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Real Analysis Chapter 2 Solutions Jonathan Conder

1. Suppose f is measurable. Then f−1(−∞) ∈ M and f−1(∞) ∈ M, because −∞ and ∞ are Borel sets. If

B ⊆ R is Borel then f−1(B) ∈M, and hence f−1(B) ∩ Y ∈M (since R is also Borel). Thus f is measurable on Y.

Conversely, suppose that f−1(−∞) ∈ M, f−1(∞) ∈ M and f is measurable on Y. Let B ⊆ R be Borel. Then

f−1(B) ∩ Y ∈ M, and f−1(B) = (f−1(B) ∩ Y ) ∪ (f−1(B) \ Y ). Clearly f−1(B) \ Y = f−1(B ∩ −∞,∞), which is

measurable because it is either ∅, f−1(−∞), f−1(−∞) or f−1(−∞)∪f−1(−∞). This implies that f−1(B) ∈M, so f is measurable.

3. If fn : X → R for all n ∈ N, then g := lim infn→∞ fn and h := lim supn→∞ fn are measurable. Therefore f := h− g is

measurable (setting f(x) = 1 whenever g(x) = h(x) ∈ −∞,∞), and hence

x ∈ X | limn→∞

fn(x) exists = x ∈ X | lim infn→∞

fn(x) = lim supn→∞

fn(x) /∈ −∞,∞ = f−1(0) = f−1([−∞, 0]) ∈M.

If fn : X → C for all n ∈ N then (fn)n∈N converges at x ∈ X iff (Re(fn))n∈N and (Im(fn))n∈N converge at x. So

x ∈ X | limn→∞

fn(x) exists = x ∈ X | limn→∞

Re(fn(x)) exists ∩ x ∈ X | limn→∞

Im(fn(x)) exists ∈M

4. If a ∈ R then there is a sequence (an)n∈N in Q ∩ (a,∞) converging to a, and f−1((a,∞]) = ∪n∈Nf−1((an,∞]) ∈ M.

Since BR is generated by such intervals (a,∞], it follows that f is measurable.

5. Suppose that f is measurable, and let E be a measurable set from the codomain of f. Then f−1(E) ∈ M, so

f−1(E) ∩A, f−1(E) ∩B ∈M. Therefore f is measurable on A and on B.

Conversely, suppose that f is measurable on A and on B. Again let E be a measurable set from the codomain of f.

Then f−1(E) ∩A, f−1(E) ∩B ∈M, so f−1(E) = (f−1(E) ∩A) ∪ (f−1(E) ∩B) ∈M and f is measurable.

6. For example set X := R and M := L. There exists a non-measurable set A ⊆ X, and for each a ∈ A the set a is

measurable. Hence χaa∈A is a family of measurable functions, but its supremum is χA, which is not measurable

because χ−1A ([1,∞]) = A.

8. Since f is measurable iff −f is measurable, we may assume that f is increasing. Let a ∈ R and x ∈ f−1([a,∞)). If

y ∈ [x,∞) then f(y) ≥ f(x) ≥ a and hence y ∈ f−1([a,∞)). This shows that f−1([a,∞)) is an interval, so it is Borel

measurable and hence f is Borel measurable.

9. (a) If x, y ∈ [0, 1] and x < y, then g(x) = f(x) + x ≤ f(y) + x < f(y) + y = g(y) and hence g is injective. Since

g(0) = f(0) = 0 and g(1) = f(1) + 1 = 2, the intermediate value theorem implies that g maps [0, 1] onto [0, 2].

Let x ∈ [0, 2] and ε ∈ (0,∞). Then there exists x0 ∈ [0, 1] such that g(x0) = x. Define x1 := maxx0 − 12ε, 0

and x2 := minx0 + 12ε, 1. Then g(x1) ≤ x ≤ g(x2), and at least one of the inequalities is strict. Define

δ := min(x − g(x1), g(x2) − x \ 0). Given y ∈ [0, 2] and |y − x| < δ, it is straightforward to check that

g(x1) ≤ y ≤ g(x2). Indeed, if x = g(x1) then g(x1) = g(0) = 0, and otherwise g(x1) ≤ x− δ. Similarly x2 = 1 or

x+ δ ≤ g(x2). Since g is increasing, it is clear that x1 ≤ h(y) ≤ x2. This implies that x0 − 12ε ≤ h(y) ≤ x0 + 1

2ε,

so |h(y)− h(x)| = |h(y)− x0| < ε. Therefore h is continuous on [0, 2].

(b) Note that C = ∑

n∈N 3−nan | (an)n∈N is a sequence in 0, 2, which implies that

g(C) =

f

(∑n∈N

3−nan

)+∑n∈N

3−nan

∣∣∣∣∣ (an)n∈N is a sequence in 0, 2

1


=

∑n∈N

2−nan2

+∑n∈N

3−nan


=

∑n∈N

(2−n−1 + 3−n)an


.

Set C0 := [0, 2], and for each n ∈ N construct Cn from Cn−1 by removing an open interval of length 3−n from

the middle of each interval comprising Cn. This works because Cn−1 is the union of 2n−1 intervals of length

21−n + 31−n > 3−n (indeed, 20 + 30 = 2 and 12(21−n + 31−n − 3n) = 2−n + 3−n). Set C ′ := ∩n∈NCn, so that

m(C ′) = limn→∞

m(Cn) = limn→∞

2n(2−n + 3−n) = 1 + limn→∞

(2

3)n = 1.

Let x ∈ g(C) and N ∈ N. There exists a sequence (an)n∈N in 0, 2 such that x =∑

n∈N(2−n−1 +3−n)an. Clearly

0 ≤ x−N∑n=1

(2−n−1 + 3−n)an ≤∞∑

n=N+1

2(2−n−1 + 3−n) =2−N−1

1− 2−1+ 2

3−N−1

1− 3−1= 2−N + 3−N .

