-
CHAPTER 1
Measure Theory and Integration
One fundamental result of measure theory which we will
encounterin this Chapter is that not every set can be measured.
More specifically,if we wish to define the measure, or size of
subsets of the real line insuch a way that the measure of an
interval is the length of the intervalthen, either we find that
this measure does not extend all subsets of thereal line or some
desirable property of the measure must be sacrificed.An example of
a desirable property that might need to be sacrificedis this: the
(extended) measure of the union of two disjoint subsetsmight not be
the sum of the measures. The Banach-Tarski paradoxdramatically
illustrates the difficulty of measuring general sets; it isstated
in Section ???????? without proof. It is with this in mind thatour
discussion of measure and integration begins, in this Section,
bymaking precise what constitutes a measurable set and a
measurablefunction and then an integrable function.
1. Algebras of Sets
Let Ω be a set and P(Ω) denote the power set of Ω which is
theset of all subsets of Ω. The reader is referred to Chapter 0,
Subsection1 if any notations or set theoretic notions are
unfamiliar.
Notation: N0 is the set of nonnegative integers; Q is the set
ofrational numbers; Z is the set of all integers.
Suppose Ω1 and Ω2 are two sets and f : Ω1 → Ω2 is a function
withdomain Ω1 and range in Ω2 (but f need not be onto). For any B ⊆
Ω2,denote by
f−1(B) = {x ∈ Ω1 : f(x) ∈ B}.the inverse image of the set B
under f . Of course f need not beone-to-one (injective) and so f−1
need not exist as a function.
Some of the elementary properties of the inverse image are
worthyof note: If B,B′ ⊆ Ω2 then
f−1(B ∩ B′) = f−1(B) ∩ f−1(B′)f−1(B ∪ B′) = f−1(B) ∪ f−1(B′)
f−1(Bc) = f−1(B)c
25
-
26 1. MEASURE THEORY AND INTEGRATION
The verification is asked for in the Exercises at the end of the
Section.Definition: If A ⊆ Ω then define the characteristic
function χA on
Ω by χA(x) = 1 if x ∈ A and χA(x) = 0 if x ∈ Ω− A.Exercise:
Determine χ−1A (B) for an arbitrary B ⊆ R. There are 4
cases, if ∅ 6= A 6= Ω.Exercise: (De Morgan’s Laws) If {Aα : α ∈
I} is a collection
of subsets of a set Ω then the operation C of complementation
has theproperties.
C (∪α∈IAα) = ∩α∈IC(Aα)C (∩α∈IAα) = ∪α∈IC(Aα)
If I = N is countable then we can write ∪n∈NAn = ∪n∈NBn whereB1
= A1, B2 = A2 ∩ Ac1, . . .Bn = An ∩ Ac1 ∩ Ac2 ∩ . . . ∩ Acn−1. The
Bnare then disjoint.
Definition: Let An, n ∈ N be a countable collection of sets in
Ω.The limit supremum of {An} is
lim supn
An = ∩n∈N ∪k≥n Ak
whereas the limit infimum of {An} islim inf
nAn = ∪n∈N ∩k≥n Ak.
Intuitively lim supn An is all points that are in infinitely
many ofthe An whereas lim infn An is all those points that are in
all exceptpossibly finitely many of the An.
If An is a decreasing sequence of sets, A1 ⊇ A2 ⊇ A3 ⊇ . . .
thenlim sup
nAn = ∩n∈NAn = lim inf
nAn
Definition: A collection F of subsets of Ω is an algebra (field)
ifa. Ω ∈ F .b. If A ∈ F then Ac ∈ F .c. If {Aα : α ∈ I} ⊆ F and if
I is finite then ∪α∈I ∈ F .
Condition c in the definition could equivalently be replaced by
eitherof the following Conditions.
c′. If A,B ∈ F then A ∪B ∈ F .c′′. If A,B ∈ F then A ∩B ∈ F
.
because Condition c′ is equivalent to c; and Condition c′′ in
conjunctionwith Condition b is equivalent to Conditions c′ and b,
as can be seenby observing that C(A ∩ B) = CA ∪ CB.
Definition: A collection F of subsets of Ω is a σ-algebra (or
σ-field)if F is an algebra and
-
1. ALGEBRAS OF SETS 27
d. If {Aα : α ∈ I} ⊆ F and if I is countable, or finite,
then∪α∈IAα ∈ F .
Of course condition d implies condition c in the definition of
analgebra. Conditions b and d together are equivalent to b and d′
where
d′. If {Aα : α ∈ I} ⊆ F and if I is countable, or finite,
then∩α∈IAα ∈ F .
Example 1.1. (i) The power set P(Ω) of a set Ω is a
σ-algebra.
(ii) {∅,Ω} is a σ-algebra.(iii) {∅, A,Ac,Ω} is a σ-algebra, if A
is any subset of Ω.Example (i) is the largest σ-algebra of subsets
of Ω; Example (ii) is
the smallest and Example (iii) is the smallest σ-algebra that
containsthe set A.
Example 1.2. Consider the set of all real intervals of the
form[a, b) along with all intervals of the form [a,∞) and (−∞, b)
where aand b are real numbers. Define F to be the set of all finite
unions ofdisjoint intervals of this form along with the null set
and R itself.
We claim that F is an algebra. Certainly ∅ ∈ F . If A ∈ F then
wemust check that Ac is in F and that is obvious if A = ∅ or A = R
and sowe may suppose A = ∪1≤j≤n[aj, bj) where −∞ ≤ a1 < b1 <
a2 < b2 <a3 < . . . < bn ≤ ∞. Then Ac is of the same
form: indeed if a1 > −∞and bn < ∞ then Ac = (−∞, a1) ∪ [b1,
a2) ∪ . . . ∪ [bn−1, an) ∪ [bn,∞).Therefore Ac ∈ F ; the cases
where a1 = −∞ or b1 = ∞ or both arehandled similarly. Therefore F
is closed under taking complements.
It remains to check that F is closed under finite unions and
itsuffices to show that if A,B ∈ F then A ∪ B ∈ F . If A = ∅ orA =
R then this is obvious and so we suppose A = ∪1≤j≤n[aj, bj) where−∞
≤ a1 < b1 < a2 < b2 < a3 < . . . < bn ≤ ∞ and
further it sufficesto consider the cases that B = [c, d) where c
< d or B = (−∞, d) orB = [c,∞) or B = R. If A ∩ B = ∅ then there
is nothing to showbut if B meets any subinterval [aj, bj) of A or
shares an endpoint thenB ∪ [aj, bj) is again an interval of the
same type, that is half open andhalf closed. (Alternatively one
argues that there are four cases: c ∈ Aor not and d ∈ A or not.)
Therefore A ∪B is in F .
Example 1.3. Sometimes it is more convenient to work with
Fconsisting of all finite disjoint intervals of the form (a, b] (as
opposedto [a, b) above) and (−∞, b] and (a,∞). Again in this case F
is analgebra but not a σ-algebra.
-
28 1. MEASURE THEORY AND INTEGRATION
Proposition 1.4. Let {Fα : α ∈ I} be a non empty collectionof
σ-algebras (resp. algebras) on a given set Ω indexed by a set I.
Thenthe intersection ∩α∈IFα is also a σ-algebra (resp.
algebra).
Proof. We need only check that ∩α∈IFα satisfies the three
prop-erties of σ-algebras The set Ω is certainly in all σ-algebras
and hencein ∩α∈IFα. If A ∈ ∩α∈IFα then Ac is in every Fα. Finally
if (An)n∈Nis a sequence of sets in ∩α∈IFα then ∪n∈NAn is in every
Fα and thiscompletes the proof in the case of σ-algebras and the
proof for algebrasis analogous. �
Suppose now that S ⊆ P(Ω) is any collection of subsets of a
setΩ. Then there is a smallest σ-algebra which contains S which we
shalldenote this σ(S). To see this observe that the set of all
σ-algebras thatcontain S is non empty because it contains P(Ω) and
so the intersectionof all σ-algebras that contain S is a σ-algebra,
by the Proposition andit certainly contains S and so it must be the
minimal such σ-algebra.
Definition: If F is a σ-algebra then a subset S ⊆ F with
theproperty that F = σ(S) is said to be a system of generators of F
.
Consider now the case that Ω = Rn, where n ≥ 1. Similar
consider-ations apply if Ω is any topological space. On Rn ∋ x =
(x1, . . . , xn), wedefine the norm |x| = (∑1≤j≤n |xj|2)1/2 and the
corresponding (usual)topology where a base for the neighborhoods of
a point x ∈ Rn is{Vr(x) : r > 0} where Vr(x) = {y ∈ Rn : |x − y|
< r} is an openball of radius r centered at x. Then a set U is
said to be open in Rn,if, for every x ∈ U there is r > 0 so that
Vr(x) ⊆ U . The set of allopen sets is denoted O and we define the
Borel subsets of Rn to beB = σ(O), that is the smallest σ-algebra
that contains O. We shallsometimes write B(Rn) for B when wishing
to emphasize that it is theBorel σ-algebra on Rn. In general, if Ω
is a topological space and Ois the set of all open subsets of Ω
then the Borel σ-algebra on Ω isσ(O).
Remark: If C denotes the set of all closed subsets of Rn
thenσ(C) = B. For recall that a set is closed if and only if its
complement isopen. Since B contains the open sets and is closed
under the operationof taking complements, it follows that B ⊇ C and
hence B ⊇ σ(C).Conversely the open sets are contained in σ(C): O ⊆
σ(C). ThereforeB = σ(O) ⊆ σ(C) and so B = σ(C). Therefore O is a
system ofgenerators of B by the definition of B and C is a second
system ofgenerators.
Example: The set S of all intervals of the form S = {(−∞, a) : a
∈R} is a system of generators of the Borel σ-algebra B of R: B =
σ(S)Indeed, since S consists solely of open sets it is obvious that
B ⊇ σ(S).
-
1. ALGEBRAS OF SETS 29
It therefore suffices to show that any open set is contained in
σ(S).Certainly if a < b then [a, b[=] − ∞, a[c∩] − ∞, b[ belongs
to σ(S).Therefore it suffices to show that any open set can be
written as a unionof at most countably many intervals of the form
[a, b). We can do this asfollows: For each n, consider all
intervals of the form [k2−n, (k+1)2−n).where k ∈ Z. Given an open
set U ⊆ R, we let An be the union of allsuch intervals that are
entirely contained in U . Then An is an increasingsequence of sets
and U = ∪n≥0An and this expresses U as a countableunion of the
desired type of intervals. It follows that B = σ(S).
Other generating sets for the Borel subsets of R are
(1) S1 = {]−∞, a] : a ∈ R}(2) S2 = {]a,∞[: a ∈ R}(3) S3 =
{[a,∞[: a ∈ R}(4) S4 = {]a, b] : a, b ∈ R, a < b}(5) S5 = {[a,
b[: a, b ∈ R, a < b}Example: For each a = (a1, a2, . . . , an) ∈
Rn let
I(a) = {x ∈ Rn : x1 < a1, x2 < a2, . . . , xn <
an}.Here x ∈ Rn has components x = (x1, x2, . . . , xn) is used.
Further letI0(a, b) = {x ∈ Rn : a1 ≤ x1 < b1, a2 ≤ x2 < b2, .
. . , an ≤ xn < bn}.
whenever a, b ∈ Rn and, presumably a1 < b1, a2 < b2, . . .
an < bnbecause I0(a, b) is void otherwise. Define S = {I(a) : a
∈ Rn} andS0 = {I0(a, b) : a, b ∈ Rn}. Therefore S and S0 are two
collections ofsubsets of Rn and we claim that σ(S) = σ(S0) = B. The
rectangu-lar sets I0(a, b) can be expressed in terms of unions,
intersections andcomplements of the sets I(a) so that σ(S) ⊇ σ(S0).
Conversely any ofthe sets I(a) is the limit of an increasing
sequence of sets I0(bn, a) sothat σ(S) = σ(S0). Also σ(S) ⊆ B since
each I(a) is open. Thereforeit remains only to show that B ⊆ σ(S)
and for this it suffices to showthat an arbitrary open set is in
σ(S0). Proceeding as in the previousexample, it can be shown that
any σ-algebra open set is a countableunion of the rectangular sets
I0(a, b) and so this completes the proof.
There will also be occasion to work in the extended real
numbersR = [−∞,∞], regarded as a two point compactification of the
real line.The open sets in this case are all those sets that are
subsets of R andare open in R as well as the sets ]a,∞] and ]−∞, b]
for arbitrary real aand b or is a union of the three types of sets.
