Chapter 4 Products of Measures
4.1 Fubini’s Theorem
The key to the construction of µ1×µ2 is found in the following function analogue of Λ-systems (cf. Lemma 2.1.12). Namely, given a spaceE, say that a collectionL of functionsf :E−→(−∞,∞] is avector semi-lattice if (a) Bothf+ andf− are in Lwheneverf ∈ L.
(b) For boundedf, g∈ L withf ≤g, g−f ∈ L.
(c) Forα, β ∈[0,∞) andf, g∈ L,αf+βg∈ L.
(d) For{fn: n≥1} ⊆ Lwith 0≤fn%f, f ∈ Lwheneverf is bounded.
A sub-collectionKof a vector semi-latticeLis called anL-system ifKis a vector semi-lattice with the additional property that
(d0) 1∈ K and, for{fn : n≥ 1} ⊆ K, 0≤fn % f =⇒ f ∈ K whenever f ∈ L.
The analogue of Lemma 2.1.12 in this context is the following.
Lemma 4.1.1. Let C be a Π-system that generates the σ-algebra B over E, and let L be a semi-lattice of functions f : E −→ (−∞,∞]. If K is an L-system for which 1Γ ∈ K wheneverΓ ∈ C, thenK contains every f ∈ L that is measurable on (E,B).
Proof: First note that {Γ ⊆ E : 1Γ ∈ K} is a Λ-system that contains C.
Hence, by Lemma 2.1.12,1Γ∈ Kfor every Γ∈ B. Combined with (c) above, this means that K contains every non-negative, measurable, simple function on (E,B).
DOI 10.1007/978-1-4614-1135-2_4, © Springer Science+Business Media, LLC 2011
D.W. Stroock, Essentials of Integration Theory for Analysis, Graduate Texts in Mathematics 262, 100
Next, suppose thatf ∈ Lis measurable on (E,B). Then, by (a), both f+ andf−are elements ofL, and so, by (c), it is enough to show thatf+, f−∈ K in order to know that f ∈ K. Thus, without loss of generality, assume that f ∈ L is a non-negative measurable function on (E,B). But in that case f is the non-decreasing limit of non-negative measurable simple functions, and therefore, by the preceding and (d0),f ∈ K.
The power of Lemma 4.1.1 to handle questions involving products is already apparent in the following.
Lemma 4.1.2. Let(E1,B1)and(E2,B2)be measurable spaces, and suppose that f is an R-valued measurable function on(E1×E2,B1× B2). Then for each x1 ∈ E1 and x2 ∈ E2, f(x1,·) and f(·, x2) are measurable functions on(E2,B2)and(E1,B1), respectively. Next, suppose thatµi,i∈ {1,2}, is a finite measure on (Ei,Bi). Then for every non-negative, measurable function f on(E1×E2,B1× B2), the functions
Z
E2
f(·, x2)µ2(dx2) and Z
E1
f(x1,·)µ1(dx1) are measurable on(E1,B1)and(E2,B2), respectively.
Proof: Clearly, by the Monotone Convergence Theorem, it is enough to check all these assertions whenf is bounded.
LetLbe the collection of all bounded functions onE1×E2, and defineKto be theB1× B2-measurable elements ofLthat have all the asserted properties.
It is clear that1Γ1×Γ2 ∈ Kfor all Γi∈ Bi, and it is easy to check thatK is an L-system. Hence, by Lemma 4.1.1 withC=
Γ1×Γ2: Γi∈ Bi fori∈ {1,2} , we are done.
Lemma 4.1.3. Given a pair (E1,B1, µ1)and (E2,B2, µ2) of finite measure spaces, there exists a unique measureν on(E1×E2,B1× B2)for which
ν(Γ1×Γ2) =µ1(Γ1)µ2(Γ2) for all Γi∈ Bi.
Moreover, for every non-negative, measurable functionfon(E1×E2,B1×B2),
(4.1.4)
Z
E1×E2
f(x1, x2)ν(dx1×dx2)
= Z
E2
Z
E1
f(x1, x2)µ1(dx1)
µ2(dx2)
= Z
E1
Z
E2
f(x1, x2)µ2(dx2)
µ1(dx1).
Proof: The uniqueness ofν is guaranteed by Theorem 2.1.13. To prove the existence ofν, define
ν1,2(Γ) = Z
E2
Z
E1
1Γ(x1, x2)µ1(dx1)
µ2(dx2)
4 Products of Measures and
ν2,1(Γ) = Z
E1
Z
E2
1Γ(x1, x2)µ2(dx2)
µ1(dx1)
for Γ ∈ B1× B2. Using the Monotone Convergence Theorem, one sees that both ν1,2 and ν2,1 are finite measures on (E1×E2,B1× B2). Moreover, by the same sort of argument as was used to prove Lemma 4.1.2, for every non- negative measurable functionf on (E1×E2,B1× B2),
Z
f dν1,2= Z
E1
Z
E2
f(x1, x2)µ1(dx1)
µ2(dx2) and
Z
f dν2,1= Z
E2
Z
E1
f(x1, x2)µ2(dx2)
µ1(dx1).