By induction on N it can be shown that∑N

n=1(2−n−1 + 3−n)an is the left endpoint of an interval from CN ,

because the N th term in the series is either 0 or 2−N + 2 · 3−N , the latter of which is the sum of length of the

intervals in CN and the length of the gaps between them. The above calculation therefore implies that x ∈ CN .It follows that x ∈ C ′, which shows that g(C) ⊆ C ′.Conversely, let x ∈ C ′, so that x ∈ Cn for all n ∈ N. For each n ∈ N define an ∈ 0, 2 depending on whether

the interval x belongs to in Cn is the left or right child of its parent in Cn−1. Then x and∑N

n=1(2−n−1 + 3−n)an

are from the same interval in CN , for all N ∈ N. In particular

limN→∞

∣∣∣∣∣x−N∑n=1

(2−n−1 + 3−n)an

∣∣∣∣∣ ≤ limN→∞

(2−N + 3−N ) = 0,

which implies that x =∑∞

n=1(2−n−1 + 3−n)an ∈ g(C). Therefore C ′ ⊆ g(C), and hence m(g(C)) = m(C ′) = 1.

(c) Since A ⊆ g(C), it is clear that B ⊆ C. Therefore m∗(B) ≤ m∗(C) = 0, so B is Lebesgue measurable. If B was

Borel measurable, then h−1(B) would be as well, because h is continuous. However h−1(B) = A, which is not

Borel. Hence B is not Borel measurable.

(d) Set F := χB and G := h. Then F is Lebesgue measurable because B ∈ L, and G is continuous by part (a). But

(F G)−1([1,∞)) = x ∈ [0, 2] | χB(h(x)) ∈ [1,∞) = x ∈ [0, 2] | h(x) ∈ B = h−1(B) = A /∈ L,

so F G is not Lebesgue measurable.

11. If n ∈ N and i ∈ Z then fn is clearly Borel measurable on [ai, ai+1], because fn|[ai,ai+1] is the sum of products of Borel

measurable functions. By an obvious generalisation of exercise 5, it follows that each fn is Borel measurable. Let

(x, y) ∈ R×Rk and ε ∈ (0,∞). Then there exists δ ∈ (0,∞) such that |f(x′, y)−f(x, y)| < ε for all x′ ∈ (x− δ, x+ δ).

Moreover, there exists N ∈ N such that 1N < δ. Let n ∈ N with n ≥ N, and choose i ∈ Z so that x ∈ [ai, ai+1]. Since

1n < δ it is clear that ai, ai+1 ∈ (x− δ, x+ δ). Therefore

|fn(x, y)− f(x, y)| =∣∣∣∣f(ai+1, y)(x− ai)− f(ai, y)(x− ai+1)− f(x, y)(ai+1 − ai)

ai+1 − ai

∣∣∣∣2


=

∣∣∣∣f(ai+1, y)(x− ai)− f(ai, y)(x− ai+1)− f(x, y)(x− ai) + f(x, y)(x− ai+1)

ai+1 − ai

∣∣∣∣=

∣∣∣∣(f(ai+1, y)− f(x, y))(x− ai)− (f(ai, y)− f(x, y))(x− ai+1)

ai+1 − ai

∣∣∣∣≤∣∣∣∣(f(ai+1, y)− f(x, y))(x− ai)

ai+1 − ai

∣∣∣∣+

∣∣∣∣(f(ai, y)− f(x, y))(x− ai+1)

ai+1 − ai

∣∣∣∣= |f(ai+1, y)− f(x, y)| · x− ai

ai+1 − ai+ |f(ai, y)− f(x, y)| · ai+1 − x

ai+1 − ai≤ ε · x− ai

ai+1 − ai+ ε · ai+1 − x

ai+1 − ai= ε.

This implies that (fn)n∈N converges to f pointwise, so f is Borel measurable. Clearly every function on R that is

continuous in each variable is Borel measurable. Let k ∈ N, and suppose that every function on Rk that is continuous

in each variable is Borel measurable. Also let g be a function on Rk+1 that is continuous in each variable. Then g(x, ·)is a function on Rk that is continuous in each variable, and hence Borel measurable, for each x ∈ R. From above,

it follows that g is Borel measurable. By induction, for each k ∈ N every function on Rk that is continuous in each

variable is Borel measurable.

13. Let E ∈M. By Fatou’s lemma∫Ef =

∫fχE =

∫lim infn→∞

fnχE ≤ lim infn→∞

∫fnχE = lim inf

n→∞

∫Efn

and similarly∫Ec f ≤ lim infn→∞

∫Ec fn. But fχE + fχEc = f and fnχE + fnχEc = fn for all n ∈ N, which implies

that∫Ec f =

∫f −

∫E f and (for sufficiently large n ∈ N)

∫Ec fn =

∫fn −

∫E fn. Therefore∫

f −∫Ef =

∫Ec

f ≤ lim infn→∞

∫Ec

fn = lim infn→∞

(∫fn −

∫Efn

)=

∫f − lim sup

n→∞

∫Efn,

so lim supn→∞∫E fn ≤

∫E f ≤ lim infn→∞

∫E fn and hence

∫E f = limn→∞

∫E fn.

Define F := (−∞, 0) and for each n ∈ N set Fn := F ∪ [n, n + 1). Then χF and each χFn are in L+, the sequence

(χFn)n∈N converges to χF pointwise and∫χF =∞ = limn→∞

∫χFn . However,

∫[0,∞) χF = 0 6= 1 = limn→∞

∫[0,∞) χFn .

14. Clearly λ(E) ≥ 0 for all E ∈M. Moreover, λ(∅) =∫∅ f dµ =

∫fχ∅ dµ =

∫0 dµ = 0. If Enn∈N is a pairwise disjoint

subcollection of M then (fχ∪Nn=1En)N∈N is a sequence of measurable functions increasing to fχ∪n∈NEn , so

λ(∪n∈NEn) =

∫∪n∈NEn

f dµ

=

∫fχ∪n∈NEn dµ

= limN→∞

∫fχ∪Nn=1En

dµ

= limN→∞

∫f

N∑n=1

χEn dµ

= limN→∞

N∑n=1

∫fχEn dµ

3


= limN→∞

N∑n=1

∫En

f dµ

=

∞∑n=1

∫En

f dµ

=∞∑n=1

λ(En)

by the monotone convergence theorem. Therefore λ is a measure. Now let g ∈ L+. If g is simple with standard

representation∑N

n=1 anχEn , then∫g dλ =

N∑n=1

anλ(En) =N∑n=1

an

∫En

f dµ =N∑n=1

an

∫fχEn dµ =

∫ N∑n=1

anfχEn dµ =

∫fg dµ.

