5B Iterated Integrals
Tonelli’s Theorem
Relook at Example5.24in the previous section and notice that the value of the iterated integral was unchanged when we switched the order of integration, even though switching the order of integration led to different intermediate results. Our next result states that the order of integration can be switched if the function being integrated is nonnegative and the measures areσ-finite.
5.28 Tonelli’s Theorem
Suppose (X,S,µ) and (Y,T,ν) are σ-finite measure spaces. Suppose f:X×Y→[0,∞]isS ⊗ T-measurable. Then
x 7→
Z
Yf(x,y)dν(y)is anS-measurable function onX, (a)
y7→
Z
X f(x,y)dµ(x)is aT-measurable function onY, (b)
and Z
X×Yfd(µ×ν) = Z
X Z
Yf(x,y)dν(y)dµ(x) = Z
Y Z
X f(x,y)dµ(x)dν(y). Proof We begin by considering the special case wheref =χEfor someE∈ S ⊗ T. In this case, Z
YχE(x,y)dν(y) =ν([E]x)for everyx∈X
and Z
XχE(x,y)dµ(x) =µ([E]y)for everyy∈Y.
Thus (a) and (b) hold in this case by5.20.
First assume thatµandνare finite measures. Let M=nE∈ S ⊗ T :
Z X
Z
YχE(x,y)dν(y)dµ(x) = Z
Y Z
XχE(x,y)dµ(x)dν(y)o. IfA∈ S andB∈ T, thenA×B∈ Mbecause both sides of the equation defining Mequalµ(A)ν(B).
LetAdenote the set of finite unions of measurable rectangles inS ⊗ T. Then 5.13(b) implies that every element ofAis a disjoint union of measurable rectangles inS ⊗ T. The previous paragraph now impliesA ⊂ M.
The Monotone Convergence Theorem (3.11) implies thatM is closed under countable increasing unions. The Bounded Convergence Theorem (3.26) implies thatMis closed under countable decreasing intersections (this is where we use the assumption thatµandνare finite measures).
We have shown thatMis a monotone class that contains the algebraAof all finite unions of measurable rectangles inS ⊗ T [by5.13(a),Ais indeed an algebra].
The Monotone Class Theorem (5.17) implies thatMcontains the smallestσ-algebra containingA. In other words,McontainsS ⊗ T. Thus
5.29
Z X
Z
YχE(x,y)dν(y)dµ(x) = Z
Y Z
XχE(x,y)dµ(x)dν(y) for everyE∈ S ⊗ T.
Now relax the assumption that µ and ν are finite measures. Write X as an increasing union of sets X1 ⊂ X2 ⊂ · · · inS with finite measure, and write Y as an increasing union of setsY1⊂ Y2⊂ · · · inT with finite measure. Suppose E∈ S ⊗ T. Applying the finite-measure case to the situation where the measures and theσ-algebras are restricted toXj andYk, we can conclude that5.29holds with E replaced by E∩(Xj×Yk)for all j,k ∈ Z+. Fix k ∈ Z+ and use the Monotone Convergence Theorem (3.11) to conclude that5.29holds withEreplaced byE∩(X×Yk)for all k ∈ Z+. One more use of the Monotone Convergence Theorem then shows that
Z
X×YχEd(µ×ν) = Z
X Z
YχE(x,y)dν(y)dµ(x) = Z
Y Z
XχE(x,y)dµ(x)dν(y) for allE ∈ S ⊗ T, where the first equality above comes from the definition of (µ×ν)(E)(see5.25).
Now we turn from characteristic functions to the general case of anS ⊗ T- measurable function f: X×Y → [0,∞]. Define a sequence f1,f2, . . .of simple S ⊗ T-measurable functions fromX×Yto[0,∞)by
fk(x,y) =
m
2k if f(x,y)<kandmis the integer with f(x,y)∈h2mk,m+1 2k
,
k if f(x,y)≥k.
Note that
0≤ f1(x,y)≤ f2(x,y)≤ f3(x,y)≤ · · · and lim
k→∞fk(x,y) = f(x,y) for all(x,y)∈X×Y.
Eachfkis a finite sum of functions of the formcχE, wherec∈RandE∈ S ⊗ T. Thus the conclusions of this theorem hold for each functionfk.