Again we denote the opensets by O and we define the Borel σ-algebra
of R to be B(R) = σ(O).We have A ∈ B(R) if and only if A−{−∞,∞},
regarded as a subset ofR is in B(R), for it is straightforward to
check that the set of such setsis closed under complementation and
countable union. The generating
-
30 1. MEASURE THEORY AND INTEGRATION
sets listed above for B(R) are also generating sets for B(R) if
one adjoinsthe singleton set {∞} (resp. {−∞}).
Monotone Classes We say that a sequence (Ap)p∈N of subsets of
aset Ω is increasing if A1 ⊆ A2 ⊆ A3 ⊆ . . . and decreasing if A1 ⊇
A2 ⊇A3 ⊇ . . . and monotone if it is either increasing or
decreasing. Everymonotone sequence has a limit: A = ∪p∈NAp if
(Ap)p∈N is increasingand A = ∩p∈NAp in the decreasing case. A set M
of subsets of Ω issaid to be a monotone class if, for every
monotone sequence in M , thelimit is also in M .
Every σ-algebra is a monotone class. For recall that an
increasingsequence of sets can be written as the union of disjoint
subsets. Thecomplement of a decreasing sequence forms an increasing
sequence andso a σ-algebra is a monotone class. On the other hand
if F0 is both analgebra and a monotone class then it must also be a
σ-algebra.
If {Mα : α ∈ I} is a nonempty collection of monotone classesthen
∩α∈IMα is also a monotone class. Since the power set P(Ω) is
amonotone class it follows that every subset S ⊆ P(Ω) is contained
ina smallest monotone class which we denote M(S)
Theorem 1.5. (Monotone Class Lemma) If F0 is an alge-bra of
subsets of a set Ω then σ(F0) = M(F0).
Proof: It has already been remarked that a σ-algebra is a
monotoneclass and so it follows that a M(F0) ⊆ σ(F0).
To prove the converse it suffices to show that M(F0) is an
algebra.Introduce the notation
MA = {B ∈ M(F0) : A ∩B,A ∩ Bc and Ac ∩ B are in M(F0)}It is not
difficult to check that MA is a monotone class. Now if wesuppose
that A ∈ F0 then MA ⊇ F0. By minimality MA = M(F0)if A ∈ F0. Next
we observe that, by symmetry that B ∈ MA impliesA ∈ MB. Therefore
if A ∈ F0 and B ∈ MA then A ∈ MB. Thisshows that, whenever B ∈
M(F0) = MA, then we have MB ⊇ F0.But then, by minimality again we
must have MB = M(F0) even forB ∈ M(F0). Therefore for every A,B ∈
M(F0)
A ∩B; Ac ∩ B, and A ∩ Bc are in M(F0).We check now that M(F0) is
an algebra. Certainly ∅ ∈ F0 ⊆ M(F0)and if A ∈ M(F0) then Ac ∈
M(F0) for take B = Ω. Finally ifA,B ∈ M(F0) then Ac, B ∈ M(F0) so
that Ac ∩Bc ∈ M(F0) and wemay take the complement to get A∪B ∈
M(F0). This proves M(F0)is an algebra and as a monotone class it is
therefore a σ-algebra. 2
-
1. ALGEBRAS OF SETS 31
We shall have occasion to use the Monotone Class Lemma
whenverifying uniqueness of measures.
Exercises:
(1) Show that, if B,B′ ⊆ Ω2 thenf−1(B ∩ B′) = f−1(B) ∩
f−1(B′)f−1(B ∪ B′) = f−1(B) ∪ f−1(B′)
f−1(Bc) = f−1(B)c
(2) Show that, in general, lim infn An ⊆ lim supnAn(3) Recall
the definition of the limit supremum and limit infimum
of a sequence of real numbers an.
lim supn∈N
an = inf{sup{ak : k ≥ n} : n ≥ 1} and
lim infn∈N
an = sup{inf{ak : k ≥ n} : n ≥ 1}
where the “value” ∞ (resp. −∞) is possible if the sequenceis
unbounded above (resp. below). (See page two of Ash’sbook Measure,
Integration, and Functional Analysis.) Howis lim supχAn related to
χB where B = lim supAn? Whathappens if we replace lim supχAn by lim
inf χAn?
(4) Suppose that An = {x ∈ R3 : |x− (0, 0, (−1)n/n)| < 1}.
Findlim supn An and lim infn An. Compare this with Exercise 3,page
3 of Ash’s book.
(5) Show that F of Example 1.2 is not a σ-algebra.(6) Suppose
that E ⊆ Ω is an arbitrary subset of a set Ω and S is
a collection of subsets of Ω. Define E ∩ S = {E ∩A : A ∈ S}.Show
thata. If F0 is an algebra of subsets of Ω then E ∩ F0 is also
an
algebra of subsets of E.b. σ(E ∩ S) = E ∩ σ(S). That is the
smallest σ-algebra in
P(E) containing E ∩ S is the intersection of E with thesmallest
σ-algebra in P(Ω) containing S.
c. Conclude further that, if F is a σ-algebra of subsets of
Ω,then E ∩ F is a σ-algebra of subsets of E.
(Reference: Halmos’s Measure Theory)
-
32 1. MEASURE THEORY AND INTEGRATION
2. Measurability
It is an objective of this Chapter to define the integral of a
function.If f is a nonnegative function defined on an interval in
the real linethen the integral should be able to tell us the area
under the graph off just as the Riemann integral does in calculus.
It will tell us muchmore but even in this limited context we shall
discover that f mustbe restricted: the notion of area under the
graph of f will not makesense for every f and we will be forced to
restrict the class of functionsconsidered. We introduce in this
Section the concept of a “measurable”function. We shall see that f
must be measurable if we are to makesense of the notion of area
under the graph. Of course this difficultyarises already in the
case of Riemann integration: the Riemann integralis not defined for
arbitrary functions f . We shall clarify this remarkand discuss the
Riemann integral and its relation to the “Lebesgueintegral,”
introduced here, at the end of this Chapter.
Definition: Let Ω be a set and F be a σ-algebra of subsets of
Ω.The we shall refer to (Ω,F) as a measurable space. The sets in F
willbe referred to as measurable sets.
Definition: Suppose (Ω1,F1) and (Ω2,F2) are two
measurablespaces. Then a mapping f : Ω1 → Ω2 is measurable, with
respectto F1 and F2 if f−1(B) ∈ F1 whenever B ∈ F2. We shall
sometimessay f is measurable function from (Ω1,F1) to (Ω2,F2) to
clarify thechoice of σ-algebras F1 and F2. Unless otherwise
specified a functionf : Ω1 → Rn is said to be measurable or Borel
measurable if f is mea-surable from (Ω1,F1) to (Rn,B).
An elementary example of a measurable function is given by
thecharacteristic function χA of a set A . Then χA is measurable as
amapping from (Ω,F) to (R,B) if and only if A ∈ F , that is if and
onlyif A is a measurable set. Further examples of measurable
functions willbe apparent once some elementary properties have been
established.
Lemma 2.1. Suppose that (Ω1,F1), (Ω2,F2) and (Ω3,F3) arethree
measurable spaces and that f : Ω1 → Ω2 and g : Ω2 → Ω3 are
twomappings. If f and g are both measurable then the composed
functiong ◦ f is measurable from (Ω1,F1) to (Ω3,F3).
Proof: We must show that, for an arbitrary set C ∈ F3, (g
◦f)−1(C) ∈ F1. It suffices to show that
(2.1) (g ◦ f)−1(C) = f−1(g−1(C))
-
2. MEASURABILITY 33
because g−1(C) ∈ F2 because g is measurable and so f−1(g−1(C)) ∈
F1because f is measurable. The verification of (2.1) is a
straightforwardexercise in checking the equality of sets. 2
Proposition 2.2. Suppose that f : Ω1 → Ω2 is a function andF1
and F2 are σ-algebras on Ω1 and Ω2 respectively. Suppose
furtherthat S is a system of generators of F2, which is to say F2 =
σ(S). Thenf is measurable from (Ω1,F1) to (Ω2,F2) if and only if
f−1(B) ∈ F1for every B ∈ S.
Proof : It is obvious that, if f is measurable then f−1(B) ∈ F1
forevery B ∈ S simply because S ⊆ F2. Conversely, suppose f−1(B) ∈
F1for every B ∈ S. Let T = {B ∈ F2 : f−1(B) ∈ F1} so that T ⊇ S.
Weshall show that T is a σ-algebra which implies that T = F2 and
so, bythe definition of T , f is measurable. This will complete the
proof.
We check therefore that T is a σ-algebra. Certainly ∅ ∈ T ,
becausef−1(∅) = ∅ ∈ F1. We check next that if B ∈ T then f−1(B) ∈
F1and so f−1(Bc) = f−1(B)c ∈ F1 because F1 is closed under
takingcomplements. (Recall f−1(Bc) = f−1(B)c by an Exercise.) We
checkfinally that if (Bn)n∈N is a sequence of sets in T so that
f−1(Bn) ∈ Tfor each n ∈ N then
f−1(∪n∈NBn) = ∪n∈Nf−1(Bn) ∈ F1because F1 is a σ-algebra. This
shows that T is closed under countableunions and is therefore a
σ-algebra and the proof is complete. 2
Remark: This set T in the proof constitute “good” sets and
theargument that there are many good sets is an instance of what
Ashcalls the good sets principle in his text page 5.
Remark: The above proof uses a special case of the
followingobservation: If f is a mapping f : Ω1 → Ω2 and if (Bα)α∈I
is a collectionof subsets of Ω2 indexed by a set I then
f−1(∪α∈IBα) = ∪α∈If−1(Bα)f−1(∩α∈IBα) = ∩α∈If−1(Bα).
On the other hand if (Aα)α∈I is a collection of subsets of Ω1
then
f(∪α∈IAα) = ∪α∈If(Aα)but it may happen that f(∩α∈IAα) 6=
∩α∈If(Aα).
Corollary 2.3. A continuous function f : Rm → Rn is
Borelmeasurable.
Of course the understood σ-algebras here are the Borel
σ-algebraB(Rm) and B(Rn).
-
34 1. MEASURE THEORY AND INTEGRATION
Proof: Since the B(Rn) is generated by the open sets, it
sufficesto show that f−1(U) ∈ B(Rm) for an arbitrary open set U .
The resultwill then follow from the Proposition. However f is
continuous meansthat f−1(U) is open whenever U is and so f−1(U) ∈
B(Rm). 2
The same reasoning applies in a more general setting.
Corollary 2.4. If f is a continuous function from one
topolog-ical space (X,Σ) to another (Y, T ) then f is measurable
from (X,B(X))to (Y,B(Y )) where B(X) is the Borel σ-algebra which
is generated bythe open sets Σ of X and similarly B(Y ) is
generated by the open setsT .
Proposition 2.5. Suppose that (Ω,F) is a measurable spaceand f1,
f2, . . . fm are m Borel measurable real valued functions, fj
:(Ω,F) → (R,B), for 1 ≤ j ≤ m. Suppose that g : (Rm,B(Rm))
→(R,B(R)) is measurable. Then
φ(x) = g(f1(x), f2(x), . . . , fm(x))
defines a measurable function φ : (Ω,F) → (R,B(R)).We shall set
aside the proof of the Proposition until later and con-
sider its consequences.
Corollary 2.6. If fj : (Ω,F) → (R,B), j = 1, 2 are
Borelmeasurable functions, then f1 + f2 is also Borel
measurable.
Proof. This is an application of the Proposition with g : R2 →
Rdefined by g(x1, x2) = x1 + x2. Because g is continuous it is
Borelmeasurable. �
Corollary 2.7. If fj : (Ω,F) → (R,B), j = 1, 2 are
Borelmeasurable functions, then the product f1f2 is also Borel
measurable.In particular, if k is a real constant then kf2 is Borel
measurable.
Proof. In this case g(x, y) = xy. The special case follows
bydefining f1(x) = k which makes f1 measurable because it is
continuous.
�
Corollary 2.8. If f : (Ω,F) → (R,B), is a Borel
measurablefunction then |f | is also Borel measurable.
Proof. In this case g(x) = |x|. �Corollary 2.9. If fj : (Ω,F) →
(R,B), j = 1, 2 are Borel
measurable functions, and if f2(x) 6= 0 for all x ∈ Ω then the
ratiof1/f2 is also Borel measurable.
