17.
Iterated integrals
In your multivariable calculus course you learned how to evaluate
double and triple integrals as “iterated integrals” — that is, by in-
tegrating with respect to one variable at a time. Also, you learned
that you could perform these “partial integrations” in any order1. The
Tonelli and Fubini Theorems are quite general justifications of those
techniques for multiple integrals. We’ll start with a very general it-
erated integral, decomposing an integral in n + m variables into an
integral in the first n variables followed by one in the last m variables.
And we’ll show that this order can be reversed without changing the
result.
Notation and Terminology 17.1. Throughout this section we iden-
tify Rn+m with Rn × Rm . Also, for a function f on Rn+m and y ∈ Rm
we write f (·, y) for the function x 7→R f (x, y), and similarly for f (x, ·).
If f is in L+ or L1 on Rn+m , then f (x, y) dx means regard f as a
function of the first n coordinates x, holding RR the last m coordinates y
fixed, and integrate as a function
R of x, and f (x, y) dx dy means first
evaluate the above integral f (x, y) dx, getting a function of y, then
integrate this function; thus there are implied parentheses:
Z Z
f (x, y) dx dy.
Theorem 17.2 (Tonelli’s Theorem). If f ∈ L+ (Rn+m ), then:
m
R is measurable for almost all y ∈ R ,
(i) f (·, y)
y→
(ii) RR 7 f (x, y) dx isRmeasurable, and
(iii) f (x, y) dx dy = f .
R RR
Similarly for f (x, ·), x 7→ f (x, y) dy, and f (x, y) dy dx.
Proof. This is a hard one, requiring many steps.
Step 1. Let F denote the set of functions in L+ satisfying (i)–(iii). We
start by establishing certain properties of F. It is completely routine
to verify that if f ∈ F then cf ∈ F for every nonnegative real number
c. We’ll use the Monotone Convergence Theorem to show that F is
closed under series
P and increasing limits: P for the first, let f1 , f2 , · · · ∈ F
and put f = i fi . Then f (·, y) = i fi (·, y) is measurable for almost
all y since each fi (·, y) isR measurable for
Palmost
R all y. By the Monotone
Convergence Theorem f (x, y) dx = i fi (x, y) dx for almost all y.
1
provided you took care with the “limits of integration”
1
2
Again by the Monotone Convergence Theorem (twice),
ZZ X ZZ XZ Z
f (x, y) dx dy = fi (x, y) dx dy = fi = f.
i i
For increasing limits the argument is the same: if f1 , f2 , · · · ∈ F and fi ↑
f , then f (·, y) = limi fi (·, y) is measurable for almost all y since each
fi (·, y) is measurable
R for almost
R all y. By the Monotone Convergence
Theorem f (x, y) dx = limi fi (x, y) dx for almost all y, and
ZZ ZZ Z Z
f (x, y) dx dy = lim fi (x, y) dx dy = lim fi = f.
i i
Step 2. Thus it now suffices to show that F contains the characteristic
function χA of every measurable set A, equivalently
(iv) Ay := {x | (x, y) ∈ A} ∈ M for almost all y,
(v) yR 7→ m(Ay ) is measurable, and
(vi) m(Ay ) dy = m(A),
since
χA (·, y) = χAy ,
Z Z
χA (x, y) dx = χAy = m(Ay ),
ZZ Z
χA (x, y) dx dy = m(Ay ) dy, and
Z
χA = m(A).
Let C denote the family of measurable sets satisfying (iv)–(vi). We will
show that C = M in a sequence of steps for different classes of A. First
S the properties of F imply that C is closed under countable
note that
unions i Ai provided either A1 , A2 , . . . are disjoint or A1 ⊂ A2 ⊂ · · · ,
since χSi Ai is the sum of the χAi in the first case and the increasing
limit of the χAi in the second case.
Step 3. Let A be a box. Then A = B × C for boxes B ⊂ Rn and
C ⊂ Rm . If y ∈ Rm then
(
B if y ∈ C
Ay =
∅ if y ∈
/ C,
so Ay ∈ M. Then m(Ay ) = m(B)χC (y), so y 7→ m(Ay ) is measurable,
and Z Z
m(Ay ) dy = m(B)χC = m(B)m(C) = m(A).
3
Step 4. It now follows that C contains every open set, because open
sets are countable disjoint unions of boxes.