Finally, sinceν1,2(Γ1×Γ2) =µ(Γ1)µ(Γ2) =ν2,1(Γ1×Γ2) for all Γi∈ Bi, we see that both ν1,2 andν2,1 fulfill the requirements placed on ν. Hence, not only does ν exist, but it is also equal to bothν1,2 and ν2,1; and so the preceding equalities lead to (4.1.4).
Recall that a measure space (E,B, µ) is said to beσ-finite ifEis the count- able union of B-measurable sets having finiteµ-measure.
Theorem 4.1.5 (Tonelli’s Theorem). Let(E1,B1, µ1)and(E2,B2, µ2)be σ-finite measure spaces. Then there is a unique measureνon(E1×E2,B1×B2) such that ν(Γ1×Γ2) = µ1(Γ1)µ2(Γ2) for all Γi ∈ Bi. In addition, for every non-negative measurable functionf on(E1×E2,B1× B2),R
f(·, x2)µ2(dx2) andR
f(x1,·)µ1(dx1)are measurable on(E1,B1)and (E2,B2), respectively, and(4.1.4)continues to hold.
Proof: Choose {Ei,n : n≥1} ⊆ Bi fori ∈ {1,2} such that µi(Ei,n)<∞ for each n≥1 andEi =S∞
n=1Ei,n. Without loss of generality, assume that Ei,m∩Ei,n=∅ form6=n. For eachn∈Z+, defineµi,n(Γi) =µi(Γi∩Ei,n), Γi∈ Bi; and, for (m, n)∈Z+2,letν(m,n)on (E1×E2,B1× B2) be the measure constructed fromµ1,m andµ2,n as in Lemma 4.1.3.
Clearly, by Lemma 4.1.2, for any non-negative measurable function f on (E1×E2,B1× B2),
Z
E2
f(·, x2)µ2(dx2) =
∞
X
n=1
Z
E2,n
f(·, x2)µ2,n(dx2)
is measurable on (E1,B1); and, similarly, R
E1f(x1, ·)µ1(dx1) is measurable on (E2,B2). Finally, the map Γ ∈ B1× B2 7−→ P∞
m,n=1ν(m,n)(Γ) defines a measure ν0 on (E1×E2,B1× B2), and it is easy to check that ν0 has all the required properties. At the same time, if ν is any other measure on (E1×E2,B1× B2) for whichν(Γ1×Γ2) =µ1(Γ1)µ2(Γ2), Γi∈ Bi, then, by the
uniqueness assertion in Lemma 4.1.3, for each (m, n)∈Z+2,ν coincides with ν(m,n)onB1× B2
E1,m×E2,n
and is therefore equal toν0onB1× B2. The measure ν constructed in Theorem 4.1.5 is called theproduct of µ1
and µ2and is denoted by µ1×µ2.
Theorem 4.1.6 (Fubini’s Theorem). Let(E1,B1, µ1)and(E2,B2, µ2)be σ-finite measure spaces andf a measurable function on (E1×E2,B1× B2).
Thenf isµ1×µ2-integrable if and only if Z
E1
Z
E2
|f(x1, x2)|µ2(dx2)
µ1(dx1)<∞ if and only if
Z
E2
Z
E1
|f(x1, x2)|µ1(dx1)
µ2(dx2)<∞.
Next, set
Λ1=
x1∈E1: Z
E2
|f(x1, x2)|µ2(dx2)<∞
and
Λ2=
x2∈E2: Z
E1
|f(x1, x2)|µ1(dx1)<∞
; and define fi onEi,i∈ {1,2}, by
f1(x1) = R
E2f(x1, x2)µ2(dx2) if x1∈Λ1
0 otherwise
and
f2(x2) = R
E1f(x1, x2)µ1(dx1) if x2∈Λ2
0 otherwise.
Then fi is an R-valued, measurable function on Ei,Bi
. Finally, if f is µ1×µ2-integrable, thenµi(Λi{) = 0,fi∈L1(µi;R), and
Z
Ei
fi(xi)µi(dxi) = Z
E1×E2
f(x1, x2) µ1×µ2
(dx1×dx2) fori∈ {1,2}.
Proof: The first assertion is an immediate consequence of Theorem 4.1.5.