The Monotone Convergence Theorem implies that Z
Y f(x,y)dν(y) = lim
k→∞
Z
Yfk(x,y)dν(y) for everyx∈X. Thus the functionx7→R
Yf(x,y)dν(y)is the pointwise limit on Xof a sequence ofS-measurable functions. Hence (a) holds, as does (b) for similar reasons.
The last line in the statement of this theorem holds for eachfk. The Monotone Convergence Theorem now implies that the last line in the statement of this theorem holds forf, completing the proof.
See Exercise1in this section for an example (with finite measures) showing that Tonelli’s Theorem can fail without the hypothesis that the function being integrated is nonnegative. The next example shows that the hypothesis ofσ-finite measures also cannot be eliminated.
5.30 Example Tonelli’s Theorem can fail without the hypothesis of σ-finite SupposeBis theσ-algebra of Borel subsets of[0, 1],λis Lebesgue measure on ([0, 1],B), andµis counting measure on([0, 1],B). LetDdenote the diagonal of [0, 1]×[0, 1]; in other words,
D={(x,x):x ∈[0, 1]}.
Then Z
[0, 1]
Z
[0, 1]χD(x,y)dµ(y)dλ(x) = Z
[0, 1]1dλ=1,
but Z
[0, 1]
Z
[0, 1]χD(x,y)dλ(x)dµ(y) = Z
[0, 1]0dµ=0.
The following useful corollary of Tonelli’s Theorem states that we can switch the order of summation in a double-sum of nonnegative numbers. Exercise2asks you to find a double-sum of real numbers in which switching the order of summation changes the value of the double sum.
5.31 double sums of nonnegative numbers
If{xj,k :j,k∈Z+}is a doubly indexed collection of nonnegative numbers, then
∑
∞ j=1∑
∞ k=1xj,k=∑
∞ k=1∑
∞ j=1xj,k.Proof Apply Tonelli’s Theorem (5.28) to µ×µ, whereµ is counting measure onZ+.
Fubini’s Theorem
Historically, Fubini’s Theorem (proved in 1907) came before Tonelli’s Theorem (proved in 1909).
However, presenting Tonelli’s Theorem first, as is done here, seems to lead to simpler proofs and better understanding. The hard work here went into proving Tonelli’s Theorem;
thus our proof of Fubini’s Theorem consists mainly of bookkeeping details.
Our next goal is Fubini’s Theorem, which has the same conclusions as Tonelli’s Theorem but has a different hypothesis.
Tonelli’s Theorem requires the function being integrated to be nonnegative. Fu- bini’s Theorem instead requires the inte- gral of the absolute value of the function to be finite. When using Fubini’s The- orem to evaluate the integral of f, you will usually first use Tonelli’s Theorem as applied to|f|to verify the hypothesis of Fubini’s Theorem.
As you will see in the proof of Fubini’s Theorem, the function in5.32(a) is defined only for almost everyx∈Xand the function in5.32(b) is defined only for almost everyy∈Y. For convenience, you can think of these functions as equaling0on the sets of measure0on which they are otherwise undefined.
5.32 Fubini’s Theorem
Suppose (X,S,µ) and (Y,T,ν) are σ-finite measure spaces. Suppose f: X×Y → [−∞,∞] is S ⊗ T-measurable and R
X×Y|f|d(µ×ν) < ∞.
Then Z
Y|f(x,y)|dν(y)<∞for almost everyx∈X
and Z
X|f(x,y)|dµ(x)<∞for almost everyy∈Y.
Furthermore, x 7→
Z
Yf(x,y)dν(y)is anS-measurable function onX, (a)
y7→
Z
X f(x,y)dµ(x)is aT-measurable function onY, (b)
and Z
X×Yfd(µ×ν) = Z
X Z
Yf(x,y)dν(y)dµ(x) = Z
Y Z
X f(x,y)dµ(x)dν(y). Proof Tonelli’s Theorem (5.28) applied to the nonnegative function|f|implies that x7→R
Y|f(x,y)|dν(y)is anS-measurable function onX. Hence nx∈X:Z
Y|f(x,y)|dν(y) =∞o
∈ S. Tonelli’s Theorem applied to|f|also tells us that
Z X
Z
Y|f(x,y)|dν(y)dµ(x)<∞ because the iterated integral above equalsR
X×Y|f|d(µ×ν). The inequality above implies that
µn
x∈X:Z
Y|f(x,y)|dν(y) =∞o
=0.
