8.11 The One Dimensional Lebesgue Stieltjes Integral
Let F be an increasing function defined on ℝ. Let μ be the Lebesgue Stieltjes measure
defined in Theorems 7.10.1 and 7.8.3. The conclusions of these theorems are reviewed
here.
Theorem 8.11.1Let F be an increasing function defined on ℝ, an integratorfunction. There exists a function μ : P
(ℝ )
→
[0,∞ ]
which satisfies the followingproperties.
If A ⊆ B, then 0 ≤ μ
(A)
≤ μ
(B)
,μ
(∅)
= 0.
μ
∞
(∪k=1Ai)
≤∑_{i=1}^{∞}μ
(Ai)
μ
([a,b])
= F
(b+)
− F
(a− )
,
μ
((a,b))
= F
(b− )
− F
(a+ )
μ
((a,b])
= F
(b+)
− F
(a+)
μ
([a,b))
= F
(b− )
− F
(a− )
where
F (b+) ≡ lim F (t),F (b− ) ≡ lim F (t).
t→b+ t→a−
There also exists a σ algebra S of measurable sets on which μ is a measure which containsthe open sets and also satisfies the regularity conditions,
μ(E) = sup {μ(K ) : K is compact, K ⊆ E} (8.10)
(8.10)
μ(E ) = inf{μ(V) : V is an open set V ⊇ E } (8.11)
(8.11)
whenever E is a set in S.
The Lebesgue integral taken with respect to this measure, is called the Lebesgue Stieltjes
integral. Note that any real valued continuous function is measurable with respect to S. This
is because if f is continuous, inverse images of open sets are open and open sets are in S.
Thus f is measurable because f^{−1}
((a,b))
∈S. Similarly if f has complex values this
argument applied to its real and imaginary parts yields the conclusion that f is measurable.
This will be denoted here by
∫
f dμ
but it is often the case that it is denoted as
∫
fdF
In the case of most interest, where F
(x)
= x, how does the Lebesgue integral compare
with the Riemann integral? The short answer is that if f is Riemann integrable, then it is also
Lebesgue interable and the two integrals coincide. It is customary to denote the Lebesgue
integral in this context as
∫ b
fdm
a
Theorem 8.11.2Suppose f is Riemann integrable on an interval
[a,b]
. Thenf is also Lebesgue integrableand the two integrals are the same.
Proof:It suffices to consider the case that f is nonnegative. Otherwise, one simply
considers 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 u_{n} and an increasing sequence of lower step functions denoted as l_{n} such
that
∫ ∫ ∫ |∫ ∫ |
bl dt ≤ bfdt ≤ bu dt,|| bu dt− bl dt|| < 2−n
a n a a n || a n a n ||
Since f must be bounded, it can be assumed that
|un (t)|,|ln(t)| < M
for some constant M. Let g
(t)
= lim_{n→∞}u_{n}
(t)
and h
(t)
= lim_{n→∞}l_{n}
(t)
. Then from the
dominated convergence theorem (Why?) one obtains
∫ ∫ ∫ ∫
b b b b
a fdt = nli→m∞ a lndt = nli→m∞ a lndm = a hdm
∫ b ∫ b ∫ b ∫ b
≤ gdm ≤ lim undm = lim undt = fdt
a n→∞ a n→∞ a a
Also, from the construction, h
(t)
≤ f
(t)
≤ g
(t)
. Then from the above
∫ b
|g (t)− h (t)|dm = 0
a
It follows that g is measurable (why?) and f
(t)
= g
(t)
for m a.e. t. (why?) By completeness
of the measure, it follows that f is Lebesgue measurable and ∫_{a}^{b}fdm = ∫_{a}^{b}gdm = ∫_{a}^{b}hdm.
(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 integral with this one
also. For f a continuous function, how does the Lebesgue Stieltjes integral compare
with the Darboux Stieltjes integral? To answer this question, here is a technical
lemma.
