This document introduces product measures and establishes foundational results about them. It defines the Cartesian product of two measurable spaces and shows how to define a measure on the product space that is the "product" of measures on the factor spaces. The key results are: 1) Tonelli's Theorem, which states that for a non-negative measurable function on a product space with σ-finite factors, the integral over the product equals the iterated integrals over the factors. 2) Fubini's Theorem, which extends this to integrable functions that may be positive or negative, showing the iterated integrals are equal almost everywhere. 3) Definitions of sections of sets and functions that are used to relate

1. Let X and Y be two sets; then the Cartesian product Z = X x Y is
the set of all ordered pairs ( x , y) with x E X and y E Y. We shall first
show that the Cartesian productof two measurable spaces ( X , X)
and ( Y, Y) can be made into a measurable space in a natural fashion.
Next we shall show that if measures are given on each of the factor
spaces, we can define a measure on the productspace. Finally, we
shall relatejntegration with respectto the productmeasure and iterated
integration with respect to the measures in the factor spaces. The
model to be kept in mind throughout this discussion is the plane, which
we regard as the productR x R.
10.1 DEFINITION. If (X, X) and ( Y, Y) are measurable spaces, then a
set of the form A x B with A E X and B E Y is called a measurable
rectangle, or simply a rectangle, in Z = X x Y, We shall denote the
collection of all finite unions of rectangles by Zo.
It is an exercise to show that every set in Zo can be expressed as a
finite disjoint union of rectangles in Z (see Exercise 10.D).
10.2 LEMMA. Thecollection Zo is an algebra of subsets of Z .
PROOF. It is clear that the union of a finite number of sets in Zo
also belongs to Zo. Similarly, it follows from the first part of Exercise
10.E that the complement of a rectangle in Z belongs to Zo. Apply
De Morgan's laws to see that the complement of any set in Zo belongs
to Zo. Q.E.D.
11 3
1 14 The Elements of Integration
10.3 DEFINITION. If (X, X) and ( Y, Y) are measurable spaces, then
Z = X x Y denotes the a-algebra of subsets of Z = X x Y generated
by rectangles A x B with A E X and B E Y. We shall refer to a set in
Z as a Z-measurable set, or as a measurable subset of Z.
If (X, X, p) and (Y, Y, v) are measure spaces, it is natural to attempt
to define a measure r on the subsets of Z = X x Y which is the
"product" of p and v in the sense that
(Recall the convention that O(fco) = 0.) We shall now show that
this can always be done.
10.4 PRODUCTM EASURTEH EOREM.I f (X, X, p) and (Y, Y, v) are
measure spaces, then there exists a measure r defined on Z = X x Y
such that
for all A E X and B E Y. If these measure spaces are a-finite, then there
is a unique measure r with property (10.1).
PROOF. Supposethat the rectangle A x B is the disjoint union of a

2. sequence (A, x Bj) of rectangles; thus
for all x E X, y E Y. Hold x fixed, integrate with respect to v, and
apply the Monotone Convergence Theorem to obtain
A further application of the Monotone Convergence Theorem yields
Now let E E ZO; without loss of generality we may assume that
Product Measures 1 15
where the sets A, x B, are mutually disjoint rectangles. If we define
no(E) by
the argument in the previous paragraph implies that no is well-defined
and countably additive on 2,. By Theorem 9.7, there is an extension
of 7r0 to a measure n on the a-algebra Z generated by Zo. Since n is
an extension of no, it is clear that (10.1) holds.
If (X, X, p) and (Y, Y, v) are a-finite, then no is a 5-finite measure
on the algebra Zo and the uniqueness of a measure satisfying (10.1)
follows from the uniqueness assertion of the Hahn Extension Theorem
9.8. Q.E.D.
Theorem 10.4 establishes the existence of a measure n on the U-algebra
Z generated by the rectangles (A x B : A E X, B E Y) and such that
(10.1) holds. Any such measure will be called a productof p and v.
If p and v are both a-finite, then they have a unique product. In the
general casethe extension procedurediscussed in the previous chapter
leads to a uniquely determined productmeasure. However, it will be
seen in Exercise 10.S that it is possible for two distinct measures on Z
to satisfy (10.1) if p and v are not a-finite.
In order to relate integration with respect to a productmeasure and
iterated integration, the notion of a section is useful.
10.5 DEFINITION. If E is a subset of Z = X x Y and x E X, then
the x-section of E is the set
Ex = {YE Y : (x, y) E E)
Similarly, if y E Y, then the y-section of E is the set
Iff is a function defined on Z to ii, and x E X, then the x-section off
is the functionf, defined on Y by
The Elements of Integration
Similarly, if y E Y, then the y-section off is the function f Y defined
on X by
f y ( x ) = f ( x , ~ ) , XEX.
10.6 LEMMA. (a) If E is a measurable subsetof 2, then every section
of E is measurable.
(b) Iff is a measurable function on Z to 8, then every section off is
measurable.