Otherwise, there exists an increasing sequence (gn)n∈N of simple functions in L+ which converges pointwise to g, so

that (fgn)n∈N increases pointwise to fg and hence∫g dλ = lim

n→∞

∫gn dλ = lim

n→∞

∫fgn dµ =

∫fg dµ,

by two applications of the monotone convergence theorem.

15. Since∫f1 < ∞, the functions fnn∈N and f can be adjusted on a set of measure zero (namely f−1

1 (∞)) so that

they map into [0,∞). This does not affect their integrals. Clearly (f1−fn)n∈N increases pointwise to f1−f. Moreover

f1 − fn ∈ L+ for all n ∈ N. By the monotone convergence theorem∫

(f1 − f) = limn→∞∫

(f1 − fn). Therefore∫f =

∫f +

∫(f1 − f)−

∫(f1 − f)

=

∫f1 − lim

n→∞

∫(f1 − fn)

= limn→∞

(∫f1 −

∫(f1 − fn)

)= lim

n→∞

(∫fn +

∫(f1 − fn)−

∫(f1 − fn)

)= lim

n→∞

∫fn,

since∫

(f1 − f) ≤∫f1 <∞, and similarly

∫(f1 − fn) <∞ for all n ∈ N.

16. For each n ∈ N define En := x ∈ X | f(x) > n−1. Clearly (fχEn)n∈N increases pointwise to f, so by the monotone

convergence theorem (∫Enf)n∈N increases to

∫f. In particular, given ε ∈ (0,∞) there exists n ∈ N such that∫

Enf >

∫f − ε. Since

∫f <∞ it is clear that µ(En) <∞.

17. Let (fn)n∈N be an increasing sequence in L+, and set f := limn→∞ f. Then f ∈ L+, and by Fatou’s lemma∫f =

∫lim infn→∞

fn ≤ lim infn→∞

∫fn.

Since fn ≤ f and hence∫fn ≤

∫f for all n ∈ N, it is clear that lim supn→∞

∫fn ≤

∫f. Therefore

lim supn→∞

∫fn = lim inf

n→∞

∫fn = lim

n→∞

∫fn,

so∫f = limn→∞

∫fn.

4


18. Let g ∈ L+ ∩L1, and (fn : X → R)n∈N be a sequence of measurable functions such that fn ≥ −g for all n ∈ N. Define

h := lim infn→∞ fn. Clearly h ≥ −g, so h−(x) = max−h(x), 0 ≤ g(x) for all x ∈ X. It follows that h− ∈ L1 and

g − h− ∈ L+. Similarly f−n ∈ L1 and g − f−n , fn + g ∈ L+ for all n ∈ N. Therefore, by Fatou’s lemma∫h+

∫g =

∫h+ −

∫h− +

∫g

=

∫h+ +

∫(g − h−)

=

∫(h+ g)

=

∫lim infn→∞

(fn + g)

≤ lim infn→∞

∫(fn + g)

= lim infn→∞

(∫f+n +

∫(g − f−n )

)= lim inf

n→∞

(∫f+n +

∫g −

∫f−n

)= lim inf

n→∞

∫fn +

∫g.

Since∫g <∞, it follows that

∫lim infn→∞ fn =

∫h ≤ lim infn→∞

∫fn.

Let (fn : X → R)n∈N be a sequence of nonpositive measurable functions. Define h := lim supn→∞ fn. Then h ≤ 0 and

(−fn)n∈N is a sequence in L+, so by Fatou’s lemma

−∫h =

∫h− −

∫h+ =

∫−h =

∫lim infn→∞

−fn ≤ lim infn→∞

∫−fn = lim inf

n→∞

∫f−n = lim inf

n→∞−∫fn = − lim sup

n→∞

∫fn.

Therefore lim supn→∞∫fn ≤

∫h =

∫lim supn→∞ fn.

19. (a) There exists N ∈ N such that |f(x)− fn(x)| ≤ 1 for all x ∈ X and n ∈ N with n ≥ N. In particular

|f | = |f − fN + fN | ≤ |f − fN |+ |fN | ≤ 1 + |fN |,

and hence∫|f | dµ ≤

∫1+|fN | dµ = µ(X)+

∫|fN | dµ <∞. This implies that f ∈ L1(µ). Similarly 1+|f | ∈ L1(µ).

Since |fn| ≤ 1 + |f | for all n ∈ N with n ≥ N, the dominated convergence theorem implies that

limn→∞

∫fn dµ = lim

n→∞

∫fN+n dµ =

∫limn→∞

fN+n dµ =

∫f dµ.

(b) For each n ∈ N define fn := 2−nχ[−2n,2n]. Clearly (fn)n∈N converges uniformly to 0, but for each n ∈ N∫fn dµ = 2−nµ([−2n, 2n]) = 2 6= 0 =

∫0 dµ,

which implies that fn ∈ L1(µ) (where µ is the Lebesgue measure) and limn→∞∫fn dµ 6=

∫0 dµ.

20. It suffices to show that limn→∞∫

Re(fn) =∫

Re(f) and limn→∞∫

Im(fn) =∫

Im(f). Since limn→∞Re(fn) = Re(f)

and limn→∞ Im(fn) = Im(f) pointwise almost everywhere, while |Re(fn)| ≤ |fn| and | Im(fn)| ≤ |fn| for all n ∈ N,we may assume without loss of generality that f and each fn are real-valued. If N,n ∈ N such that n ≥ N, then

inf

∫gm +

∫fm

∞m=N

≤∫gn +

∫fn ≤ sup

∫gm

∞m=N

+

∫fn,

5


which implies that

inf

∫gm +

∫fm

∞m=N

≤ inf

sup

∫gm

∞m=N

+

∫fn

∞n=N

= sup

∫gm

∞m=N

+ inf

∫fm

∞m=N

.