Step 5. Next we show that if A, B ∈ C with A ⊂ B and m(A) < ∞
(in particular, if A is bounded), then B \ A ∈ C: we have
(B \ A)y = By \ Ay ,
which is measurable for almost all y. Since
Z
∞ > m(A) = m(Ay ) dy,
for almost all y we have m(Ay ) < ∞, hence m((B \ A)y ) = m(Ay ) −
m(Ay ). Thus y 7→ m((B \ A)y ) is measurable. Since y 7→ m(Ay ) is
integrable, we have
Z Z
m((B \ A)y ) dy = m(By ) − m(Ay ) dy
Z Z
= m(By ) dy − m(Ay ) dy
= m(B) − m(A) = m(B \ A).
Step 6. Next we show T that if A1 , A2 , · · · ∈ C with A1 bounded and
A1 ⊃ A2 ⊃ · · · , then i Ai ∈ C, which will imply that C contains all
bounded Gδ sets, since every bounded Gδ set can be expressed as a
countable decreasing intersection
S of bounded open sets. For each i ∈ N
put Bi = A1 \ Ai . S Then i Bi ∈ C, being an increasing union of
elements of C. Since i Bi is a bounded subset of A1 ,
\ [
Ai = A1 Bi ∈ C.
i i
Step 7. We now show that C contains every bounded null set A.
Choose a bounded null Gδ set B ⊃ A. Then B ∈ C, so
Z
0 = m(B) = m(By ) dy,
hence m(By ) = 0 for almost all y. Since Ay ⊂ By for all y ∈ Rm ,
m(Ay ) = 0 for almost all y. Thus Ay ∈ M for almost all y, y 7→ m(Ay )
is measurable, and
Z
m(Ay ) dy = 0 = m(A).
4
Step 8. Next we show that C contains every bounded measurable set
A. Choose a bounded Gδ set B ⊃ A such that m(B \ A) = 0. Then
B, B \ A ∈ C and B \ A is a bounded subset of B, so
A = B \ (B \ A) ∈ C.
Step 9. Finally, it follows that C = M, because every measurable set
is a countable increasing union of bounded measurable sets.
A similar argument proves the corresponding result with the roles of
x and y reversed.
Exercise 17.3. Let A and B be measurable subsets of R. Prove that
A × B is measurable in R2 , and m(A × B) = m(A)m(B).
Hint: the first part is not as trivial as it sounds; first of all, take
the statements out of order by using Tonelli’s Theorem to show that if
A × B is measurable then its measure is m(A)m(B), then for the first
part show that it suffices to consider A × R, and for this show that
the family of all sets A for which A × R is measurable is a σ-algebra
containing every Borel set and every null set. For the null sets, show
that it’s enough to consider Borel null sets.
Exercise 17.4. Let f be a measurable function on R, and define g
on R2 by g(x, y) = f (x). Prove that g is measurable. Hint: use the
definition of measurable function.
Exercise 17.5. Let f : R → [0, ∞) be measurable, and put
G = {(x, y) ∈ R2 | 0 ≤ y ≤ f (x)}.
Prove that G is measurable and
Z
m(G) = f.
Hint: for the first part, show that the function g : R2 → R defined by
g(x, y) = f (x) − y is measurable. For the second part, use Tonelli’s
Theorem to integrate with respect to y first.
Theorem 17.6 (Fubini’s Theorem). Same as Tonelli, but L1 instead
of L+ .
Proof. Let f ∈ L1 . Then f + ∈ L1 , so by Tonelli’s Theorem
Z ZZ
+
f = f + (x, y) dx dy < ∞,
so Z
y 7→ f + (x, y) dx is integrable,
5
R
hence f + (x, y) dx < ∞ for almost all y. Thus f + (·, y) ∈ L1 for almost
all y, and similarly for f − . Then
f (·, y) = f + (·, y) − f − (·, y) ∈ L1 for almost all y,
and then
Z Z Z
y 7→ f (x, y) dx = f (x, y) dx − f − (x, y) dx
+
is integrable.
Then, by Tonelli’s Theorem again,
ZZ ZZ ZZ
f (x, y) dx dy = +
f (x, y) dx dy − f − (x, y) dx dy
Z Z Z
+ −
= f − f = f.
Here’s how these theorems are used to evaluate multiple integrals:
Corollary 17.7. If either f ∈ L+ or f ∈ L1 , and if (i1 , . . . , in ) is a
rearrangement of (1, . . . , n), then
Z Z Z
f = · · · f (x1 , . . . , xn ) dxi1 · · · dxin .