3. PROOF. (a) If E = A x B and x E X, then the x-section of E is B
if x E A, and is 0 if x # A. Therefore, the rectangles are contained in
the collection E of sets in Z having the property that each x-section is
measurable. Since it is easily seen that E is a U-algebra (see Exercise
10.I), it follows that E = 2 .
(b) Let x E Xand a E R, then
If f is 2-measurable, then f, is Y-measurable. Similarly, f Y is Xmeasurable.
Q.E.D.
We interpolate an important result, which is often useful in measure
and probability theory, and which will be used below. We recall (see
Exercise 2.V) that a monotone class is a nonempty collection M of sets
which contains the union of each increasing sequence in M and the
intersection of each decreasing sequence in M. It is easy (see Exercise
2.W) to show that if A is a nonempty collection of subsets of a set S,
then the U-algebra S generated by A contains the monotone class M
generated by A. We now show that if A is an algebra, then S = M.
10.7 MONOTONCE LASSLEMMA.1 f.A is an algebra of sets, then the
a-algebra Sgenerated by A coincides with the monotone class M generated
by A.
PROOF. We have remarked that M G S. To obtain the opposite
inclusion it suffices to prove that M is an algebra.
If EE M, define M(E) to be the collection of FE M such that
E F, E n F, F E all belong to M. Evidently 0, E E M(E) and it is
Product Measures 1 17
readily seen that M(E) is a monotone class. Moreover, F E M(E) if
and only if E E M(F).
If E belongs to the algebra A, then it is clear that A s M(E).
But since M is the smallest monotone class containing A, we must have
M(E) = MforEinA. T h e r e f o r e , i f E ~ A a n d F ~ M , t h e n F ~ M ( E ) .
We infer that if E E A and F E M, then E E M(F) so that A s M(F) for
any FE M. Using the minimality of M once more we conclude that
M(F) = M for any FE M. Thus M is closed under intersections and
relative complements. But since X E M it is plain that M is an algebra ;
since it is a monotone class, it is indeed a a-algebra. Q.E.D.
It follows from the Monotone Class Lemma that if a monotone class
contains an algebra A, then it contains the a-algebra generated by A.
10.8 LEMMA. Let (X, X, ,u) and ( Y, Y, v) be a-finite measure spaces.
If E E Z = X x Y, then the functions defined by
are measurable, and

4. J fdp = n(E) = J gdv.
X Y
PROOF. First we shall supposethat the measure spaces are finite
and let M be the collection of all E E Z for which the above assertion
is true. We shall show that M = Z by demonstrating that M is a
monotone class containing the algebra Zo. In fact, if E = A x B
with A E X and B E Y, then
Since an arbitrary element of Zo can be written as a finite disjoint
union of rectangles, it follows that Z, G M.
We now show that M is a monotone class. Indeed, let (En) be a
monotone increasing sequence in M with union E, Therefore
1 1 8 The Elements of Integration
are measurable and
Jt is clear that the monotone increasing sequences (A) and (gn) converge
to the functions f and g defined by
If we apply the fact that n is a measure and the Monotone Convergence
Theorem, we obtain
so that EE M. Since n is finite measure, it can be proved in the same
way that if (F,) is a monotone decreasing sequence in M7 then F = n F,
belongs to M. Therefore M is a monotone class, and it follows from
the Monotone Class Lemma that M = 2.
If the measure spaces are a-finite, let Z be the increasing union of a
sequence of rectangles (Z,) with n(Zn) < +a and apply the previous
argument and the Monotone Convergence Theorem to the sequence
(E n Zn) Q.ED. .
10.9 TONELLI'ST HEOREM.L et (X, X, ,u) and ( Y, Y, v) be a-finite
measure spaces and let F be a nonnegative measurable function on
Z = X x Y to R. Then the functions defined on X and Y by
are measurable and
(1 0.5) IX fd,u = J' Fdn = IYgdv.
z
In other symbols,
PROOF. If F is the characteristic function of a set in 2 , the assertion
follows from the Lemma 10.8. By linearity, the present theorem holds
Product Measures 1 19
for a measurable simple function. If F is an arbitrary nonnegative