Therefore

lim infn→∞

(∫gn +

∫fn

)≤ lim sup

n→∞

∫gn + lim inf

n→∞

∫fn =

∫g + lim inf

n→∞

∫fn.

Similarly

lim infn→∞

(∫gn −

∫fn

)≤∫g + lim inf

n→∞−∫fn =

∫g − lim sup

n→∞

∫fn.

Since gn + fn, gn − fn ∈ L+ for all n ∈ N, Fatou’s lemma implies that∫g +

∫f =

∫(g + f) =

∫lim infn→∞

(gn + fn) ≤ lim infn→∞

∫(gn + fn) = lim inf

n→∞

(∫gn +

∫fn

)≤∫g + lim inf

n→∞

∫fn

and∫g −

∫f =

∫(g − f) =

∫lim infn→∞

(gn − fn) ≤ lim infn→∞

∫(gn − fn) = lim inf

n→∞

(∫gn −

∫fn

)≤∫g − lim sup

n→∞

∫fn.

Since∫g <∞, it follows that lim supn→∞

∫fn ≤

∫f ≤ lim infn→∞

∫fn, and hence

∫f = limn→∞

∫fn.

21. Suppose that limn→∞∫|fn − f | = 0. For every ε ∈ (0,∞), there exists N ∈ N such that∣∣∣∣∫ |fn| − ∫ |f |∣∣∣∣ =

∣∣∣∣∫ (|fn| − |f |)∣∣∣∣ ≤ ∫ ∣∣|fn| − |f |∣∣ ≤ ∫ |fn − f | = ∣∣∣∣∫ |fn − f | − 0

∣∣∣∣ < ε

for all n ∈ N with n ≥ N (by the reverse triangle inequality). Therefore limn→∞∫|fn| =

∫|f |.

Conversely, suppose that limn→∞∫|fn| =

∫|f |. For each n ∈ N it is clear that |fn|+ |f | ∈ L1 and |fn−f | ≤ |fn|+ |f |,

so that |fn − f | ∈ L1. Moreover (|fn − f |)n∈N converges to 0 ∈ L1 pointwise almost everywhere. Also (|fn|+ |f |)n∈Nconverges to 2|f | ∈ L1 pointwise almost everywhere, and

limn→∞

∫(|fn|+ |f |) = lim

n→∞

∫|fn|+

∫|f | = 2

∫|f | =

∫2|f |.

Therefore, by the previous exercise, limn→∞∫|fn − f | =

∫0 = 0.

23. (a) Let x ∈ [a, b], and suppose that H(x) = h(x). Fix ε ∈ (0,∞). Since

limδ→0+

(sup f([a, b] ∩ [x− δ, x+ δ])− inf f([a, b] ∩ [x− δ, x+ δ])) = H(x)− h(x) = 0,

there exists η ∈ (0,∞) such that | sup f([a, b] ∩ [x− δ, x+ δ])− inf f([a, b] ∩ [x− δ, x+ δ])| < ε for all δ ∈ (0, η),

in particular for δ := η2 . If y ∈ [a, b] and |x− y| < δ then y ∈ [a, b] ∩ [x− δ, x+ δ], so

f(x)− f(y) ≤ sup f([a, b] ∩ [x− δ, x+ δ])− inf f([a, b] ∩ [x− δ, x+ δ]) < ε

and similarly f(y)− f(x) < ε, so |f(x)− f(y)| < ε. This shows that f is continuous at x.

Conversely, suppose that f is continuous at x. Let ε ∈ (0,∞). There exists η ∈ (0,∞) such that |f(x)−f(y)| < ε3

for all y ∈ [a, b] with |x− y| < η. Therefore f(x)− ε3 < f(y) < f(x) + ε

3 for all y ∈ [a, b] ∩ (x− η, x+ η), so

sup f([a, b] ∩ [x− δ, x+ δ])− inf f([a, b] ∩ [x− δ, x+ δ]) ≤ f(x) +ε

3− f(x) +

ε

3< ε

for all δ ∈ (0, η). This shows that

H(x)− h(x) = limδ→0+

(sup f([a, b] ∩ [x− δ, x+ δ])− inf f([a, b] ∩ [x− δ, x+ δ])) = 0.

6


(b) Choose a nested sequence (Pn)n∈N of partitions of [a, b] such that (SPnf)n∈N converges to Iba(f). For each n ∈ N

let En be the set of endpoints of the intervals comprising Pn, so that m(En) = 0. Define E := ∪n∈NEn, so that

m(E) = 0. Let x ∈ [a, b] \ E, and choose δ ∈ (0,∞) such that

sup f([a, b] ∩ [x− δ, x+ δ]) < H(x) +ε

2.

There exists N ∈ N such that SPnf < Iba(f) + εδ

2 for all n ∈ N with n ≥ N. Fix n ∈ N with n ≥ N. There is an

interval [a′, b′] in Pn such that x ∈ (a′, b′) (because x /∈ En). If [a′, b′] ⊆ [x− δ, x+ δ] then

GPn(x) = sup f([a′, b′]) ≤ sup f([a, b] ∩ [x− δ, x+ δ]) < H(x) +ε

2< H(x) + ε.

Otherwise a′ < x− δ or x+ δ < b′, so that [x− δ, x] or [x, x+ δ] is contained in (a′, b′). Construct a new partition

P ′n of [a, b] from Pn by inserting x and one of x− δ or x+ δ between a′ and b′. In the former case

SP ′nf − SPnf = sup f([a′, x− δ])(x− δ − a′) + sup f([x− δ, x])δ + sup f([x, b′])(b′ − x)−GPn(x)(b′ − a′)

< GPn(x)(x− δ − a′) +(H(x) +

ε

2

)δ +GPn(x)(b′ − x)−GPn(x)(b′ − a′)

= GPn(x)(x− δ − a′ + b′ − x− b′ + a′) +(H(x) +

ε

2

)δ

= GPn(x)(−δ) +(H(x) +

ε

2

)δ

=(H(x)−GPn(x) +

ε

2

)δ,

which still holds for the latter case, by a similar calculation. It follows that

GPn(x) <1

δ(SPnf − SP ′nf) +H(x) +

ε

2<

1

δ

(Iba(f) +

εδ

2− SP ′nf

)+H(x) +

ε

2≤ H(x) + ε.