-
2. MEASURABILITY 35
Proof. In this case we define
g(x, y) =
{
x/y if y 6= 00 if y = 0
so that g is defined on all of R2 but of course it is not
continuous.We shall check however that g is measurable, as a
mapping from(R2,B(R2)) to (R,B). It suffices to show that g−1(−∞,
a) ∈ B(R2)for any real a by Proposition 2.2. If a < 0 then
g−1(−∞, a) = {(x, y) ∈R2 : x/y < a} is easily seen to be open in
R2 and hence in B(R2). Ifa ≥ 0 then g−1(−∞, a) = {(x, y) ∈ R2 : x/y
< a} ∪ {(x, y) : y = 0}which is the union of an open and a
closed set and is therefore in B(R2).This proves that g is
measurable and so the above Proposition applieswhich completes the
proof. �
Exercise: Show that if fj : (Ω,F) → (R,B), j = 1, 2 are
Borelmeasurable functions then max{f1, f2} and min{f1, f2} are also
mea-surable. Remark: max{a, b} = (|a − b| + a + b)/2 for any reals
a andb.
It remains to establish the Proposition.Proof of the
Proposition: It suffices to show that the map-
ping, f say, defined by f(x) = (f1(x), f2(x), . . . fm(x)) is
measurableas a mapping from (Ω,F) to (Rm,B(Rm) because the
compositionof measurable functions is measurable. To verify f is
measurable,it suffices to show that, for an arbitrary a = (a1, a2,
. . . , am) ∈ Rm,f−1({x1 < a1, x2 < a2, . . . , xm < am})
∈ F by Proposition 2.2. How-ever
f−1({x1 < a1, x2 < a2, . . . , xm < am}) = ∩1≤j≤mf−1j
(−∞, aj)and as the intersection of m measurable sets, this set is,
itself measur-able. 2
Definition 2.10. A function f : (Ω,F) → R is said to besimple if
there exist finitely many sets Aj ∈ F , 1 ≤ j ≤ m and scalarsλj so
that
f(x) =∑
1≤j≤mλjχAj(x)
The set of all simple functions is denotes S(Ω,F) or S when the
contextis clear.
Observe that a simple function is measurable because it is the
sumof measurable functions. Also a simple function takes on only
finitelymany values, λj, 1 ≤ j ≤ m and possibly 0. Conversely a
measurablefunction that takes on finitely many values is simple. To
see this, sim-ply define, for each λj in the image of f , Aj =
f
−1({λj}). The set S
-
36 1. MEASURE THEORY AND INTEGRATION
is closed under addition, scalar multiplication and
multiplication andtherefore forms a linear algebra over R of
functions. (Recall that a lin-ear algebra is a vector space with a
multiplication operation (f, g) 7→ fgthat is associative and
distributes over addition from the left and rightand, for any
scalar α, α(fg) = (αf)g = f(αg). Reference: Naimark’sNormed
Algebras §7.) It is further worth noting that the
representationf(x) =
∑
1≤j≤m λjχAj(x) in the definition of f ∈ S is not
unique.Proposition 2.11. Let fn, n ∈ N be a sequence of Borel
mea-
surable real valued functions defined on a measurable space
(Ω,F). Sup-pose that
limn∈N
fn(x) = f(x) exists in R for every x ∈ Ω.
Then the function f : (Ω,F) → R, so defined, is measurable.This
result says, briefly, that the pointwise limit of measurable
func-
tions is measurable.Proof: We will show that
(2.2) f−1((−∞, a]) = ∩p∈N ∪n∈N ∩j≥nf−1j ((−∞, a+ 1/p])for every
a ∈ R The set of all sets of the form (−∞, a] generate theBorel
σ-algebra B because any open set is generated. Therefore, wewill
have shown f is measurable by Proposition 2.2.
To verify (2.2), let x ∈ f−1((−∞, a]) so that f(x) ≤ a. Then,for
any p ∈ N there is n so that fj(x) ≤ a + 1/p for all j ≥ n:x ∈ f−1j
((−∞, a + 1/p]) for all j ≥ n or x ∈ ∩j≥nf−1j ((−∞, a + 1/p]).But p
was arbitrary and so x belongs to the right side of (2.2) and
thisshows that f−1((−∞, a]) is a subset or equal to the right side
of (2.2).
Conversely suppose that x belongs to the right hand side of
(2.2).Then, for every p ∈ N, there exists n ∈ N so that fj(x) ≤ a +
1/p forall j ≥ n. Taking limits in this last expression as j → ∞ we
see thatf(x) ≤ a+ 1/p. Since p is arbitrary f(x) ≤ a and this shows
the rightside of (2.2) is in f−1((−∞, a]) which verifies (2.2).
This proves that fis measurable. 2.
The Extended Reals R: We introduce the extended real line R
=[−∞,∞], sometimes referred to as the two point compactification
ofthe real line, and this is just R with two points ±∞ adjoined: R
= R∪{∞}∪{−∞}. This will be a convenience when discussing
convergence.We introduce the following topology on R. The
neighborhoods of ∞(resp. −∞) are those sets which contain an
interval of the form (a,∞]for some a ∈ R (resp. [∞, a)) and the
neighborhoods of x ∈ R are theusual: those sets that contain an
interval of the form {y : |y − x| < δ}
-
2. MEASURABILITY 37
for some δ > 0. Of course the topology that R inherits as a
subset of Ris its usual topology. As a consequence of these
definitions we see thatR is homeomorphic to the compact interval
[−1, 1] with the (usual)topology it inherits as a subset of R.
Indeed
(2.3) Φ(x) =
x√x2+1
if x ∈ R1 if x = ∞−1 if x = −∞
is continuous from R to [−1, 1] with inverse
Φ−1(x) =
x√1−x2 if −1 < x < 1
∞ if x = 1−∞ if x = −1
which is also continuous.
Corollary 2.12. Let fn, n ∈ N be a sequence of Borel mea-surable
extended real valued functions defined on a measurable space(Ω,F).
Suppose that
limn∈N
fn(x) = f(x) exists in R for every x ∈ Ω.
Then the function f : (Ω,F) → R, so defined, is
measurable.Proof. An extended real valued function g is Borel
measurable
if and only if Φ ◦ g is Borel measurable for Φ as in (2.3)
because Φand Φ−1 are continuous and hence measurable. Therefore the
previousProposition 2.11 applied to Φ◦fn implies implies the
present result �
Measurable functions can be written as the pointwise limit of
simplefunctions. We begin by considering nonnegative bounded
functions andlater we will extend to all measurable functions.
Proposition 2.13. Suppose f is be a nonnegative,
bounded,measurable function defined on a measurable space (Ω,F).
Then thereis a sequence (fn)n∈N of simple functions such that
(1) (fn)n∈N is increasing.(2) 0 ≤ fn ≤ f , for every n ∈ N(3)
(fn)n∈N converges to f pointwise and even uniformly.
Proof. Let ǫ > 0 be given. We shall construct a simple
functiongǫ so that gǫ ≤ f and |f(x) − gǫ(x)| < ǫ for all x ∈ Ω.
Since f isbounded we have 0 ≤ f ≤ M for some positive constant M .
Wechoose a0 = 0 < a1 < a2 < . . . < am < am+1] so
that aj+1 − aj < ǫ, for0 ≤ j ≤ m+ 1 and M < am+1 < M + ǫ
We define
Aj = {x ∈ Ω : aj ≤ f(x) < aj+1} = f−1([aj, aj+1))
-
38 1. MEASURE THEORY AND INTEGRATION
for 0 ≤ j ≤ m. Then Aj is measurable and Aj ∩ Ak = ∅ if j 6= k
and∪0≤j≤mAj = Ω. We define
gǫ(x) =∑
0≤j≤majχAj
Then gǫ ≤ f < gǫ + ǫ, in fact, for any x ∈ Ω, x ∈ Aj for some
j and sogǫ(x) = aj and aj ≤ f(x) < aj+1.
Define f1 = g1, f2 = max{f1, g1/2}, . . . fn = max{fn−1, g1/n}
Itfollows that fn ≤ f < fn + 1/n Also that fn is measurable as
themaximum of two simple functions. And (fn)n∈N is increasing and
so fnhas the required properties. �
Next consider the case that f is not necessarily nonnegative but
isstill bounded.
Proposition 2.14. Suppose f is a real valued, bounded,
mea-surable function defined on a measurable space (Ω,F). Then
there is asequence (fn)n∈N of simple functions such that
(1) |fn| ≤ |f |, for every n ∈ N(2) (fn)n∈N converges to f
pointwise and even uniformly.
Remark: The proof is based on the observation that any
realvalued, measurable function f = f+ − f− where f+ = max{f, 0}
=(|f | + f)/2 is nonnegative and measurable and f− = −min{f, 0}
=max{−f, 0} is also nonnegative and measurable.
Proof of the Proposition 2.14 : Let gn be a sequence of
non-negative simple functions convergent to f+ = max{f, 0} as
guaranteedby the preceding Proposition. Similarly let hn be simple
functions con-vergent to f− and define fn = gn − hn. Then |fn| =
|gn − hn| ≤gn + hn ≤ f+ + f− = |f |. Moreover |f − fn| = |f+ − gn −
(f− − hn)| ≤|f+−gn|+ |f−−hn| and since f+−gn and f−−hn go to zero
uniformlyon Ω, fn converges uniformly to f . 2
Finally we consider the general case when f need not be
bounded.
Theorem 2.15. Let f be a nonnegative extended value measur-able
function, f : (Ω,F) → R+. Then there is an increasing sequencefn of
simple functions (measurable on (Ω,F)) so that
(1) 0 ≤ fn ≤ f , for every n ∈ N(2) (fn)n∈N converges to f
pointwise.
Theorem 2.16. Let f be a measurable function, f : (Ω,F) →R. Then
there is a sequence fn of simple functions so that
(1) |fn| ≤ |f |, for every n ∈ N(2) (fn)n∈N converges to f
pointwise.
-
2. MEASURABILITY 39
Remark: It is not true in general that the convergence is
uniformin this more general setting. Indeed, if f is unbounded then
it cannotbe approximated uniformly by bounded functions. For
suppose fn be asequence of functions, uniformly convergent to a
function f and supposeeach fn is bounded (with bound depending on
n). If |f − fn| < 1 then||f | − |fn|| < 1 which says that |f
| is bounded. In the setting of theTheorem above, simple functions
are bounded and so we cannot expectuniform convergence.
Proof of Theorem 2.15: For each p ∈ N, define hp = min{f, p}so
that hp is a bounded, nonnegative measurable function and so itis
possible to choose a simple function gp ≥ 0 so that hp − 1p ≤ gp
≤hp by Proposition 2.13. Then we define f1 = g1, f2 = max{f1, g2},.
. . fp = max{fp−1, gp} so that fp is simple, nonegative and
increasing.We also see by induction on p, fp ≤ hp ≤ f . To check
the pointwiseconvergence, suppose x0 ∈ Ω. If f(x0) < ∞ then we
may choose p0 ∈ Nso that p0 ≥ f(x0). Then for p ≥ p0, f(x0) =
hp(x0) so that f(x0)− 1p =hp(x0)− 1p ≤ gp ≤ fp(x0) ≤ f(x0) and so
f(x0) = limp∈N fp(x0). On theother hand, if f(x0) = ∞ then hp(x0) =
p and gp(x0) ≥ p − (1/p) sothat fp(x0) ≥ gp(x0) ≥ p− (1/p).
Therefore limp fp(x0) = ∞ = f(x0).2
Proof of Theorem 2.16 The proof of Proposition 2.14 applieshere
except the uniform convergence there must be replaced by point-wise
convergence. Let gn be a sequence of nonnegative simple
functionsconvergent to f+ = max{f, 0} as guaranteed by the
preceding Theo-rem. Similarly let hn be simple functions convergent
to f
− and definefn = gn − hn. Then |fn| = |gn − hn| ≤ gn + hn ≤ f+ +
f− = |f |.Moreover, if f(x) ≥ 0 then hn(x) = 0 and
limn∈N
fn(x) = limn∈N
gn(x) = f+(x) = f(x)
Similarly f(x) < 0, then gn(x) = 0 and
limn∈N
fn(x) = limn∈N
−hn(x) = −f−(x) = f(x)
Note that limn gn(x) = ∞ and limn hn(y) = ∞ are both possible,
butonly if x 6= y. 2
The set of simple functions is an algebra as we have seen.
Moreoverif φ : R → R is Borel measurable then φ ◦ f is simple
whenever f is.In particular |f | is simple if f is and max{f, g} =
1
2(f + g + |f − g|) is
simple whenever f and g are. Similarly min{f, g} is simple.
-
40 1. MEASURE THEORY AND INTEGRATION
3. Measures and Premeasures
A measure, or more generally a set function is a mapping from
aσ-algebra into R = [−∞,∞]. Arithmetic on R is defined as
follows.Significantly ∞−∞ is not defined but
∞+ a = ∞ if a 6= −∞−∞+ a = −∞ if a 6= ∞
a · ∞ = ∞ if a > 0
Definition: A measure µ on a measurable space (Ω,F) is a
map-ping of F to R+ = [0,∞] such that
(1) µ(∅) = 0(2) µ(∪α∈IAα) =
∑
α∈I µ(Aα)
whenever {Aα ∈ F : α ∈ I} is a countable collection of sets in F
whichare pairwise disjoint which means that Aα ∩ Aβ = ∅ whenever α
6= β.