Recall that f+and f− are nonnegativeS ⊗ T-measurable functions such that
|f| = f++f−and f = f+−f−(see3.17). Applying Tonelli’s Theorem to f+ and f−, we see that
5.33 x7→
Z
Yf+(x,y)dν(y) and x 7→
Z
Yf−(x,y)dν(y)
areS-measurable functions fromXto[0,∞]. Because f+ ≤ |f|andf−≤ |f|, the sets{x ∈ X : R
Y f+(x,y)dν(y) =∞}and{x ∈ X :R
Yf−(x,y)dν(y) = ∞} haveµ-measure0. Thus the intersection of these two sets, which is the set ofx ∈X such thatR
Y f(x,y)dν(y)is not defined, also hasµ-measure0.
Subtracting the second function in5.33from the first function in5.33, we see that the function that we define to be0for thosex ∈Xwhere we encounter∞−∞(a set ofµ-measure0, as noted above) and that equalsR
Y f(x,y)dν(y)elsewhere is an S-measurable function onX.
Now Z
X×Yfd(µ×ν) = Z
X×Yf+d(µ×ν)− Z
X×Yf−d(µ×ν)
= Z
X Z
Yf+(x,y)dν(y)dµ(x)− Z
X Z
Y f−(x,y)dν(y)dµ(x)
= Z
X Z
Y f+(x,y)−f−(x,y)dν(y)dµ(x)
= Z
X Z
Yf(x,y)dν(y)dµ(x),
where the first line above comes from the definition of the integral of a function that is not nonnegative (note that neither of the two terms on the right side of the first line equals∞becauseR
X×Y|f|d(µ×ν)<∞) and the second line comes from applying Tonelli’s Theorem to f+and f−.
We have now proved all aspects of Fubini’s Theorem that involve integrating first overY. The same procedure provides proofs for the aspects of Fubini’s theorem that involve integrating first overX.
Area Under Graph
5.34 Definition region under the graph;Uf
SupposeXis a set and f: X→[0,∞]is a function. Then theregion under the graphof f, denotedUf, is defined by
Uf ={(x,t)∈X×(0,∞): 0<t< f(x)}.
The figure indicates why we callUf the region under the graph of f, even in cases when X is not a subset of R.
Similarly, the informal termareain the next paragraph should remind you of the area in the figure, even though we are really dealing with the measure ofUf in a product space.
The first equality in the result below can be thought of as recovering Riemann’s conception of the integral as the area under the graph (although now in a much more general context with arbitraryσ-finite measures). The second equality in the result below can be thought of as reinforcing Lebesgue’s conception of computing the area under a curve by integrating in the direction perpendicular to Riemann’s.
5.35 area under the graph of a function equals the integral
Suppose (X,S,µ) is a σ-finite measure space and f: X → [0,∞] is an S-measurable function. LetBdenote theσ-algebra of Borel subsets of(0,∞), and letλdenote Lebesgue measure on (0,∞),B. ThenUf ∈ S ⊗ Band
(µ×λ)(Uf) = Z
X fdµ= Z
(0,∞)µ({x∈X:t< f(x)})dλ(t). Proof Fork∈Z+, let
Ek =
k2[−1
m=0
f−1 [mk,m+1k )× 0,mk and Fk = f−1([k,∞])×(0,k).
Then Ek is a finite union of S ⊗ B-measurable rectangles and Fk is an S ⊗ B- measurable rectangle. Because
Uf = [∞ k=1
(Ek∪Fk), we conclude thatUf ∈ S ⊗ B.
Now the definition of the product measureµ×λimplies that (µ×λ)(Uf) =
Z X
Z
(0,∞)χU
f(x,t)dλ(t)dµ(x)
= Z
X f(x)dµ(x),
which completes the proof of the first equality in the conclusion of this theorem.
Tonelli’s Theorem (5.28) tells us that we can interchange the order of integration in the double integral above, getting
(µ×λ)(Uf) = Z
(0,∞) Z
XχU
f(x,t)dµ(x)dλ(t)
= Z
(0,∞)µ({x∈X:t< f(x)})dλ(t),
which completes the proof of the second equality in the conclusion of this theorem.
Markov’s inequality (4.1) implies that if f andµare as in the result above, then µ({x∈X: f(x)>t})≤
R
X fdµ t for allt>0. Thus ifR
X fdµ<∞, then the result above should be considered to be somewhat stronger than Markov’s inequality (becauseR
(0,∞)1
tdλ(t) =∞).