Since x ∈ (a′, b′), there exists η ∈ (0,∞) such that [x− η, x+ η] ⊆ (a′, b′). This implies that

H(x) = infζ∈(0,∞)

sup f([a, b] ∩ [x− ζ, x+ ζ]) ≤ sup f([a, b] ∩ [x− η, x+ η]) ≤ sup f([a′, b′]) = GPn(x).

Therefore |GPn(x)−H(x)| < ε, so (GPn(x))n∈N converges to H(x) and hence (GPn)n∈N converges to H pointwise

almost everywhere. Since f is bounded and m([a, b]) <∞, the dominated convergence theorem implies that∫[a,b]

H dm = limn→∞

∫GPn dm = lim

n→∞SPnf = I

ba(f).

A similar argument implies that (gPn)n∈N converges to h pointwise almost everywhere, for all nested sequences

(Pn)n∈N of partitions of [a, b] such that (sPnf)n∈N converges to Iba(f). Therefore∫

[a,b] h dm = Iba(f).

25. (a) By the monotone convergence theorem and Theorem 2.28,∫f = lim

n→∞

∫ 1

1/nx−1/2 dx = lim

n→∞2x1/2

∣∣∣11/n

= limn→∞

(2− 2n−1/2) = 2. (1)

Therefore∫|g| =

∑∞n=1 2−n

∫f(x − rn) dx =

∑∞n=1 2−n

∫f = 2, by the monotone convergence theorem. It

follows that g ∈ L1(m), and g <∞ almost everywhere by Proposition 2.20.

7


(b) Let E ⊆ R be a null set and suppose that h ∈ L1(m) is equal to g on Ec. If I ⊆ R is an interval with at least

two points, there exists n ∈ N such that rn is an interior point of I. For each k ∈ N note that (rn, rn + k−1) ∩ Ihas positive measure, so there exists xk ∈ ((rn, rn + k−1) ∩ I) \ E. Clearly limk→∞ xk = rn, in which case

limk→∞ 2−nf(xk−rn) = 2−n limk→∞(xk−rn)−1/2 =∞. But 2−nf(xk−rn) ≤ g(xk) = h(xk) for all k ∈ N, which

implies that h is unbounded on I. This shows that h is unbounded on every interval, so it is clearly everywhere

discontinuous.

(c) By part (a) g2 <∞ almost everywhere. If I ⊆ R is an interval with at least two points, there exists n ∈ N such

that rn is an interior point of I. There exists δ ∈ (0, 1) such that (rn, rn + δ) ⊆ I, and∫Ig2 ≥

∫ rn+δ

rn

g2 ≥∫ rn+δ

rn

2−2nf(x− rn)2 dx = 2−2n

∫ δ

0f2 = 2−2n

∫ δ

0x−1 dx =∞,

where the last step follows from an argument similar to (1) (and is even in the undergraduate calculus textbooks).

26. Let x ∈ R and ε ∈ (0,∞). For each n ∈ N define fn := |f |χ[x−2−n,x+2−n], so that (fn)n∈N is a sequence in L1(m)

which is dominated by |f | ∈ L1(m). Moreover (fn)n∈N converges to 0 pointwise almost everywhere Therefore

limn→∞

∫fn =

∫0 = 0,

by the dominated convergence theorem. Choose n ∈ N such that∣∣∫ fn∣∣ < ε, and let y ∈ (x− 2−n, x+ 2−n). Then

|F (x)− F (y)| =∣∣∣∣∫ x

−∞f(t) dt−

∫ y

−∞f(t) dt

∣∣∣∣=

∣∣∣∣∫ fχ(−∞,x] −∫fχ(−∞,y]

∣∣∣∣=

∣∣∣∣∫ f · (χ(−∞,x] − χ(−∞,y])

∣∣∣∣≤∫|f | · |χ(−∞,x] − χ(−∞,y]|

=

∫|f | · χ[minx,y,maxx,y]

≤∫|f | · χ[x−2−n,x+2−n]

=

∫fn

=

∣∣∣∣∫ fn

∣∣∣∣< ε.

This shows that then F is continuous at x, and hence F is continuous on R.

28. (a) Fix x ∈ [0,∞) and define f : [1,∞)→ (0, 1] by f(n) := (1 + xn)−n. Then

(log f)′(n) = − log(

1 +x

n

)− n(−xn−2)

1 + xn

=x

n+ x− log

(1 +

x

n

)for all n ∈ [1,∞). Note that

exp

(x

n+ x

)=∞∑k=0

1

k!

(x

n+ x

)k≤ 1 +

∞∑k=1

(x

n+ x

)k= 1 +

xn+x

1− xn+x

= 1 +x

n+xn

n+x

= 1 +x

n

8


and hence xn+x ≤ log

(1 + x

n

)for all n ∈ [1,∞), so log f is a decreasing function. Therefore f = exp log f

is decreasing, so the sequence (fn)n∈N is decreasing as well, where each fn : [0,∞) → (0, 1] is defined by

fn(x) := (1 + xn)−n. In particular fn ≤ f2 for all n ∈ N with n ≥ 2. By the monotone convergence theorem∫ ∞

0f2 = lim

n→∞

∫ ∞0

f2χ[0,n] = limn→∞

∫ n

0

(1 +

x

2

)−2dx = lim

n→∞

∫ n

0

d

dx

(−2(

1 +x

2

)−1)dx,

so by the fundamental theorem of calculus∫ ∞0

f2 = limn→∞

(2(1 + 0)−1 − 2

(1 +

n

2

)−1)

= 2− limn→∞

2

1 + n2

= 2.

Therefore f2 ∈ L1, so (since | sin | ≤ 1) by the dominated convergence theorem

limn→∞

∫ ∞0

(1 +

x

n

)−nsin(xn

)dx =

∫ ∞0

limn→∞

(1 +

x

n

)−nsin(xn

)dx =

∫ ∞0

0 dx = 0.