A triple (Ω,F , µ) where µ is a measure on the measurable
space(Ω,F) is ameasure space or sometimes measured space to
distinguishit from a measurable space. In the special case that
µ(Ω) < ∞ thenµ is said to be a finite measure and if µ(Ω) = 1
then µ is said to be aprobability measure and (Ω,F , µ) as a
probability space.
Remark: If {xα : α ∈ I} is a set of nonnegative numbers
then∑
α∈I xα = supF∑
α∈F xα where the supremum is taken over all finitesubsets F of I
and the sum could be ∞. It is not necessary for thisdefinition that
I be countable but of course if {xα > 0 : α ∈ I}is uncountable
then the sum is necessarily ∞. (reference: GeneralTopology by
Bourbaki, Chapter IV, §4.3 and §7.1.)
A somewhat more primitive concept than a measure is a
premeasureDefinition: Let F0 be an algebra of subsets of a set Ω. A
premea-
sure µ0 is a mapping of to R+ = [0,∞] such that(1) µ(∅) = 0(2)
µ(∪α∈IAα) =
∑
α∈I µ(Aα)
whenever {Aα ∈ F0 : α ∈ I} is a countable collection of sets in
F0which are pairwise disjoint and provided ∪α∈IAα is in F0.
Of course a measure is a premeasure and a premeasure is a
measureif its domain is a σ-algebra.
Property 2 for measures and premeasures is referred to as
countableadditivity. If a set function µ0 is countably additive on
an algebra F0then
µ0(A) = µ0(A) +∑
α∈Nµ(∅) for any A ∈ F0
-
3. MEASURES AND PREMEASURES 41
Therefore if µ0(A) < ∞ we must have µ0(∅) = 0 that is,
property 2of premeasures implies property 1. Therefore property 1
could equallywell be replaced by µ0(A) < ∞ for some A ∈ F .
A measure or premeasure must be finitely additive (that is I
couldbe taken to be finite in property 2) because, by property
2
µ0(∪α∈IAα) = µ0([∪α∈IAα] ∪ [∪n∈N∅])=
∑
α∈Iµ0(Aα) +
∑
n∈Nµ0(∅) =
∑
α∈Iµ0(Aα)
An elementary consequence of this is that, whenever A ⊆ B and
A,B ∈F then µ0(A) ≤ µ(B), because µ0(B) = µ0(A) + µ0(B − A) ≥
µ0(A)(Of course, R is ordered so that ∞ ≥ a if a ∈ R.) As a
consequence,when we define a measure or premeasure µ0 to be finite
if µ0(Ω) < ∞,we assure that indeed µ0(A) < ∞ for every A.
Example 3.1. Discrete Measures: Let Ω be any set andF = P(Ω)
(the power set of Ω). For every x ∈ Ω assign a nonnegativeweight 0
≤ px ≤ ∞ and define
µ(A) =∑
x∈Apx.
Property 1 of measures is easily checked and property 2 is left
as anexercise to the reader. In the special case that px = 1 for
every x thenµ is called the counting measure on Ω.
Example 3.2. Let Ω be any nonvoid set and F = P(Ω) be thepower
set and let µ(∅) = 0 and µ(A) = ∞ if Ω ⊇ A 6= ∅. Then µ is
ameasure.
More interesting examples will be constructed by starting with
adistribution function which we now define.
Definition: A real valued function F defined on R is a
distributionfunction if
(1) F is increasing which means, whenever x < y, F (x) ≤ F
(y)(2) F is right continuous, which means that limx→a+ F (x) = F
(a),
for every real a .
The archtypal example of a distribution function is F (x) = x
butgiven any distribution function there is a corresponding
“LebesgueStieltjes” premeasure which we now introduce.
Example 3.3. Lebesgue Stieltjes Premeasures Considerthe algebra
F0 of all finite unions of half open, half closed intervals ofthe
form (a, b] or (−∞, b] or (a,∞) for any reals a and b. We saw
in§1.1 that indeed these sets do form an algebra. If F is a
distribution
-
42 1. MEASURE THEORY AND INTEGRATION
function then we define µ0((a, b]) = F (b)−F (a) and extend µ0
by finiteadditivity to F0. Later we shall see that µ0 is indeed a
premeasure, thatis we will check the countable additivity property
2. It will further beshown, in §1.4 that µ0 has an extension
defined on the Borel subsetsof the real line which is a measure.
This extension is a “LebesgueStieltjes” measure and its
construction and properties is an objective ofthis Chapter.
Some properties of measures and premeasures will now be
recorded.
Proposition 3.4. If µ and ν are measures on a measurablespace
(Ω,F) and a ≥ 0 and b ≥ 0 the aµ + bν is also a measure.Similarly
if µ0 and ν0 are premeasures on an algebra of sets F0 thenaµ0 + bν0
is also a measure.
Proof: The proof is left as an exercise.
Theorem 3.5. Suppose that µ0 is a premeasure on an algebraof
subsets F0 of a set Ω. Then
(1) If B ⊆ A and A,B ∈ F then µ0(B) ≤ µ0(A).(2) If {Aα : α ∈ I}
is a countable collection of sets in F0 and if
∪α∈IAα ∈ F0 thenµ0(∪α∈IAα) ≤
∑
α∈Iµ0(Aα)
(3) If A1 ⊆ A2 ⊆ A3 ⊆ . . . is an increasing sequence of sets in
F0and if ∪n∈NAα ∈ F0 then
limn∈N
µ0(An) = µ0(∪n∈NAn)
(4) If B1 ⊇ B2 ⊇ B3 ⊇ . . . is a decreasing sequence of sets in
F0such that ∩n∈NBn is also in F0 and if µ0(Bk) < ∞ for somek ∈ N
then
limn∈N
µ0(Bn) = µ0(∩n∈NBn)
Proof. Part 1 was discussed early in this section. For Part 2
werecall from §1.1 that ∪α∈IAα can be written as the union of
disjointssets Bn ∈ F . Indeed if α : N → I is an enumeration of I
and if withBn = Aα(n) ∩ Acα(n−1) ∩ . . . ∩ Acα(1).
µ0(∪α∈IAα) = µ0(∪n∈NBn) =∑
n∈Nµ0(Bn) ≤
∑
n∈Nµ0(Aα(n))
where the last inequality follows because Bn ⊆ Aα(n). Of course
theorder of summation is irrelevant since the terms are
nonnegative.
-
3. MEASURES AND PREMEASURES 43
Consider Part 3. We observe that, if µ0(Ak) = ∞ for some k,
thenµ0(∪n∈NAn) ≥ µ0(Ak) = ∞ and so there is nothing to check and so
wecan suppose that µ0(Ak) is finite for every k. We have, by the
additivityproperty for measures,
µ0(∪n∈NAn) = µ0(A1) + µ0(A2 − A1) + . . .+ µ0(Ap − Ap−1) + . .
.= lim
p∈Nµ0(A1) + µ0(A2 − A1) + . . .+ µ0(Ap − Ap−1)
= limp∈N
µ0(Ap)
We shall suppose that µ0(B1) < ∞; the general case is very
similar.Let B = ∩n∈NBn. Then B1−Bn is an increasing sequence of
sets whoseunion is B1 −B and so by Part 3
µ0(B1)− µ0(B) = µ0(B1 −B) = µ0(∪n∈N(B1 − Bn)= lim
n∈Nµ0(B1 −Bn)
= µ0(B1)− limn∈N
µ0(Bn)
This implies Part 4 because µ0(B1) < ∞. �
Remark: In Part 4 of the preceding Theorem, the hypothesisµ0(Bk)
< ∞ for some k cannot be dispensed with. For example if µ isthe
counting measure on the rational numbers Q then µ(−1/n, 1/n) =∞ for
every n but µ(∩n∈N(−1/n, 1/n)) = µ({0}) = 1.
The Theorem is of course true for measures in place of
premea-sures in which case F0 would be replaced by a σ-algebra F
say and sothe ∪α∈IAα and ∩n∈NBn are automatically in F and so the
statementsimplifies slightly in this case.
There is a partial converse to parts 3 and 4 of the Theorem that
ishelpful for checking the properties of premeasures.
Proposition 3.6. Suppose that F0 is an algebra of subsets ofa
set Ω and µ0 : F0 → R+ is a mapping so that
(1) µ0(∅) = 0(2) µ(∪α∈IAα) =
∑
α∈I µ(Aα)
whenever {Aα ∈ F0 : α ∈ I} is a finite collection of sets in F0
which arepairwise disjoint. Suppose in addition that either one of
the followingtwo conditions is valid.
a. Whenever A1 ⊆ A2 ⊆ A3 ⊆ . . . is an increasing sequence
ofsets in F0 such that the limit A = ∪α∈IAα is also in F0
thenlimn∈N µ0(An) = µ0(A)
b. Whenever B1 ⊇ B2 ⊇ B3 ⊇ . . . is a decreasing sequence of
setsin F0 such that the limit ∩n∈NBn = ∅ then limn∈N µ0(Bn) = 0
-
44 1. MEASURE THEORY AND INTEGRATION
Then, in either case, µ0 is a premeasure.
Proof. Let C1, C2, C3 . . . be a sequence of pairwise disjoint
setsin F0 and such that C = ∪p∈NCp is also in F0. Define An =
∪1≤p≤nCp.
Assume Condition a. It implies limn∈N µ0(An) = µ0(C) becauseAn ր
C and C ∈ F0. By finite additivity, µ0(An) =
∑
1≤p≤n µ0(Cp) sothat
∑
p∈Nµ0(Cp) = lim
n∈N
∑
1≤p≤nµ0(Cp) = µ0(C)
which verifies µ0 is countably additive and hence a premeasure
underCondition a.
Assume Condition b. We define Bn = C −An so that Bn ց ∅
andtherefore Condition b implies limn µ0(Bn) = 0 We have
µ0(C) = µ0(An) + µ0(Bn) =∑
1≤p≤nµ0(Cp) + µ0(Bn)
Now if we take the limit in n ∈ N and we have µ0(C) =∑
n µ0(Cp)which says µ0 is countably additive. �
We recall from Example 3.3 that each distribution function F
de-fines a set function µ0 so that µ0((a, b]) = F (b)−F (a) which
is a finitelyadditive set function on the set F0 of all finite
disjoint unions of inter-vals of the form (a, b] and (−∞, b]) and
(a,∞). Of course µ0((−∞, b]) =limn∈N = µ0((−n, b]) = limn∈N F (b)−F
(−n), with the value of ∞ pos-sible. In general, if A ∈ F0 then
µ0(A) = limn∈N µ0(A ∩ (−n, n]). Weare now ready to show that µ0 is
a premeasure.
Lemma 3.7. Let F be any distribution function and let µ0 bethe
corresponding finitely additive set function defined on the
algebraF0 as in Example 3.3 (so that, in particular µ0((a, b]) = F
(b)− F (a)).Then µ0 is a premeasure.
Proof: Consider first the finite case where F (x) is constant
out-side (−N,N ] for some N so that µ0 is finite. We shall use Part
b ofthe preceding proposition and so we assume that Bp ∈ F0 forms a
de-creasing sequence of sets and Bp ց ∅. Given ǫ > 0, we will
show thatµ0Bp) < ǫ for all suffiiciently large p and this will
imply the countableadditivity of µ0. We claim that there is another
sequence Cp ∈ F0 withcompact closures Cp ⊆ Bp and such that µ0(Bp −
Cp) < ǫ/2p This ispossible because for each subinterval (a, b]
of Bp we have µ0((a, b]) =F (b) − F (a) = lima′→a+ F (b) − F (a′) =
lima′→a+ µ((a′b]) by the rightcontinuity of F . We may also suppose
that Cp ⊆ (−N,N ] for all p.Since ∩p∈NCp ⊆ ∩p∈NBp = ∅ it follows
that, for some n, ∩1≤p≤nCp = ∅
-
3. MEASURES AND PREMEASURES 45
by a compactness argument. Consequently
µ0(Bn) = µ0(Bn − ∩1≤p≤nCp)= µ0(∪1≤p≤nBn − Cp)≤ µ0(∪1≤p≤nBp − Cp)
≤
∑
1≤p≤nǫ/2p < ǫ
Since (Bp)p∈N is a decreasing sequence and ǫ > 0 was
arbitrary thissays µ0(Bp) ց 0 as p → ∞ and this verifies µ0 is
countably additive,and so a premeasure, in the case F (x) is
constant outside (−N,N ] forsome N .
