1 + 4Asn,j ) − Πj=1 (1 + 4Asn,j )
n→∞
= lim
n→∞
(1 + 4Asn,1 ) +
m=2
n
X
4Asn,m ·
Πm−1
j=1 (1
+ 4Asn,j )
m=2
X
=1 + 4A0 +
4As · Π0≤z<s (1 + 4Az )
0<s≤t
=
X
4Ys .
0≤s≤t
We have used the bounded
convergence theorem in the second last equalP
ity. It is bounded by e s∈[0,a] |4As | , with counting measure 4As .
Theorem 17. [Revuz. Yor] If A is a RC function of finite variation,
Ac
A0 = A0− = 0, then Yt := Y0 Π0≤s≤t
(1
+
4A
s )e t is the only locally bounded
Rt
P
solution of the equation Yt = Y0 + 0 Ys− dAs , where Act := At − 0≤s≤t 4As .
c
Proof. Apply Theorem 14. to Ct := Y0 Π0≤s≤t (1 + 4As ) and Dt := eAt , we
have
Z
Act
t
Z
Acs
t
Y0 Π0≤s≤t (1 + 4As )e = Y0 +
e dCs +
Y0 Π0≤z<s (1 + 4Az ) dDs
0
0
Z t
X
c
= Y0 +
Y0 eAs · Π0≤z<s (1 + 4Az ) d
4As
0
0≤z≤s
t
Z
c
Y0 Π0≤z<s (1 + 4Az )eAs dAcs
+
Z0
= Y0 +
t
c
Y0 eAs · Π0≤z<s (1 + 4Az ) dAs .
0
Assume there are two locally bounded solutions with the same initial
value Y0 , and we denote their difference by Zt . See [Revuz. Yor] for details.
8