5. measurable function on Z to k, Lemma 2.11 implies that there is a
sequence (0,) of nonnegative measurable simple functions which
converges in a monotone increasing fashion on Z to F. If yn and +, are
defined by
then yn and +,, are measurable and monotone in n. By the Monotone
Convergence Theorem, (y,) converges on X to f and (i,hn) converges on
Y to g. Another application of the Monotone Convergence Theorem
implies that
f dp = lim yn dp = lim IZ 0. d~
X
The same theorem also shows that
from which (10.5) follows. Q.E.D.
It will be seen in the exercises that Tonelli's Theorem may fail if we
drop the hypothesis that F is nonnegative, or if we drop the hypothesis
that the measures p, v are a-finite.
Tonelli's Theorem deals with a nonnegative function on Z and
affirms the equality of the integral over Z and the two iterated integrals
whether these integrals are finite or equal +a. The final result
considers the casewhere the function is allowed to take both positive
and negative values, but is assumed to be integrable.
10.10 FUBINI'TS HEOREM.L et (X, X, ,u) and ( Y, Y, v) be a-Jinite
spaces and let the measure T on Z = X x Y be the product of p and v. I f
the function F on Z = X x Y to R is integrable with respect to T, then
the extended real-valued functions deJned almost everywhere by
120 The Elements of Integration
have finite integrals and
In other symbols,
PROOF. Since F is integrable with respectto n, its positive and
negative parts F + and F- are integrable. Apply Tonelli's Theorem
to F+ and F- to deduce that the carresponding f + and f- have finite
integrals with respectto p. Hence f+ and f- are finite-valued
p-almost everywhere, so their difference f is defined p-almost everywhere
and the first part of (10.9) is clear. The second part is similar.
Q.E.D.
Since we have chosenin Chapter 5 to restrict the use of the word
"integrable" to real-valued functions, we cannot conclude that the

6. functions f, g defined in (10.8) are integrable. However, they are
almost everywhere equal to integrable functions.
It will be seen in an exercise that Fubini's Theorem may fail if the
hypothesis that F is integrable is dropped.
EXERCISES
10.A. Let A s X and B s Y. If A or B is empty, then A x B = 0.
Conversely, if A x B = 0, then either A = 0 or B = 0.
10.B. Let A, s Xand Bj s Y, j = 1,2. If Al x B1 = A2 x B2
# 0, then Al =A2 and B1 = B2.
10.C. Let A, s Xand Bj s Y, j = 1,2. Then
and the sets on the right side are mutually disjoint.
Product Measures 121
10.D. Let (X, X) and (Y, Y) be measurable spaces. If Aj EX and
Bj E Y for j = 1,. . . , m, then the set
can be written as the disjoint union of a finite number of rectangles in Z.
10.E. Let Aj G X and B, s Y, j = 1,2. Then
10.F. If (R, B) denotes the measurable space consisting of real
numbers together with the Borel sets, show that every open subsetof
R x R belongs to B x B. In fact, this 5-algebra is the 5-algebra
generated by the open subsets of R x R. (In other words, B x B is
the Borel algebra of R x R .)
10.G. Let f and g be real-valued functions on X and Y, respectively;
supposethat f is X-measurable and that g is Y-measurable. If h is
defined for (x, y) in X x Y by h(x, y) = f(x)g(y), show that h is
X x Y-measurable.
10.H. IfEis asubset ofR, let y(E) = {(x, y )R~ x R : x - y E E).
If E E B, show that y(E) E B x B. Use this to prove that if ,f is a
Borel measurable function on R to R, then the function F defined by
F(x, y) = f(x - y) is measurable with respect to B x B.
10.1. Let E and F be subsets of Z = X x Y, and let x E X. Show
that (E E), = Ex Fx. If (Ea) are subsets of Z, then
10.J. Let (X, X, tc) be the measure space on the natural numbers
X = N with the counting measure defined on all subsets of X = N.
Let (Y, Y, v) be an arbitrary measure space. Show that a set E in
Z = X x Y belongs to Z = X x Y if and only if each section En of E
belongs to Y. In this casethere is a unique productmeasure T, and
122 The Elements of Integration
A function f on Z = X x Y to R is measurable if and only if each