Consider now the general case. For each N ∈ N let FN(x) = F
(x)if |x| ≤ N and FN(x) = F (−N) if x ≤ −N and FN(x) = F (N) ifx ≥
N . Then FN is a distribution function and the
correspondingfinitely additive set function µN is in fact a
premeasure by the firstpart of the proof and µN(A) = µ0(A) if A ∈
F0 and A ⊆ (−N,N ].Moreover, for any A ∈ F0, limN∈N µN(A) = µ(A) by
the discussionpreceding the statement of this result. Let (Ap)p∈N
be a sequence ofpairwise disjoint sets such that A = ∪p∈NAp is in
F0. Then, simplybecause µ0 is finitely additive we have, for any n
∈ N
µ0(A) ≥ µ0(∪1≤p≤nAp) =∑
1≤p≤nµ0(Ap) so that µ0(A) ≥
∑
p∈Nµ0(Ap)
Conversely, by the special case
µ0(A) = limN∈N
µN(A) = limN∈N
∑
p∈NµN(Ap) ≤ lim
N∈N
∑
p∈Nµ0(Ap) =
∑
p∈Nµ0(Ap)
because µN(B) ≤ µ0(B) for any B ∈ F0. Combining these two
in-equalities we have µ0(A) =
∑
p∈N µ0(Ap) and this verifies the µ0 is apremeasure and completes
the proof. 2
Thus every distribution function corresponds to a premeasure
(andindeed, we shall see, to a measure). The following result
provides apartial converse.
Definition: We shall say that µ is a Lebesgue-Stieltjes measure
ifµ is a measure on the Borel subsets B of R and µ(K) < ∞ if K
is acompact subset of R
Proposition 3.8. Suppose that µ is a Lebesgue-Stieltjes mea-sure
and C is any real constant. Define F (x) = µ((0, x]) + C if x ≥
0and F (x) = C − µ((x, 0]) if x < 0. (Here (0, 0] = ∅ by
convention.)Then F is a distribution function and µ((a, b]) = F
(b)−F (a) whenevera < b.
-
46 1. MEASURE THEORY AND INTEGRATION
Therefore every Lebesgue-Stieltjes measure corresponds to a
distri-bution function which is unique up to an additive
constant.
Proof: Since µ(A) ≤ µ(B) whenever A ⊆ B are sets in B, µ((0,
x])is increasing x ≥ 0 and µ((x, 0]) is decreasing for x ≤ 0 and so
F isincreasing on R.
Suppose a ≥ 0. Thenlim
x→a+F (x) = C + µ(∩x>a(0, x]) = C + µ((0, a]) = F (a)
so that F is right continuous on [0,∞). Next suppose a < 0.
Thenlim
x→a+F (x) = C − µ(∪x>a(x, 0]) = C − µ((a, 0]) = F (a)
so that in fact F is right continuous everwhere and is therefore
a dis-tribution function.
Next check that F (b) − F (a) = µ((a, b]) whenever a < b. If
0 ≤ athen this is obvious and similarly if b < 0. Suppose
therefore that a < 0and b ≥ 0. Then F (b)−F (a) = C+µ((0,
b])−(C−µ((a, 0])) = µ((a, b]).
The uniqueness, up to additive constants, of F with the
propertyF (b) − F (a) = µ((a, b]) follows because if G is another
distributionfunction and if G(0) = F (0) then G(b)−G(0) = µ((a, b])
= F (b)−F (0)implies G(b) = F (b) for b ≥ 0. Similarly G(0) − G(a)
= µ((a, 0]) =F (0) − F (a) so that G(a) = F (a) for a < 0. Thus
G = F . In generalG = F −F (0)+G(0) so that the choice of G is
completely determinedby the choice of G(0). 2
-
4. EXTENSIONS AND MEASURES: 47
4. Extensions and Measures:
In this Section we show that certain premeasures extend to
mea-sures and that a measure may be “completed.” As an application
weare able to construct Lebesgue Stieltjes measures. Recall the
definition:
Definition: A real valued function F defined on R is a
distributionfunction if
(1) F is increasing which means, whenever x < y, F (x) ≤ F
(y)(2) F is right continuous, which means that limx→a+ F (x) = F
(a),
for every real a .
We begin by considering a premeasure µ0 defined on an algebra
F0.Suppose that (Ap)p∈N is an increasing sequence in F0 and Ap ր A
sothat A1 ⊆ A2 ⊆ A3 ⊆ . . . and A = ∪p∈NAn but A may not be itself
inF0. We would like to extend µ0 to µ0 as µ0(A) = limp∈N µ0(Ap).
Thelimit exists in R+ but we should verify that it does not depend
on theparticular sequence and that is the object of the Lemma
below.
Lemma 4.1. Suppose that µ0 is a premeasure defined on an
al-gebra F0 and (Ap)p∈N and (Bp)p∈N are two increasing sequences of
setsin F0 with limits, A = ∪p∈NAp and B = ∪p∈NBp. If A ⊆ B
thenlimp∈N µ(Ap) ≤ limp∈N µ(Bp)
Proof: Because µ0 is a premeasure we have for any q ∈ Nµ0(Aq) =
lim
p∈Nµ0(Aq ∩ Bp) ≤ lim
p∈Nµ0(Bp)
Taking the limit in q gives the result. 2Define therefore G to
consist of all A ⊆ Ω such that there exists an
increasing sequence (Ap)p∈N ∈ F0, so that Ap ր A and define also
aset function µ0 on G by
µ0(A) = limp∈N
µ0(Ap).
The preceding Lemma assures that µ0 is well defined because it
does notdepend on the particular sequence (Ap)p∈N ∈ F0 which
approximatesA. We see further that µ0 extends µ0 because any A ∈ F0
can bewritten as the limit of a constant sequence. An = A.
Lemma 4.2. Suppose that µ0 is a finite premeasure on an
algebraF0 and µ0 is its extension to G as described above. Then G
is closedunder finite intersections and countable unions and
a. If G1, G2 are in G thenµ0(G1 ∪G2) + µ0(G1 ∩G2) = µ0(G1) +
µ0(G2)
-
48 1. MEASURE THEORY AND INTEGRATION
b. If (Gp)p∈N is an increasing sequence in G then G = ∪p∈NGn
isin G and
limp∈N
µ0(Gn) = µ0(G)
c. If Gp is a sequence in G thenµ0(∪p∈NGp) ≤
∑
p∈Nµ0(Gp)
Proof: Suppose that Ap, Bp are increasing sequences of sets inF0
and Ap ր G1 and Bp ր G2 so that G1, G2 ∈ G. It follows thatAp ∩ Bp
ր G1 ∩G2 and Ap ∪ Bp ր G1 ∪G2 so that G is closed underfinite
unions and intersections. Moreover since µ0 is a premeasure wecan
take the limit in p ∈ N in the relation µ0(Ap∪Bp)+µ0(Ap∩Bp) =µ0(Ap)
+ µ0(Bp) to verify Part a.
To verify Part b, suppose that Gp is an increasing sequence in
G,Gp ր G. For each p, suppose Apq is an increasing sequence in
F0convergent to Gp = ∪q∈NApq.
A11 A12 . . . A1q . . . ր G1A21 A22 . . . A2q . . . ր G2...
......
......
Ap1 Ap2 . . . Apq . . . ր Gp...
......
......
Define Bq = ∪1≤p≤qApq. Then Bq is an increasing sequence in F0
andit is convergent to G so that G ∈ G and so µ0(G) = limq∈N
µ0(Bq).Certainly µ0(G) ≥ µ0(Gp) for any p by the preceding Lemma.
On theother hand µ0(Gp) ≥ µ0(Bp) so that limp∈N µ0(Gp) ≥ limp∈N
µ0(Bp) =µ0(G). This establishes Part b.
To verify Part c, it suffices to check that, for any n ∈
Nµ0(∪1≤p≤nGp) ≤
∑
1≤p≤nµ0(Gp)
because we may take the limit in n ∈ N and applying Part b
above. Inthe case n = 2 we have µ0(G1∪G2) ≤ µ0(G1)+µ0(G2) by Part
a. Thecase of general n follows by induction. The proof is
complete. 2.
Thus extension µ0 is a countably additive set function and G ⊇
F0is closed under countable unions, but A ∈ G does not imply Ac ∈
G.Also unions of sets in G may not be expressible as unions of
disjointsets in G.
Definition: A mapping µ∗ on the power set P(Ω) of a set Ω to
R+is said to be an outer measure if
(1) µ∗(∅) = 0
-
4. EXTENSIONS AND MEASURES: 49
(2) If A ⊆ B ⊆ Ω then µ∗(A) ≤ µ∗(B).(3) If Aα∈I is a countable
collection of sets in Ω then
µ∗(∪α∈IAα) ≤∑
α∈Iµ∗(Aα)
Frequently an outer measure will not be a measure because it is
notadditive but it is possible that a restriction to a σ-algebra of
subsets ofΩ may indeed be a measure.
We now suppose that µ0 is a finite premeasure on an algebra
F0and that µ0 is the extension of µ0 to the G of the preceding
Lemma.Later in this Section we weaken the assumption of finite to
“σ-finite”to be defined below. Define
(4.1) µ∗(A) = inf{µ0(G) : G ⊇ A,G ∈ G}for any A ⊆ Ω.
Alternatively, in terms of µ0 itself, we define
λ(A) = inf
{
∑
p∈Nµ0(Ap) : Ap ∈ F0,∪p∈NAp ⊇ A
}
We claim µ∗(A) = λ(A). For suppose that Ap, p ∈ N is a sequence
inF0 and ∪p∈NAp ⊇ A, Define G = ∪p∈NAp so that G ∈ G and µ0(G)
=limn µ0(∪1≤p≤nAp) ≤
∑
p∈N µ0(Ap). Consequently µ∗(A) ≤ λ(A). Con-
versely suppose G ∈ G so that there is a sequence Ap ∈ F0 so
thatAp ր G and we can form a sequence Bp ∈ F0 which is pairwise
dis-joint and ∪p∈NBp = G so that µ0(G) =
∑
p∈N µ0(Bp). This shows thatµ∗(A) ≥ λ(A) and complete the
verification that λ = µ∗.
Now let us check that µ∗ is indeed an outer measure and
recordsome of its properties.
Lemma 4.3. Suppose that µ0 is a finite premeasure on an
algebraF0 of subsets of a set Ω and that µ∗ is defined as indicated
above; (see(4.1)). Then µ∗ is an outer measure and agrees with µ0
on F0 and µ0on G. Moreover
(1) µ∗(A∪B)+µ∗(A∩B) ≤ µ∗(A)+µ∗(B) for any sets A,B ⊆ Ω.In
particular µ∗(A) + µ∗(Ac) ≥ µ0(Ω).
(2) If Ap, p ∈ N is an increasing sequence of sets in Ω
thenlimp∈N µ∗(Ap) = µ∗(∪p∈NAp).
Proof: It is clear that µ∗(G) = µ0(G) if G ∈ G, by the
definition(4.1) of µ∗ and because µ0(G) ≤ µ0(G′) if G ⊆ G′ and G′ ∈
G. Similarlyone sees that, if A ⊆ B then µ∗(A) ⊆ µ∗(B), by the
definition (4.1) ofµ∗.
-
50 1. MEASURE THEORY AND INTEGRATION
Therefore, to complete the check that µ∗ is an outer measure,
wesuppose that Ap, p ∈ N is a sequence of subsets. of Ω. We
chooseGp ∈ G, Gp ⊇ Ap so that µ0(Gp) ≤ µ∗(Ap)+2−pǫ for each p ∈ N.
Then
µ∗(∪p∈NAp) ≤ µ0(∪p∈NGp) ≤∑
p∈Nµ0(Gp) ≤
∑
p∈Nµ∗(Ap) + ǫ/2
p
where we have applied Part c of the preceding Lemma. Therefore
µ∗
is an outer measure.Check next Part 1. We suppose that A,B ⊆ Ω
and ǫ > 0 is given.
Choose G1, G2 ∈ G A ⊆ G1 and B ⊆ G2 and µ0(G1) < µ∗(A) + ǫ
andµ0(G2) < µ
∗(B) + ǫ. Then
µ∗(A)+µ∗(B) > µ0(G1)+µ0(G2)−2ǫ = µ0(G1∪G2)+µ0(G1∩G2)−2ǫby
Part a of the preceding Lemma. Since G1 ∪ G2 ⊇ A ∪ B andG1 ∩G2 ⊇ A
∩B and G1 ∪G2 ∈ G and G1 ∪G2 ∈ G, we haveµ∗(A) + µ∗(B) > µ0(G1)
+ µ0(G2)− 2ǫ ≥ µ∗(A ∪B) + µ∗(A ∩B)− 2ǫand, since ǫ > 0 was
arbitrary this implies Part 1.