Indeed, if x ∈ [0,∞) then limn→∞ sin(xn) = sin(limn→∞xn) = sin(0) = 0, and

limn→∞

(1 +

x

n

)−n= exp

(limn→∞

(−n log

(1 +

x

n

)))= exp

(− limn→∞

(log(1 + x

n

)1n

))

= exp

(−x lim

n→∞

(log(1 + x

n

)− log(1)

xn

))= exp

(−x log′(1)

)= e−x.

(b) If x ∈ [0, 1] and n ∈ N with n ≥ 1 then 0 ≤ (1 + nx2)(1 + x2)−n ≤ 1 because

(1 + x2)n =n∑k=0

(n

k

)(x2)k = 1 + nx2 +

n∑k=2

(n

k

)(x2)k ≥ 1 + nx2.

Moreover∫ 1

0 1 dx = 1. Hence, by the dominated convergence theorem

limn→∞

∫ 1

0(1 + nx2)(1 + x2)−n dx =

∫ 1

0limn→∞

(1 + nx2)(1 + x2)−n dx =

∫ 1

00 dx = 0.

Indeed, (1 + nx2)(1 + x2)−n ≤ (1 + nx2)(1 + nx2 +(n2

)x4)−1 for all x ∈ [0, 1] and n ∈ N with n ≥ 2, and

limn→∞

1 + nx2

1 + nx2 +(n2

)x4

= limn→∞

1 + nx2

1 + nx2 + n(n−1)2 x4

= limn→∞

n−2 + n−1x2

n−2 + n−1x2 + n−12n x

4=

0 + 0

0 + 0 + 12x

4= 0

for all x ∈ (0, 1] (and hence almost all x ∈ [0, 1]).

(c) Define f : [0,∞) → R by f(x) := sin(x) − x. Since f(0) = 0 and f ′(x) = cos(x) − 1 ≤ 0 for all x ∈ [0,∞),

this function is nonpositive and hence sin(x) ≤ x for all x ∈ [0,∞). Moreover sin(x) ≥ 0 for all x ∈ [0, 1], so

−x ≤ sin(x) for all x ∈ [0,∞). By the monotone convergence theorem and the fundamental theorem of calculus,∫ ∞0

(1+x2)−1 dx = limn→∞

∫ ∞0

(1+x2)−1χ[0,n](x) dx = limn→∞

∫ n

0

d

dxtan−1(x) dx = lim

n→∞(tan−1(n)−tan−1(0)) =

π

2.

9


Since | sin(xn)/xn | ≤ 1 for all x ∈ [0,∞) and n ∈ N, the dominated convergence theorem implies that

limn→∞

∫ ∞0

n sin(xn

)(x(1 + x2))−1 dx = lim

n→∞

∫ ∞0

sin(xn)xn

(1 + x2)−1 dx

=

∫ ∞0

limn→∞

sin(xn)xn

(1 + x2)−1 dx

=

∫ ∞0

limn→∞

sin(xn)− sin(0)xn

(1 + x2)−1 dx

=

∫ ∞0

sin′(0)(1 + x2)−1 dx

=

∫ ∞0

cos(0)(1 + x2)−1 dx

=

∫ ∞0

(1 + x2)−1 dx

=π

2.

(d) Fix n ∈ N. By the monotone convergence theorem and the fundamental theorem of calculus,∫ ∞a

n(1 + n2x2)−1 dx = limm→∞

∫ m

a

d

dxtan−1(nx) dx = lim

m→∞tan−1(nm)− tan−1(na) =

π

2− tan−1(na).

Therefore

limn→∞

∫ ∞a

n(1 + n2x2)−1 dx =

0, a > 0

π2 , a = 0

π, a < 0.

This implies that there is no f ∈ L1 such that limn→∞∫∞a n(1 + n2x2)−1 dx =

∫f, by a similar argument to

exercise 26. So the usual convergence theorem approach would not have helped to solve this exercise..

29. If t ∈ (0,∞), then (by the monotone convergence theorem and Theorem 2.28)∫ ∞0

e−tx dx = limk→∞

∫ k

0e−tx dx = lim

k→∞

e−tx

−t

∣∣∣∣k0

= limk→∞

(−e−tk

t+e0

t

)=

1

t.

Given n ∈ N, define fn : [0,∞)× [12 , 2] → [0,∞) by fn(x, t) := xne−tx. We claim that

∫∞0 fn(x, t) dx = n! t−(n+1) for

each t ∈ [12 , 2]. By induction, we may assume that

∫∞0 fn−1(x, t) dx = (n− 1)! t−n. Note that

fn−1(x, t) = xn−1e−tx ≤ 3n−1(n− 1)! ex/3e−tx = 3n−1(n− 1)! e−(t−1/3)x ≤ 3n−1(n− 1)! e−(1/6)x

for all x ∈ [0,∞), so fn−1(−, t) ∈ L1([0,∞)). Moreover∣∣∣∣ ∂∂tfn−1(x, t)

∣∣∣∣ = |−xne−tx| = fn(x, t) ≤ 3nn! e−(1/6)x

for all x ∈ [0,∞), so by Theorem 2.27∫ ∞0

fn(x, t) dx = − ∂

∂t

∫ ∞0

fn−1(x, t) dx = − ∂

∂t(n− 1)! t−n = n(n− 1)! t−(n+1) = n! t−(n+1),

as claimed. In particular∫∞

0 xne−x dx = n!.

10


If t ∈ (0,∞), then∫∞−∞ e

−tx2 dx =√π/t. The easiest proof of this requires a theorem from later in the course, but

if you are curious you can look up an alternative proof on Wikipedia. Given n ∈ N, define fn : R × [12 , 2] → [0,∞)

by fn(x, t) := x2ne−tx2. We claim that

∫∞−∞ fn(x, t) dx = (2n)!

4nn!

√πt−(2n+1)/2 for each t ∈ [1

2 , 2]. By induction, we may

assume that∫∞−∞ fn−1(x, t) dx = (2n−2)!

4n−1(n−1)!