Proof. Every permutation of (1, . . . , n) is a finite composition of trans-
positions of adjacent elements, and Tonelli-Fubini can be applied re-
peatedly, so without loss of generality
p
if j = p + 1
ij = p + 1 if j = p
j if j 6= p, p + 1
for some p = 1, . . . , n − 1. Applying Tonelli-Fubini twice, we have
Z Z Z Z
f= f (x1 , . . . , xn )
Rn−p−1 R2 Rp−1
d(x1 , . . . , xp−1 ) d(xp , xp+1 ) d(xp+2 , . . . , xn )
Again by Tonelli-Fubini, the order of integration with respect to (xp , xp+1 )
can be reversed, so we’re done.
Example 17.8. Define f on A := (0, 1) × (0, 1) by
x−y
f (x, y) = .
(x + y)3
We’ll show that
Z 1Z 1 Z 1Z 1
1 1
f (x, y) dy dx = and f (x, y) dx dy = − .
0 0 2 0 0 2
By Fubini’s Theorem, we can conclude that f is not integrable on A.
6
For the first iterated integral, we start with
Z 1
x−y
3
dy.
0 (x + y)
Note that this integral is infinite when x = 0, but we can ignore this
because the iterated integral can be regarded as
x−y
Z Z
3
dy dx,
(0,1) (0,1) (x + y)
so without loss of generality we assume y > 0.
We have
x−y 2x − (x + y) 2x 1
3
= 3
= 3
− ,
(x + y) (x + y) (x + y) (x + y)2
so
Z 1 1 1
x−y −x 1 y 1
3
dy = + = = .
0 (x + y) (x + y)2 x + y y=0 (x + y)2 y=0 (x + 1)2
Therefore 1
1 1
x−y −1
Z Z
1
3
dy dx = = .
0 0 (x + y) x+1 0 2
For the other iterated integral, we can use symmetry:
Z 1Z 1 Z 1Z 1
x−y y−x 1
3
dx dy = − 3
dx dy = − .
0 0 (x + y) 0 0 (y + x) 2
3
Example 17.9. Define
R R f (x, y) = 1/x if 0 < R Ry < |x| < 1 and 0
otherwise. Then R R f (x, y) dx dy = 0 and R R f (x, y) dy dx does
not exist.
Exercise 17.10. Define f (x, y) = xy/(x2 + y 2 )2 away from the origin,
and let f (0, 0) = 0. Prove:
R R R R
(a) R R f (x, y) dx dy = R R f (x, y) dy dx = 0, but
(b) f is not integrable.
Exercise 17.11. Let f be integrable on (0, a) (meaning that f is mea-
surable on (0, a), and the trivial extension of f to R, with constant
value 0 outside (0, a), is integrable). Prove that
Z aZ a Z a
f (x)
dx dy = f.
0 y x 0
Hint: use Tonelli’s Theorem to prove that the function (x, y) 7→ f (x)/x
is integrable on the set
A := {(x, y) | 0 ≤ y ≤ a, y ≤ x ≤ a},
then use Fubini’s Theorem to interchange the order of integration.
7
Example 17.12. Although Lebesgue integrals subsume Riemann in-
tegrals, they do do not subsume improper integrals. For example, you
probably learned in a previous analysis course that the improper inte-
gral Z ∞
sin x
dx
x
0
diverges, but the improper integral
Z ∞
sin x
dx
0 x
converges. In the language of Lebesgue integrals, the function x 7→
sin x/x is not integrable on (0, ∞), although the limit
Z b
sin x
lim dx
b→∞ 0 x
exists. In the next exercise you’ll use Tonelli-Fubini to compute the
value of this limit.
Exercise 17.13. In this exercise you’ll show that
Z b
sin x π
lim dx = .
b→∞ 0 x 2
−xy
Let f (x, y) = e sin x and A = (0, b) × (0, ∞) (where without loss of
generality b > 0).
(a) Prove that f is integrable on A by carefully estimating |f |.
Rb
(b) Integrate f over A in one order to get 0 sin x/x dx.
(c) Integrate f over A in the other order to get
Z ∞ −by Z ∞ −by
π ye e
− sin b 2
dy − cos b dy.
2 0 1+y 0 1 + y2
(d) Show how the Dominated Convergence Theorem can be used
to help prove that both of the above integrals go to 0 as b → ∞.
(e) Indicate how this implies the desired result.
Exercise 17.14. Let 0 < a < b < ∞. Integrate e−xy over A :=
(0, ∞) × (a, b) in two different ways to help show that
Z ∞ −ax
e − e−bx b
dx = log .
0 x a