It remains to check Part 2. Certainly µ∗(An) ≤ µ∗(∪p∈NAp) for
anyn ∈ N so that limp∈N µ∗(Ap) ≤ µ∗(∪p∈NAp). To check the
converse,suppose that ǫ > 0 is arbitrary, and that, for each p ∈
N, Gp ∈ Gis chosen so that Gp ⊇ Ap and µ0(Gp) ≤ µ∗(Ap) + 2−pǫ. We
shallshow that there is a increasing sequence Hp ∈ G so that Hp ⊇
Apµ0(Hp) ≤ µ∗(Ap) +
∑
1≤q≤p 2−qǫ. The existence of such a sequence
would imply that
µ∗(∪p∈NAp) ≤ µ0(∪p∈NHp) = limp∈N
µ0(Hp) ≤ limp∈N
µ∗(Ap) +∑
1≤q≤p2−qǫ.
so that µ∗(∪p∈NAp) ≤ limp∈N µ∗(Ap) + ǫ and since ǫ is arbitrary
thiswould complete the proof. To construct the sequence Hp we
proceed byinduction on n supposing H1 ⊆ H2 ⊆ . . . ⊆ Hn have been
constructedin G and Hp ⊇ Ap and µ0(Hp) ≤ µ∗(Ap) +
∑
1≤q≤p 2−qǫ for 1 ≤ p ≤ n.
Define Hn+1 = Hn ∪Gn+1 so that Hn+1 ∈ G andµ0(Hn+1) = µ0(Hn
∪Gn+1)
= µ0(Hn) + µ0(Gn+1)− µ0(Hn ∩Gn+1)≤ µ∗(An) +
∑
1≤p≤n2−pǫ+ µ∗(An+1) + ǫ/2
n+1 − µ∗(An)
since An ⊆ Hn ∩ Gn+1. Thus µ0(Hn+1) ≤ µ∗(An+1) +∑
1≤p≤n+1 2−pǫ
and that completes the construction and therefore the proof.
2Thus every finite premeasure µ0 extends to an outer measure µ
∗.One of the less desirable features of µ∗ is that there may
exist sets A ⊆
-
4. EXTENSIONS AND MEASURES: 51
Ω so that µ∗(A) + µ∗(Ac) > µ0(Ω) (but of course 0 ≤ µ∗(A) ≤
µ0(Ω)).However if we omit such sets then we have a measure as the
followingTheorem asserts.
Theorem 4.4. Suppose that µ0 is a finite premeasure definedon an
algebra F0 of subsets of a set Ω and that µ∗ is the outer
measuredefined by (4.1). Define
F = {A ⊆ Ω : µ∗(A) + µ∗(Ac) = µ0(Ω)}Then F is a σ-algebra, F ⊇
F0 and indeed F ⊇ G and the restrictionof µ∗ to F is a measure that
extends µ0 and µ0.
Proof: It is immediately clear that ∅ ∈ F and that A ∈ F
impliesAc ∈ F . Also F0 ⊆ F because µ∗ = µ0 on F0 and µ0 is
finitely additive.
Suppose that A,B ∈ F . By Part 1 of the preceding Lemma
wehave
µ∗(A ∪ B) + µ∗(A ∩ B) ≤ µ∗(A) + µ∗(B)(4.2)µ∗(Ac ∪ Bc) + µ∗(Ac ∩
Bc) ≤ µ∗(Ac) + µ∗(Bc).
We add the two relations and recall that A,B ∈ F .µ∗(A ∪ B) +
µ∗(A ∩ B) + µ∗(Ac ∪ Bc) + µ∗(Ac ∩ Bc) ≤ 2µ∗(Ω)
[µ∗(A ∪ B) + µ∗((A ∪ B)c)] + [µ∗(A ∩ B) + µ∗((A ∩ B)c)] ≤
2µ∗(Ω)by rearranging. But we also know by Part 1 of the preceding
lemmathat µ∗(A ∪ B) + µ∗((A ∪ B)c) ≥ µ∗(Ω) and µ∗(A ∩ B) + µ∗((A
∩B)c) ≥ µ∗(Ω) and so the above bound proves we must have
equalityµ∗(A∪B)+µ∗((A∪B)c) = µ∗(Ω) and µ∗(A∩B)+µ∗((A∩B)c) =
µ∗(Ω)which says that F is closed under union and intersection. Thus
F is analgebra. We can say more because we must have equality in
equation(4.2) and so, in the case A∩B = ∅, we have µ∗(A∪B) =
µ∗(A)+µ∗(B)so that µ∗ is finitely additive on F .
Suppose now that Ap, p ∈ N is an increasing sequence in F ;
wewant to show that A = ∪p∈NAp is also in F . By the preceding
Lemmaµ∗(A) = limp∈N µ∗(Ap) and so µ∗(Ap)+µ∗(Acp) = µ
∗(Ω), for all p impliesµ∗(A)+limp∈N µ∗(Acp) = µ
∗(Ω). But Acp is a decreasing sequence of setsAcp ⊇ Ac and so
this implies µ∗(A)+µ∗(Ac) ≤ µ∗(Ω). But by Part 1 ofthe previous
Lemma µ∗(A) + µ∗(Ac) ≥ µ∗(Ω) and so we have A ∈ F .This proves that
F is a σ-algebra and since limp∈N µ∗(Ap) = µ∗(A), bythe preceding
Lemma, µ∗ is countably additive on F and therefore ameasure when
restricted to F . Finally we recall that G consists of thelimits of
all increasing sequences in F0 and of course these limits arein any
σ-algebra that contains F0 and so F ⊇ G and µ∗ agrees with µ0by the
preceding Lemma. 2
-
52 1. MEASURE THEORY AND INTEGRATION
The next step is to generalize to not necessarily finite
measures.
Definition: A premeasure µ0 on an algebra F0 of subsets of a
setΩ is said to be σ-finite if there is a sequence sets Ap ∈ F0 so
thatµ0(Ap) < ∞ and Ω = ∪p∈NAp.
Thus a finite premeasure is σ-finite and a Lebesgue-Stieltjes
mea-sure is also σ-finite since it is bounded on every bounded
interval.
Theorem 4.5. Carathéodory Extension Theorem: Sup-pose that µ0
is a σ-finite premeasure defined on an algebra F0 of sub-sets of a
set Ω. Then µ0 has an unique extension to a measure on thesmallest
σ-algebra, σ(F0) that contains F0.
Remark: The hypothesis that µ0 be σ-finite is only necessary
toprove the uniqueness of the extension of µ0 and its not required
toprove existence. See Problem 3, page 22 of Ash’s book.
Proof: Let Ep be a sequence of disjoint subsets in F0 of
finiteµ0 measure such that Ω = ∪p∈NEp. Then Ep ∩ F0 is an algebra
ofsubsets of Ep (by an Exercise in §1.1) and µp(A) = µ0(A ∩Ep)
definesa premeasure on Ep∩F0 for each p ∈ N (check this). Since µp
is a finitepremeasure, it extends as to a measure on a σ-algebra
that containsEp ∩F0 and hence σ(Ep ∩F0) = Ep ∩ σ(F0). Denote the
extension byµp as well. Define µ on σ(F0) by
µ(A) =∑
p∈Nµp(A ∩ Ep)
Then µ extends µ0 because if A ∈ F0 thenµ(A) =
∑
p∈Nµ0(A ∩ Ep) = µ0(A)
because µ0 is countably additive. Moreover if Aq is a sequence
of dis-joint subsets in σ(F0) thenµ(∪q∈NAq) =
∑
p∈Nµ0(∪q∈NAq ∩ Ep)
=∑
p∈N
∑
q∈Nµ0(Aq ∩ Ep)
=∑
q∈N
∑
p∈Nµ0(Aq ∩ Ep) =
∑
q∈Nµp(Aq ∩ Ep) =
∑
q∈Nµ(Aq)
because the order of summation does not matter if all terms are
non-negative. Therefore µ is σ-additive and therefore a measure on
σ(F0).
We have shown that µ0 has an extension µ defined on σ(F0) andit
remains to show µ is unique. Suppose therefore that ν is
another
-
4. EXTENSIONS AND MEASURES: 53
measure defined on σ(F0) so that µ0(A) = ν(A) for all A ∈ F0.
Define,for fixed p ∈ N , M = {A ∩ σ(F0) : µ(A∩Ep) = ν(A∩Ep)}. Then
Mcontains F0; moreover it is a monotone class because the limit of
anyincreasing or decreasing sequence of sets in M is also in M. By
theMonotone Class Lemma the smallest monotone class that contains
F0is σ(F0) and so µ(A∩Ep) = ν(A∩Ep) for every A ∈ σ(F0). But p ∈
Nwas arbitrary and so
µ(A) =∑
p
µ(Ep ∩ A) =∑
µ(Ep ∩ A) =∑
p
ν(Ep ∩ A) = ν(A)
if A ∈ σ(F0) by countable additivity and so µ = ν which
verifiesuniqueness. 2.
Corollary 4.6. If F is a distribution function defined on Rthen
there is one and only measure µ on the Borel subsets B(R) so
thatµ((a, b]) = F (b)− F (a) for every half closed interval (a,
b].
Proof: This follows from the Carathéodory Extension Theorem
byway of Lemma 3.7.
Exercise Construct a set which is unbounded (and measurable)but
has finite Lebesgue measure.
Example: Consider the rationals Q and define an algebra F0
asfollows. All intervals of the form (a, b] where a < b are
rationals aswell as intervals of the form (a,∞) and (−∞, b] with a,
b ∈ Q belongto F0 as well as all finite disjoint unions of such
intervals. Finally wesuppose the full space Q and ∅ belong to F0.
It is not difficult to checkthat F0 is closed under finite
intersections and complements and so itis an algebra. Define µ0 on
F0 by µ(A) = ∞ if A 6= ∅ and µ(∅) = 0.Then µ0 is a premeasure. It
is however not σ-finite. It can be shownthat every premeasure
defined on an algebra has an extension to thesmallest σ-algebra
containing the original algebra but, as we shall seethe extension
need not be unique. Indeed in the present case σ(F0)obviously
contains the singleton sets because {b} = ∩n(b− 1/n, b] andso σ(F0)
contains all countable sets and so σ(F0) = P(Q) since Q
iscountable. Define µ1 on P(Q), µ1(A) = ∞ if A 6= ∅ and µ1(∅) =
0.Then µ1 is a measure which extends µ0. Define µ2 to be the
countingmeasure on P(Q). Again µ2 extends µ0 and yet µ2 6= µ1.
Observefurther that µ2 is σ-finite even though it extends a
premeasure that isnot σ-finite.
Proposition 4.7. Assume the hypotheses of the
CarathéodoryExtension Theorem 4.5 so that µ0 is a σ-finite
premeasure defined onan algebra F0 and µ is its extension to the
σ-algebra F = σ(F0). Then,
-
54 1. MEASURE THEORY AND INTEGRATION
for any set A ∈ F such that µ(A) < ∞ and any ǫ > 0 there
is B ∈ F0so that µ(A∆B) < ǫ.
Proof: In the case that µ is finite we recall the construction
ofLemma 4.2 of G ⊇ F0 and of µ0(G) = sup{µ0(Ap) : Ap ր G}.
Recallfurther that if A ∈ F then µ(A) = µ∗(A) = inf{µ0(G) : G ∈ G,
G ⊇A}. It follows that A∆Ap = (A−Ap)∪ (Ap−A) ⊆ (G−Ap)∪ (G−A)has
arbitrarily small measure provided G ∈ G is chosen appropriatelyand
then Ap ∈ F0 approximates G closely enough. This proves theresult
in the case µ is finite.
In the general case let Ep be a sequence of disjoint subsets in
F0of finite µ0 measure such that Ω = ∪p∈NEp. Define µp by µp(A)
=µ(A∩Ep) so that µp is a finite measure on Ep. By the first part of
theproof, if A ∈ F , and µ(A) < ∞ and ǫ > 0, then there exist
sets Ap ⊆ Epwith Ap ∈ F0 with µp((A ∩ Ep)∆Ap) < ǫ2−p. Consider B
= ∪p∈NAp.Then
µ(A∆B) =∑
p∈Nµ((A∆B) ∩ Ep) =
∑
p∈Nµ((A ∩ Ep)∆Ap) ∩ Ep) < ǫ
Here we have used the observations that B∩Ep = Ap and that
(C∆D)∩E = (C ∩ E)∆(D ∩ E), for any sets C,D and EOf course B may
not itself be in F0 but it can be approximated byBN = ∪1≤p≤NAp
which is in F0 and BN ր B We havelim
N→∞µ(A∆BN) = lim
N→∞µ((A−BN) ∪ (BN − A))
= limN→∞
µ(A− BN) + µ(BN − A) = µ(A∆B) < ǫ.
by Theorem 3.5 and because µ(A) < ∞. Therefore for N large
enoughwe have µ(A∆BN) < 2ǫ say. Since ǫ > 0 is arbitray the
proof iscomplete. 2
Exercise: Show that, for any Borel subset A of the real line
withfinite Lebesgue measure and for any ǫ > 0, there is a
compact (that isclosed and bounded) set K so that the Lebesgue
measure of A∆K isat most ǫ.