7. section f, is Y-measurable. Moreover, f is integrable with respectto T
if and only if the series
is convergent, in which case
10.K. Let X and Y be the unit interval [0, 11 and let X and Y be the
Bore1 subsets of [0, 11. Let p be Lebesgue measure on X and let v be
the counting measure on Y. If D = {(x, y) : x = y}, show that D is a
measurable subsetof Z = X x Y, but that
Hence Lemma 10.8 may fail unless both of the factors are required to
be a-finite.
10.L. If F is the characteristic function of the set D in the Exercise
10.K, show that Tonelli's Theorem may fail unless both of the factors
are required to be a-finite.
10.M. Show that the example considered in Exercise 10.J demonstrates
that Tonelli's Theorem holds for arbitrary (Y, Y, v) when
(X, X, p) is the set N of natural numbers with the counting measure on
arbitrary subsets of N.
10.N. If am, 2 0 for m, n E N, then
10.0. Let a,, be defined for m, n E N by requiring that a,, = + 1,
a,,,,, = -1,anda,, = Oifm # n o r m # n + 1 . Showthat
so the hypothesis of integrability in Fubini's Theorem cannot be
dropped,
Product Measures 1 23
1O.P. Let f be integrable on (X, X, ,u) , let g be integrable on ( Y, Y, Y) ,
and define h on Z by h(x, y) = f(x) g(y). If n is a productof p and Y,
show that h is n-integrable and
10.Q. Supposethat (X, X, ,u) and (Y, Y, Y) are a-finite, and let
E, F belong to X x Y. If v(E,) = v(F,) for all x E X, then n(E) = n(F).
10.R. Let f and g be Lebesgue integrable functions on (R, B) to R.
From Exercise 10.H it follows that the function mapping (x, y) into
f(x - y)g(y) is measurable with respectto B x B. If X denotes
Lebesgue measure on B, use Tonelli's Theorem and the fact that
to show that the function h defined for x E R by
is finite almost everywhere. Moreover,
The function h defined above is called the convolution off and g and is
usually denoted by f * g.
10.S. Let X = R, X be the a-algebra of all subsets of R and let p be
defined by p(A) = 0 if A is countable, and p(A) = +co if A is uncountable.
We shall constructdistinct products of p with itself.

8. (a) If E E Z = X x X, define n(E) = 0 in case E can be written as
the union E = G u H of two sets in Z such that the x-projection of G
is countable and the y-projection of H is countable. Otherwise, define
n(E) = +a. It is evident that n is a measure on Z. If n(E) = 0,
then E is contained in the union of a countable set of lines in the plane.
If A, B E X, show that n(A x B) = p(A) p(B). Hence n is a product
of p with itself.
124 The Elements of Integration
(b) If EE 2 , define p(E) = 0 in case E can be written as the union
E = G u H u K of three sets in Z such that the x-projection of G is
countable, the y-projection of H is countable, and the projection of K
on the line with equation y = x is countable. Otherwise, define
p(E) = +co. Now p is a measure on 2, and if p(E) = 0, then E is
contained in the union of a countable set of lines. Show that
p(A x B) = p(A) p(B) for all A, B E X; hence p is a productof p with
itself.
(c) Let E = {(x, y) : x + y = 0); show that E E Z . However,
p(E) = 0, whereas r(E) = + co .