Lemma 8.11.3Let D be a countable subset of ℝ and suppose a,b
∕∈
D. Also suppose f isa continuous function defined on
[a,b]
. Then there exists a sequence of functions
{sn}
of theform
m∑n ( n )
sn (x) ≡ f zk−1 X [znk−1,znk)(x)
k=1
such that each z_{k}^{n}
∕∈
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
an interval of length 2ε, let I_{k} be an interval centered at d_{k} which has length ε∕2^{k}. Therefore,
the sum of the lengths of these intervals is no more than
∑∞
ε-= ε.
k=1 2k
Thus D cannot contain an interval of length 2ε. Since ε is arbitrary, D cannot contain any
interval.
Since f is continuous, it follows from Theorem 2.7.4 on Page 72 that f is uniformly
continuous. Therefore, there exists δ > 0 such that if
| |
||m∑n (n ) ||
|| f zk−1 X [znk−1,znk)(x) − f (x)|| < 1∕n.■
k=1
Proposition 8.11.4Let 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 F and
[a,b]
⊇
[c,d]
,
∫ b ∫
f X[c,d]dF = fX [c,d]dμ
a
Here μ is the Lebesgue Stieltjes measure defined above.
Proof:Since F is an increasing function it can have only countably many discontinuities.
The reason for this is that the only kind of discontinuity it can have is where F
(x+ )
> F
(x− )
.
Now since F is increasing, the intervals
(F (x− ),F (x+ ))
for x a point of discontinuity are
disjoint and so since each must contain a rational number and the rational numbers are
countable, and therefore so are these intervals.
Let D denote this countable set of discontinuities of F. Then if l,r
∕∈
D,
[l,r]
⊆
[a,b]
, it
follows quickly from the definition of the Darboux Stieltjes integral that
∫
b
a X[l,r)dF = F (r)− F (l) = F (r− )− F (l− )
∫
= μ ([l,r)) = X [l,r)dμ.
Now let
{s }
n
be the sequence of step functions of Lemma 8.11.3 such that these
step functions converge uniformly to f on
||∫ b( ) || ∫ b 1
|| X [c,d]f − X[c,d]sn dF ||≤ X[c,d]|f − sn|dF < -(F (b)− F (a)).
|a | a n
Also if s_{n} is given by the formula of Lemma 8.11.3,
∫ ∫ ∑mn ∑mn∫
X [c,d]sndμ = f (znk− 1)X[zn ,zn)dμ = f (znk−1)X[zn ,zn)dμ
k=1 k−1 k k=1 k−1 k
∑mn ( n ) ( n n ) ∑mn ( n )( n ( n ))
= f zk− 1 μ [zk−1,zk) = f zk− 1 F (zk− )− F zk−1−
k=1 k=1
mn
= ∑ f (zn )(F (zn) − F (zn ))
k=1 k−1 k k−1
m∑n ∫ b ∫ b
= f (znk−1)X[zn ,zn)dF = sndF.
k=1 a k−1 k a
Therefore,
||∫ ∫ b || |∫ ∫ |
|| X [c,d]fdμ− X [c,d]f dF||≤ || X[c,d]fdμ − X[c,d]sndμ||
| a | | |
| | | |
||∫ ∫ b || ||∫ b ∫ b ||
+ || X[c,d]sndμ− a sndF ||+ ||a sndF − a X[c,d]fdF ||
1- 1-
≤ nμ([c,d])+ n (F (b)− F (a))
and since n is arbitrary, this shows
∫ ∫ b
fdμ − f dF = 0.■
a
In particular, in the special case where F is continuous and f is continuous,
∫ b ∫
fdF = X[a,b]fdμ.
a
Thus, if F
(x)
= x so the Darboux Stieltjes integral is the usual integral from calculus,
∫ ∫
b
a f (t)dt = X[a,b]fdμ
where μ is the measure which comes from F
(x)
= x as described above. This measure is often
denoted by m. Thus when f is continuous
∫ b ∫
f (t)dt = X[a,b]fdm
a
and so there is no problem in writing
∫ b
f (t) dt
a
for either the Lebesgue or the Riemann integral. Furthermore, when f is continuous, you can
compute the Lebesgue integral by using the fundamental theorem of calculus because in this
case, the two integrals are equal.