Definition 4.8. A measure space (Ω,F , µ) is said to be
com-plete (or µ is said to be complete) if, whenever A ∈ F has
measureµ(A) = 0, then every subset B ⊆ A is also in F .
Therefore µ is complete if every subset of a set of measure zero
isa set of measure zero.
Suppose now that (Ω,F , µ) is any measure space. DefineN = {N ⊆
Ω : there exists M ∈ F ,M ⊇ N and µ(M) = 0}.
-
4. EXTENSIONS AND MEASURES: 55
Define further Fµ to consist of all sets of the form A∪N where A
∈ Fand N ∈ N .
Exercise Show that
F = {A∆N : A ∈ F and N ∈ N}is equal to Fµ. Suggestion: Begin by
showing that, for N ⊆ M ,
A ∪N = (A−M)∆(M ∩ (A ∪N)A∆N = (A−M) ∪ [M ∩ (A∆N)]
Theorem 4.9. Suppose that (Ω,F , µ) is a measure space andFµ is
defined as above. Define µ on Fµ by µ(A ∪ N) = µ(A), for allA ∪ N ∈
Fµ so that where A ∈ F and N ∈ N . Then Fµ is a σ-algebra, µ is
well defined and a measure that extends µ and (Ω,Fµ, µ)is a
complete measure space.
Proof: It is clear that Fµ contains ∅. Moreover if Ap ∪ Np is
asequence in Fµ (so that AP ∈ F and Np ∈ N ) then ∪p∈NAp ∪ Np
=(∪p∈NAp) ∪ (∪p∈NNp) is in Fµ also. Suppose next that A ∪ N ∈
Fµwith A ∈ F and N ∈ N and suppose that N ⊆ M where M ∈ F andµ(M) =
0. To show that Fµ contains the complement of A ∪ N wenote that (A
∪ N)c = Ac ∩ N c = (Ac ∩ M c) ∪ Ac ∩ (N c − M c). NowN c −M c = N c
∩M ⊆ M and so Ac ∩ (N c −M c) ∈ N and so (A∪N)cis in Fµ and we have
Fµ is a σ-algebra.
Check next that µ is well defined. Suppose therefore that A1∪N1
=A2 ∪ N2 where A1, A2 ∈ F and N1, N2 ∈ N . Then A1 = (A1 ∩ A2) ∪(A1
− A2). and µ(A1 − A2) = 0 because A1 − A2 ⊆ (A1 ∪N1)− A2 =(A2 ∪ N2)
− A2 ⊆ N2. Therefore µ(A1) = µ(A1 ∩ A2); and by asymmetric argument
µ(A2) = µ(A1 ∩ A2). This shows µ(A1 ∪ N1) =µ(A1) = µ(A2) = µ(A2 ∪
N2) so that µ is well defined. Obviously µextends µ
Check next that µ is countably additive. Suppose that Ap∪Np is
asequence of pairwise disjoint sets in Fµ (so that AP ∈ F and Np ∈
N )then
µ(∪p∈NAp ∪Np) = µ ((∪p∈NAp) ∪ (∪p∈NNp))= µ(∪p∈NAp) =
∑
p∈Nµ(Ap) =
∑
p∈Nµ(Ap ∪Np)
because Ap forms a sequence of pairwise disjoint sets in F .
Thereforeµ is a measure. To check that it is complete, suppose that
A∪N ∈ Fµwhere N ∈ N and A ∈ F and µ(A∪N) = µ(A) = 0. Then A∪N ∈
Nand so any subset of A ∪ N is also in N and so has zero µ
measure.Thus µ is complete. 2
-
56 1. MEASURE THEORY AND INTEGRATION
We have constructed Lebesgue-Stieltjes measure on the Borel
setsB(R) corresponding to a distribution function F and of course
Lebesguemeasure corresponds to the case F (x) = x. Each of these
measures hasa (unique) completion; and usually “Lebesgue Stieltjes
measure” refersto this completion.
Historical Note: Constantin Carathéodory (September 13,1873 to
February 2, 1950) was born in Berlin of Greek parents and grewup in
Brussels. He is known for his work in the calculus of
variations,conformal representations and the foundations of
thermodynamics be-sides measure theory; he was also an accomplished
linguist, speakingsix modern and several ancient languages. He
joined the faculty of theill-fated Greek University of Smyrna (in
present day Izmir, Turkey) in1920 and stayed until the Greek
expulsion in 1922. He was Professorof Mathematics at the University
of Munich 1924-1950.
-
5. THE INTEGRAL 57
5. The Integral
In this section we introduce the integral and derive some of
itselementary properties.
Suppose that (Ω,F , µ) is a measure space and S = S(Ω,F) is
theset of all simple functions; see Definition 2.10. Define S0 to
be thesubset of S which consists of all those f ∈ S so that µ({x ∈
Ω : f(x) 6=0}) < ∞. (Observe that {x ∈ Ω : f(x) 6= 0} is
measurable because fis.) Consequently f =
∑
1≤j≤n λjχAj for some scalars λj and Aj ∈ Fwith µ(Aj) < ∞. We
define the integral of f with respect to µ as
∫
f dµ =∑
1≤j≤nλjµ(Aj)
Since the representation of f is not unique it is necessary to
check thatthe integral is well defined.
Suppose therefore that f ∈ S0 and so f takes on finitely many
nonzero distinct values {fi : 1 ≤ i ≤ n}. If Ai = {x ∈ Ω : f(x) =
fi}then f =
∑
1≤i≤n fiχAi . We want to show that if there is any
otherrepresentation f =
∑
1≤j≤m λjχBj then
∑
1≤i≤nfiµ(Ai) =
∑
1≤j≤mλjµ(Bj)
It is convenient to define f0 = 0 and A0 = (∪1≤j≤mBj) −
(∪1≤i≤nAi).Then, for arbitrary i, 0 ≤ i ≤ n
fiχAi =∑
1≤j≤mλjχBj∩Ai
Since the Ai are disjoint, if one shows that fiµ(Ai) =∑
1≤j≤m λjµ(Bj∩Ai) then that suffices (for we can sum over i.) We
also note that∪1≤j≤mBj ∩ Ai = Ai. For i > 0 this is because fi
6= 0 and for i = 0it is by the choice of A0. We now subdivide Ai
into disjoint subsets asfollows. For each subset I of {1 ≤ j ≤ m}
let
CI = [(∩j∈IBj)− (∪j /∈IBj)] ∩ Ai =[
(∩j∈IBj) ∩(
∩j /∈IBcj)]
∩ AiIf J is another such subset of {1 ≤ j ≤ m} then CJ is
disjoint fromCI . Some of the CI may be empty but their union (over
all 2
m − 1choices of I which are nonempty) is Ai. Suppose now that x
∈ CI .Then we must have fi =
∑
j∈I λj. But we also have Ai = ∪ICI wherethe sum is over all 2m −
1 nonvoid subsets I of {1 ≤ j ≤ m}. Take theintersection with Bj to
get Bj ∩ Ai = ∪IBj ∩ CI = ∩I∋jCI where now
-
58 1. MEASURE THEORY AND INTEGRATION
the intersection is over all subsets I which contain j.
Therefore∑
1≤j≤mλjµ(Bj ∩ Ai) =
∑
1≤j≤m
∑
I∋jλjµ(CI)
=∑
I
∑
j∈Iλjµ(CI) =
∑
I
fiµ(CI) = fiµ(Ai)
where the interchange of summation is justified by checking that
indexset {(j, I) : 1 ≤ j ≤ m, I ∋ j} is the same set as {(j, I) : I
6= ∅, j ∈ I}.
Theorem 5.1. Suppose f, g ∈ S0(Ω,F , µ) and λ is a real.
Thena.∫
Ωλf dµ = λ
∫
Ωf dµ;
b.∫
Ωf + g dµ =
∫
Ωf dµ+
∫
Ωg dµ;
c.∫
Ωf dµ ≥ 0 if f ≥ 0;
d.∫
Ωf dµ ≤
∫
Ωg dµ if f ≤ g;
e. |∫
Ωf dµ| ≤
∫
Ω|f | dµ.
Proof. The proof of Parts a and c are left as elementary
exercises.For Part b we have f =
∑
1≤i≤n fiχAi and g =∑
1≤j≤m gjχBj . Thenf + g =
∑
1≤k≤m+n hkχCk where hk = fk and Ck = Ak, if 1 ≤ k ≤ nand hk =
gk−n and Ck = Bk−n, if n < k ≤ n+m. Therefore∫
Ω
f + g dµ =∑
1≤k≤n+mhkµ(Ck)
=∑
1≤k≤nfkµ(Ak) +
∑
n ǫ}) = 0.
-
5. THE INTEGRAL 59
If (Ω,F , µ) is a probability space and A = Ω then fp is said to
convergein probability.
Proposition 5.2. Suppose that fp is a sequence of real
valued,measurable functions defined on a measure space (Ω,F , µ)
and conver-gent pointwise to f . Then f is measurable and, for
every A ∈ F offinite measure, fp converges in measure to f on
A.
Proof. We have already seen in Proposition 2.11 that the
point-wise limit of a sequence of real valued, measurable functions
is mea-surable. We may suppose that f = 0 because otherwise we
replace fpby fp − f . For ǫ > 0, we claim that
∪m∈N ∩p≥m {x ∈ A : |fp(x)| ≤ ǫ} = AIn fact, for any x ∈ A, limp
fp(x) = 0 implies that there is some m forwhich |fp(x)| ≤ ǫ for
every p ≥ m, that is there is m so that x ∈ Amwhere Am = ∩p≥m{x ∈ A
: |fp(x)| ≤ ǫ}. Since x ∈ A was arbitrary thisverifies the claim.
Therefore A−Am is a decreasing sequence of sets offinite measure
converging to the empty set and so limm∈N µ(A−Am) = 0by Theorem
3.5. But
A− Am = {x ∈ A : |fp(x)| > ǫ, for some p ≥ m}and the result
follows. �
Thus pointwise convergence implies convergence in measure on
setsof finite measure.
Exercise: Construct a sequence of measurable functions on a
prob-ability space which converges to 0 in probability but does not
convergepointwise.
Theorem 5.3. Let fp be a decreasing sequence of
nonnegativefunctions in S0(Ω,F , µ) which converges pointwise to 0.
Then
limp∈N
∫
Ω
fp dµ = 0.
This result will allow us to show later that the integral can
beextended to a wider class of functions.
Proof. We have f1 ≥ f2 ≥ . . . ≥ fp ≥ . . . ≥ 0. Since f1 is
finitevalued, there is a constant M > 0 so that f1 ≤ M and hence
fp ≤ Mfor all p. Also, the set A = {x : f1(x) > 0} has finite
measure andfp(x) = 0, for every p ∈ N if x ∈ Ac. Suppose the ǫ >
0. Define
Ap = {x : fp(x) > ǫ}Then Ap is a decreasing sequence of sets
of finite measure; indeed Ap ⊆A. We know from the preceding
Proposition that limp∈N µ(Ap) = 0.
-
60 1. MEASURE THEORY AND INTEGRATION
For any p we have∫
Ω
fp dµ =
∫
Ω
χA−Apfp dµ+
∫
Ω
χApfp dµ
≤∫
Ω
χA−Apǫ dµ+
∫
Ω
χApM dµ
≤ ǫµ(A− Ap) +Mµ(Ap) ≤ ǫµ(A) +Mµ(Ap)Since limp∈N µ(Ap) = 0, it
follows that, provided p is large enough,∫
Ωfp dµ ≤ ǫ(µ(A)+M), for example. Since ǫ > 0 was
arbitrary
∫
Ωfp dµ
must converge to zero. �
Corollary 5.4. Let fp be an increasing sequence of functionsin
S0(Ω,F , µ) which converges pointwise to a function f also in
S0.Then
limp∈N
∫
Ω
fp dµ =
∫
Ω
f dµ
Proof. Apply the Theorem to the sequence f − fp. �We are now
prepared to extend the integral to a broader family
of functions. It will be convenient to focus first on
nonnegative func-tions. Let S+0 = S+0 (Ω,F , µ) be the subset of
S0(Ω,F , µ) consistingof nonnegative functions. Thus S+0 is the set
of all nonegative simplefunctions f such that {x : f(x) 6= 0} has
finite measure. Define furtherD+(Ω,F , µ) = D+ by
D+ = {f : Ω → R+ : f = supH for some countable H ⊆ S+0 }.Later
we shall see that for f ∈ D+, then we can extend the concept
ofintegral as
∫ ∗
Ω
f dµ = sup{∫
Ω
h dµ : h ∈ S+0 , h ≤ f}
but first we explore the properties of D+. A cautionary
observation isthat elements of D+ may take on the value ∞ unlike
those in S+0 .