√πt−(2n−1)/2. Note that

fn−1(x, t) = x2n−2e−tx2 ≤ 3n−1(n− 1)! ex

2/3e−tx2

= 3n−1(n− 1)! e−(t−1/3)x2 ≤ 3n−1(n− 1)! e−(1/6)x2

for all x ∈ R, so fn−1(−, t) ∈ L1(R). Moreover∣∣∣∣ ∂∂tfn−1(x, t)

∣∣∣∣ = |−x2ne−tx2 | = fn(x, t) ≤ 3nn! e−(1/6)x2

for all x ∈ R, so by Theorem 2.27∫ ∞−∞

fn(x, t) dx = − ∂

∂t

∫ ∞−∞

fn−1(x, t) dx

= − ∂

∂t

(2n− 2)!

4n−1(n− 1)!

√πt−(2n−1)/2

=(2n− 2)!(2n− 1)

2 · 4n−1(n− 1)!

√πt−(2n+1)/2

=(2n)!

4n · 4n−1(n− 1)!

√πt−(2n+1)/2

=(2n)!

4nn!

√πt−(2n+1)/2,

as claimed. In particular∫∞−∞ x

2ne−x2dx = (2n)!

4nn!

√π.

30. For each k ∈ N, we claim that (1− k−1x)k ≤ e−x for all x ∈ (0, k). If so, xn(1− k−1x)kχ(0,k)(x) ≤ xne−x for all k ∈ Nand x ∈ [0,∞), and by the previous exercise we may apply the dominated convergence theorem to show that

limk→∞

∫ k

0xn(1− k−1x)k dx =

∫ ∞0

limk→∞

xn(1− k−1x)kχ(0,k) dx =

∫ k

0xne−x dx = n!

(to prove that limk→∞(1 − k−1x)k = e−x, take logarithms and apply l’Hopital’s rule). Now we prove the claim. It

suffices to show that k log(1− k−1x) + x ≤ 0 for all x ∈ (0, k). This is certainly true for x = 0. Moreover,

d

dx(k log(1− k−1x) + x) =

k(−k−1)

1− k−1x+ 1 = 1− 1

(1− k−1x)=−k−1x

1− k−1x< 0

for all x ∈ (0, k). By the mean value theorem k log(1− k−1x) + x = (k log(1− k−1x) + x)− (k log(1− 0) + 0) < 0 for

all x ∈ (0, k), which proves the claim.

33. There clearly exists a subsequence (∫fnk

)k∈N of (∫fn)n∈N such that limk→∞

∫fnk

= lim infn→∞∫fn. Moreover

(fnk)k∈N converges to f in measure, because for every ε ∈ (0,∞)

limk→∞

µ(x ∈ X | |fnk(x)− f(x)| ≥ ε) = lim

n→∞µ(x ∈ X | |fn(x)− f(x)| ≥ ε) = 0.

In particular (fnk)k∈N is Cauchy in measure, so it has a subsequence (fnki

)i∈N which converges to a measurable function

g pointwise almost everywhere. Clearly (fnki)i∈N also converges to g+ pointwise almost everywhere. Moreover, f = g+

almost everywhere because (fnki)i∈N converges in measure to both f and g+ (thus µ(x ∈ X | |f(x)−g+(x)| ≥ ε) < δ

for all ε, δ ∈ (0,∞)). Therefore, by Fatou’s lemma∫f =

∫g+ ≤ lim inf

i→∞

∫fnki

= limi→∞

∫fnki

= limk→∞

∫fnk

= lim infn→∞

∫fn.

11


34. (a) It suffices to show that limn→∞∫

Re(fn) =∫

Re(f) and limn→∞∫

Im(fn) =∫

Im(f). Since

x ∈ X | |Re(fn)(x)− Re(f)(x)| ≥ ε ∪ x ∈ X | | Im(fn)(x)− Im(f)(x)| ≥ ε ⊆ x ∈ X | |fn(x)− f(x)| ≥ ε

for all n ∈ N and ε ∈ (0,∞), while |Re(fn)| ≤ |fn| and | Im(fn)| ≤ |fn| for all n ∈ N, we may assume

without loss of generality that f and each fn are real-valued. Note that (fn)n∈N is Cauchy in measure, so it has

a subsequence which converges pointwise almost everywhere to a measurable function which equals f almost

everywhere. Therefore f ∈ L1. Since (g + fn)n∈N and (g − fn)n∈N are sequences of non-negative measurable

functions which converge in measure to g + f and g − f respectively, the previous exercise implies that∫g +

∫f =

∫(g + f) ≤ lim inf

n→∞

∫(g + fn) = lim inf

n→∞

(∫g +

∫fn

)=

∫g + lim inf

n→∞

∫fn

and ∫g −

∫f =

∫(g − f) ≤ lim inf

n→∞

∫(g − fn) = lim inf

n→∞

(∫g −

∫fn

)=

∫g − lim sup

n→∞

∫fn

Since∫g <∞, it follows that lim supn→∞

∫fn ≤

∫f ≤ lim infn→∞

∫fn, and hence

∫f = limn→∞

∫fn.

(b) Note that (|fn − f |)n∈N converges to 0 in measure, because

x ∈ X |∣∣|fn(x)− f(x)| − 0

∣∣ ≥ ε = x ∈ X | |fn(x)− f(x)| ≥ ε

for all n ∈ N and ε ∈ (0,∞). Moreover |fn − f | ≤ |fn|+ |f | ≤ 2g ∈ L1 for all n ∈ N. Therefore, by part (a),

limn→∞

‖fn − f‖1 = limn→∞

∫|fn − f | =

∫0 = 0.

This implies that (fn)n∈N converges to f in L1.

35. Suppose that (fn)n∈N converges to f in measure. For every ε ∈ (0,∞), limn→∞ µ(x ∈ X | |fn(x)− f | ≥ ε) = 0. In

particular, for every ε ∈ (0,∞) there exists N ∈ N such that

0− ε < µ(x ∈ X | |fn(x)− f | ≥ ε) < 0 + ε = ε

for all n ∈ N with n ≥ N.