Moreover, because Λi ∈ Bi, it is easy (cf. Lemma 4.1.2) to check thatfi is an R-valued, measurable function on (Ei,Bi). Finally, iff isµ1×µ2-integrable,
4 Products of Measures
then, by the first assertion,µi(Λ{i) = 0 andfi∈L1(µi;R). Hence, by Theorem 4.1.5 applied to f+and f−, we see that
Z
E1×E2
f(x1, x2) (µ1×µ2)(dx1×dx2)
= Z
Λ1×E2
f+(x1, x2) (µ1×µ2)(dx1×dx2)
− Z
Λ1×E2
f−(x1, x2) (µ1×µ2)(dx1×dx2)
= Z
Λ1
Z
E2
f+(x1, x2)µ1(dx2)
µ1(dx1)
− Z
Λ1
Z
E2
f−(x1, x2)µ2(dx2)
µ1(dx1)
= Z
Λ1
f1(x1)µ1(dx1) = Z
E1
f1(x1)µ1(dx1), and the same line of reasoning applies to f2.
One may well wonder why I have separated the statement in Tonelli’s The- orem from the statement in Fubini’s Theorem. The reason is that Tonelli’s Theorem requires no a priori information about integrability. Thus, for ex- ample, Tonelli’s Theorem allowed me to show that f is µ1×µ2-integrable if and only if
Z
E2
Z
E1
|f(x1, x2)|µ1(dx1)
µ2(dx2)
∧ Z
E1
Z
E2
|f(x1, x2)|µ2(dx2)
µ1(dx1)
<∞.
Exercises for §4.1
Exercise 4.1.7. Let (E,B, µ) be a σ-finite measure space. Given a non- negative measurable functionf on (E,B), define
Γ(f) =
(x, t)∈E×[0,∞) :t≤f(x) and
Γ(fb ) =
(x, t)∈E×[0,∞) :t < f(x) .
Show that both Γ(f) andΓ(f) are elements ofb B × B[0,∞) and, in addition, that
(∗) µ×λR Γ(fb )
= Z
E
f dµ=µ×λR Γ(f) .
Clearly (∗) can be interpreted as the statement that the integral of a non- negative function is the area under its graph.
Hint: In proving measurability, consider the function (x, t)∈E×[0,∞)7−→
f(x)−t∈(−∞,∞], and get (∗) as an application of Tonelli’s Theorem.
Exercise 4.1.8. Let (E1,B1, µ1) and (E2,B2, µ2) beσ-finite measure spaces and assume that, fori∈ {1,2},Bi=σ(Ei;Ci), whereCiis a Π-system contain- ing a sequence {Ei,n : n≥1} for which Ei =S∞
n=1Ei,n and µi(Ei,n)<∞, n≥1. Show that ifνis a measure on (E1×E2,B1×B2) with the property that ν(Γ1×Γ2) =µ1(Γ1)µ2(Γ2) for all Γi∈ Ci, thenν =µ1×µ2. Use this fact to show that, for anyM, N ∈Z+,λRM+N =λRM×λRN onBRM+N =BRM× BRN. Exercise 4.1.9. Let (E1,B1, µ1) and (E2,B2, µ2) beσ-finite measure spaces.
Given Γ∈ B1× B2, define Γ(1)(x2)≡
x1∈E1: x1, x2
∈Γ for x2∈E2 and
Γ(2)(x1)≡
x2∈E2: x1, x2
∈Γ for x1∈E1.
If i 6= j, check both that Γ(i)(xj) ∈ Bi for each xj ∈ Ej and that xj ∈ Ej 7−→ µi Γ(i)(xj)
∈ [0,∞] is measurable on (Ej,Bj). Finally, show that µ1×µ2(Γ) = 0 if and only ifµi Γ(i)(xj)
= 0 for µj-almost everyxj ∈Ej, and conclude that µ1 Γ(1)(x2)
= 0 forµ2-almost every x2 ∈E2 if and only ifµ2 Γ(2)(x1)
= 0 forµ1-almost everyx1∈E1. In other words, Γ∈ B1× B2
hasµ1×µ2-measure 0 if and only ifµ1-almost everyvertical slice(µ2-almost every horizontal slice) has µ2-measure (µ1-measure) 0. In particular, µ2- almost every horizontal slice has µ1-measure 0 if and only ifµ1-almost every vertical slice has µ2-measure 0.
Exercise 4.1.10. The condition that the measure spaces of which one is taking a product beσ-finite is essential if one wants to carry out the program in this section. To see this, let E1=E2= (0,1) andB1=B2 =B(0,1). Take µ1 to be the counting measure (i.e.,µ(Γ) = card(Γ)) andµ2 to be Lebesgue measure λ(0,1) on (E(0,1),B(0,1)). Show that there is a set Γ∈ B1× B2 such that
Z
E2
1Γ(x1, x2)µ2(dx2) = 0 for everyx1∈E1 but
Z
E1
1Γ(x1, x2)µ1(dx1) = 1 for everyx2∈E2.
Notice that what fails is not so much the existence statement as the uniqueness in Lemma 4.1.3.
4 Products of Measures