Proposition 5.5. Properties of D+(Ω,F , µ)(1) S+0 ⊆ D+.(2) A
function f : Ω → R+ is in D+ if and only if f is the
pointwise limit of an increasing sequence gp ∈ S+0 .(3) A
function f : Ω → R+ is in D+ if and only if f is nonnegative
and measurable and {x ∈ Ω : f(x) > 0} is the union of at
mostcountably many sets of finite measure.
(4) Suppose f : Ω → R+ is measurable and f ≤ g where g ∈ D+.Then
f ∈ D+.
(5) If f, g ∈ D+ then f + g ∈ D+.
-
5. THE INTEGRAL 61
(6) If f, g ∈ D+ then fg ∈ D+. Here it is understood the ∞·0 =
0,that is if f(x) = ∞ and g(x) = 0 then fg(x) = 0.
(7) If f ∈ D+ and λ ≥ 0 then λf ∈ D+.(8) If f, g ∈ D+ then
sup{f, g} and inf{f, g} are also in D+.
Proof: If f ∈ S+0 then f ∈ D+ because f = supH where H =
{f},that is H consists of only one element. Thus S+0 ⊆ D+ which is
Part(a).
Suppose f ∈ D+ so that f = supH where H is a countable set,and
suppose H = {hp : p ∈ N} is an enumeration of H. Defineg1 = h1, g2
= sup{g1, h2}, and in general gp = sup{gp−1, hp} for p ≥ 2so that
gp is an increasing sequence in S+0 . Since hp ≤ gp ≤ f we havef =
sup{gp}. Conversely if f is the pointwise limit of an
increasingsequence gp in S+0 then f ∈ D+ because we can choose H =
{gp}.
If f ∈ D+ then f ≥ 0 and f is the pointwise limit of an
increasingsequence gp of functions in S+0 . As the limit of
measurable functions fis measurable by Proposition 2.11.
Moreover
{x ∈ Ω : f(x) > 0} = ∪p∈N{x ∈ Ω : gp(x) > 0}
where {x ∈ Ω : gp(x) > 0} is an increasing sequence of sets
of finitemeasure.
Conversely suppose that f is a nonnegative measurable
functionand {x ∈ Ω : f(x) > 0} = ∪p∈NAp where Ap is a sequence
of setsof finite measure. We may assume that Ap is an increasing
sequence.Because f ≥ 0 is measurable there is an increasing
sequence gp ∈ Sof nonnegative functions which converges pointwise
to f by Theorem2.15. The sequence χApgp belongs to S+0 , is
increasing and convergespointwise to f because, if x /∈ ∪pAp then
f(x) = 0 = gp(x) and ifx ∈ Aq, for some q then x ∈ Ap if p ≥ q and
f(x) = limp gp(x) =limp χAp(x)gp(x). This shows that f ∈ D+ and
verifies Part 3.
If f is measurable and 0 ≤ f ≤ g where g ∈ D+ then {x : f(x)
>0} ⊆ {x : g(x) > 0}. It follows that {x : f(x) > 0} can
be written asthe countable union of sets of finite measure because
{x : g(x) > 0}can, Therefore by Part 3. f ∈ D+.
Let f and g be in D+ so that, by Part 2, there are
increasingsequences fp and gp, p ∈ N in S+0 so that f = limp fp and
g = limp gp.Consequently f + g = limp fp + gp and this implies that
f + g ∈ D+,again by Part 2. This verifies Part (5). The proof of
Part (6) is similarbecause fg = limp fpgp. (Here we use the
convention 0 · ∞ = 0.) Thisverifies Part 6.
Exercise: Verify Properties 7 and 8 of D+. 2
-
62 1. MEASURE THEORY AND INTEGRATION
For f in D+, define Sf = {g ∈ S+0 : g ≤ f} and∫ ∗
Ω
f dµ = sup{∫
Ω
g dµ : g ∈ Sf}
Of course Sf is a very large set but we could actually use a
much smallerset as the following result shows.
Theorem 5.6. If f ∈ D+ then there is an increasing sequencegp in
S+0 so that f = supp gp. Moreover for any such sequence,
∫ ∗
Ω
f dµ = supp
∫ ∗
Ω
gp dµ
Proof. We have already seen that there exists an increasing
se-quence gp in S+0 so that f = sup gp: this is Property 2 of D+.
For anyp,
∫ ∗
Ω
gp dµ ≤ sup{∫
Ω
h dµ : h ∈ Sf}
simply because gp ∈ Sf . We may take the supremum over p to
get
sup
{∫
Ω
gp dµ : p ∈ N}
≤ sup{∫
Ω
h dµ : h ∈ Sf}
=
∫ ∗
Ω
f dµ
To prove the reverse inequality, we shall show that if h ∈ Sf
then∫
Ωh dµ ≤ supp{
∫
Ωgp dµ. Taking the supremum over all such h will
complete the argument. Observe that h = supp inf{h, gp} so that
h isthe pointwise limit of an increasing sequence in S+0 . By
Corollary 5.4,
∫
Ω
h dµ = limp∈N
∫
Ω
inf{h, gp} dµ ≤ limp∈N
∫
Ω
gp dµ
This completes the proof. �
Notation: We have S+0 ⊆ D+ (Property 1 of D+). Indeed anyh ∈ S+0
is the supremum of the constant sequence hp = h and
∫ ∗
Ω
h dµ = supp
∫
Ω
hp dµ =
∫
Ω
h dµ
Therefore the two integrals∫ ∗Ωh dµ =
∫
Ωh dµ coincide on the intersec-
tion S+0 of their domains. Later we shall construct an extension
of theintegral on S0 and that integral (which will be finite valued
on its do-main) will be denoted
∫
Ω. The extension
∫ ∗Ωdefined on D+ is distinct
in that it may take on the value ∞.Proposition 5.7. If f, g ∈
D+(Ω,F , µ) and λ > 0. Then
a.
∫ ∗
Ω
λf dµ = λ
∫ ∗
Ω
f dµ;
-
5. THE INTEGRAL 63
b.
∫ ∗
Ω
f + g dµ =
∫ ∗
Ω
f dµ+
∫ ∗
Ω
g dµ;
c.
∫
Ω
f dµ ≤∫
Ω
g dµ if f ≤ g.
Proof. Suppose that f ∈ D+ and f = supp fp where fp is
anincreasing sequences in S+0 . Then, for λ > 0. λfp is an
increasingsequence in S+0 convergent pointwise to λf so that λf ∈
D+ and
∫ ∗
Ω
λf dµ = supp
∫
Ω
λfp dµ = λ supp
∫
Ω
fp dµ = λ
∫ ∗
Ω
f dµ
by the Theorem 5.1. This proves Part a because∫ ∗
=∫
.Suppose now that g ∈ D+ also and g = supp gp where gp is an
increasing sequence in S+0 . Then f + g = sup fp + gp and∫ ∗
Ω
f + g dµ = supp
∫
Ω
fp + gp dµ = supp
∫
Ω
fp dµ+
∫
Ω
gp dµ
= supp
∫
Ω
fp dµ+ supp
∫
Ω
gp dµ
=
∫ ∗
Ω
f dµ+
∫ ∗
Ω
g dµ
again by Theorem 5.1 and this proves Part b.If we further assume
that f ≤ g then fp ∈ Sg so that
∫
Ω
fp dµ ≤ suph∈Sg
∫
Ω
h dµ =
∫ ∗
Ω
g dµ
Taking the supremum over p gives Part c. �
Historical Notes. Henri Léon Lebesgue 1875-1941 was born
inBeauvais France and received a teaching diploma from the École
Nor-male Supérieure in Paris in 1897. From 1899-1902 he was a
professor atthe Lycée Centrale at Nancy where he wrote several
papers includinghis “Sur une généralisation de l’intégrale
définie” in which he intro-duces his novel approach to integration
on the real line. Lebesgue’sdoctoral dissertation, Intégrale,
longueur, aire , presented to the Fac-ulty of Science in Paris in
1902, and the 130 page work was publishedin the Annali di
Matematica in the same year. contains a chapter onBorel measure and
measure theory and ends with a discussion of thePlateau
problem.
-
64 1. MEASURE THEORY AND INTEGRATION
6. Convergence of Integrals
In this section we extend the integral to real valued functions
andshow that the integral of a sequence of pointwise convergent
functionswill converge in many circumstances.
Theorem 6.1. Monotone Convergence Theorem: Let fpbe an
increasing sequence in D+ and f = supp fp. Then f ∈ D+ and
∫
Ω
f dµ = sup
∫
Ω
fp dµ
Proof. For each p there is an increasing sequence h(p)n in S+0
so
that fp = supn h(p)n .
h(1)1 h
(1)2 . . . h
(1)p . . . ր f1
h(2)1 h
(2)2 . . . h
(2)p . . . ր f2
......
......
......
h(p)1 h
(p)2 . . . h
(p)p . . . ր fp
......
......
......
We define
gp = sup{h(1)p , h(2)p , . . . , h(p)p }so that, gp ∈ S+0 for
all p ∈ N and
(1) gp is an increasing sequence;(2) gp ≤ fp;(3) f = limp
gp.
Assuming these three for the moment we can now derive the
result. Wehave f ∈ D+ as the pointwise limit of a sequence in S+0
and gp ≤ fp ≤ fso that
∫
Ω
gp dµ ≤∫ ∗
Ω
fp dµ ≤∫ ∗
Ω
f dµ
If we take the supremum over p, we see sup∫ ∗Ωfp dµ =
∫ ∗Ωf dµ.
Check the three properties. The sequence gp increases because
h(n)p
increases as p does if n is fixed. We also have gp ≤ fp because
h(n)p ≤ fnand fn ≤ fp if n ≤ p. Finally we show that, for an
arbitrary x ∈ Ω,f(x) = supp gp(x). Certainly gp ≤ fp ≤ f and so we
need only showthat if λ < f(x) then there is n so that gn(x)
> λ. Choose p so large
that fp(x) > λ. Choose n ∈ N so large that h(p)n (x) > λ
and n ≥ p.Then gn(x) ≥ h(p)n (x) > λ. This completes the proof.
�
-
6. CONVERGENCE OF INTEGRALS 65
Example It is not true that, if fn is a decreasing sequence in
D+which is pointwise convergent to f then
∫
Ωfn dµ converges to
∫
Ωf dµ.
For example fn to be the characteristic function of [−n, n]c
convergespointwise to 0 but
∫
Rfn dµ = ∞. However, if we further suppose that
∫
Ωfn0 dµ < ∞ for some n0 then the integrals converge as can be
seen by
applying the Monotone Convergence theorem to the sequence fn0 −
fnwhich is in D+ by property 4 of Proposition 5.5 D+.
Corollary 6.2. If gn, n ∈ N is a sequence in D+ then f =∑
n∈N gn is also in D+ and∫
Ω
∑
n∈Ngn dµ =
∑
n∈N
∫
Ω
gn dµ
Proof. Define fk =∑
1≤n≤k gn so that fk ∈ D+ and apply theTheorem. Because
∫
Ωfk dµ =
∑
1≤n≤k∫
Ωgn dµ the result follows. �
We shall need the concept of lim infn an and lim supn an in the
casethat an is a sequence in R. For such a sequence, define bk =
infn≥k anfor each k ∈ N; bk is well defined and an increasing
sequence and itslimit is lim infn an. Similarly, if ck = supn≥n ak
then ck is a decreasingsequence with limit lim supn an.
Lemma 6.3. Suppose that fp : Ω → R, p ∈ N is a sequence ofreal
valued functions in D+. Then lim infp fp ∈ D+.
lim infp
∫
Ω
fp dµ ≥∫
Ω
lim infp
fp dµ
Proof. For each n ∈ N, define gn = infp≥n fp. Then gn is
measur-able as the limit of the measurable sequence hN = infp:N≥p≥n
fp. andgn is in D+ by Property 4 of Proposition 5.5 because gn ≤
fn. The se-quence gn increases to lim infp fp which, by the
Monotone ConvergenceTheorem is in D+, and
∫
Ω
lim infp∈N
fp dµ = limn∈N
∫
Ω
gn dµ.
On the other hand gn ≤ fp for all p ≥ n so that∫
Ω
gn dµ ≤∫
fp dµ for all p ≥ n
If we take the infimum over p and then the limit over