Conversely, suppose that, for every ε ∈ (0,∞), there exists N ∈ N such that µ(x ∈ X | |fn(x) − f | ≥ ε) < ε

for all n ∈ N with n ≥ N. Let ε ∈ (0,∞) and δ ∈ (0,∞). Define η := minε, δ. There exists N ∈ N such that

µ(x ∈ X | |fn(x)− f | ≥ η) < η for all n ∈ N with n ≥ N. Therefore

µ(x ∈ X | |fn(x)− f | ≥ ε) ≤ µ(x ∈ X | |fn(x)− f | ≥ η) < η ≤ δ

for all n ∈ N with n ≥ N, which implies that limn→∞ µ(x ∈ X | |fn(x) − f | ≥ ε) = 0. This shows that (fn)n∈N

converges to f in measure.

37. (a) Let x ∈ X be a point where (fn)n∈N converges to f. Then

limn→∞

φ(fn(x)) = φ( limn→∞

fn(x)) = φ(f(x)),

so (φ fn)n∈N converges to φ f on the same set that (fn)n∈N converges to f.

12


(b) Suppose that (fn)n∈N converges to f uniformly, and let ε ∈ (0,∞). Since φ is uniformly continuous, there exists

δ ∈ (0,∞) such that |φ(w)− φ(z)| < ε for all w, z ∈ C with |w− z| < δ. Moreover, there exists N ∈ N such that

|fn(x) − f(x)| < δ for all x ∈ X and n ∈ N with n ≥ N. Therefore |φ(fn(x)) − φ(f(x))| < ε for all x ∈ X and

n ∈ N with n ≥ N. This shows that (φ fn)n∈N converges to φ f uniformly.

Now suppose that (fn)n∈N converges to f almost uniformly. For every ε ∈ (0,∞) there exists E ∈M such that

µ(E) < ε and (fn)n∈N converges to f uniformly on Ec, and hence (φ fn)n∈N converges to φ f uniformly on

Ec by the previous argument. This shows that (φ fn)n∈N converges to φ f almost uniformly.

Finally, suppose that (fn)n∈N converges to f in measure. Let ε ∈ (0,∞). There exists δ ∈ (0,∞) such that

|φ(w)− φ(z)| < ε for all w, z ∈ C with |w − z| < δ. Moreover

limn→∞

µ(x ∈ X | |fn(x)− f(x)| ≥ δ) = 0.

Clearly x ∈ X | |φ(fn(x))− φ(f(x))| ≥ ε ⊆ x ∈ X | |fn(x)− f(x)| ≥ δ for all n ∈ N, so

limn→∞

µ(x ∈ X | |φ(fn(x))− φ(f(x))| ≥ ε) = 0

and hence (φ fn)n∈N converges to φ f in measure.

(c) For each n ∈ N define a measurable function fn : R→ C by fn(x) := 2−n. Also define φ : C→ C by

φ(z) :=

1, Re(z) > 0

−1, Re(z) ≤ 0.

For all x ∈ R (fn(x))n∈N converges to 0, but (φ(fn(x)))n∈N = (1)n∈N does not converge to to φ(0) = −1.

Now define f : R→ C by f(x) := x, and for each n ∈ N define fn : R→ C by fn(x) := x+ 2−n. Clearly (fn)n∈N

is a sequence of measurable functions converging uniformly, almost uniformly and in measure to the measurable

function f. Define φ : C→ C by φ(z) := z2. Let E ⊆ R and suppose that (φfn)n∈N converges to φf uniformly

on E. Then there exists N ∈ N such that |φ(fn(x))− φ(f(x))| < 1 for all x ∈ E and n ∈ N with n ≥ N Since

|φ(fN (x))− φ(f(x))| = |(x+ 2−N )2 − x2| = |21−Nx+ 2−2N |

for all x ∈ E, it follows that E ⊆ (−2N−1 − 2−N−1, 2N−1 − 2−N−1). In particular µ(Ec) = ∞, so (φ fn)n∈N

does not converge to φ f uniformly or almost uniformly. If ε ∈ (0,∞) and n ∈ N, then

[2n−1ε,∞) ⊆ x ∈ R | |φ(fn(x))− φ(f(x))| ≥ ε

because |φ(fn(x))− φ(f(x))| = 21−nx+ 2−2n ≥ ε+ 2−2n for all x ∈ [2n−1ε,∞). Therefore

limn→∞

µ(x ∈ R | |φ(fn(x))− φ(f(x))| ≥ ε) =∞,

so (φ fn)n∈N does not converge to φ f in measure.

39. Let (fn)n∈N be a sequence of functions which converges to f almost uniformly. For each n ∈ N there exists En ∈ M

such that µ(En) < 2−n and (fn)n∈N converges to f uniformly (hence pointwise) on Ecn. Define E := ∩n∈N ∪∞k=n Ek.

Then µ(E) = limn→∞ µ(∪∞k=nEk) = 0, since µ(∪∞k=nEk) ≤ 21−n for all n ∈ N. Moreover, if x ∈ Ec there exists n ∈ Nsuch that x ∈ Ecn and hence limk→∞ fk(x) = f(x). Therefore (fn)n∈N converges to f pointwise almost everywhere.

Let ε ∈ (0,∞) and take E ∈M such that µ(E) < ε and (fn)n∈N converges to f uniformly on Ec. There exists N ∈ Nsuch that |fn(x)− f(x)| < ε for all x ∈ Ec and n ∈ N with n ≥ N. It follows that x ∈ X | |fn(x)− f(x)| ≥ ε ⊆ E,and hence µ(x ∈ X | |fn(x)− f(x)| ≥ ε) ≤ µ(E) < ε. By exercise 35, (fn)n∈N converges to f in measure.

13


40. Let (fn)n∈N be a sequence of complex-valued measurable functions that converge pointwise to some f : X → C on a

set A ⊆ X with µ(Ac) = 0. Suppose there exists g ∈ L1(µ) such that |fn| ≤ g for all n ∈ N. Then |f(x)| ≤ g(x) for

all x ∈ A. Fix k ∈ N, and for each n ∈ N define Ek,n := ∪∞m=nx ∈ A | |fm(x) − f(x)| ≥ 2k−1. If x ∈ Ek,1 there

exists m ∈ N such that |fm(x)− f(x)| ≥ 2k−1 and hence 2g(x) ≥ |fm(x)|+ |f(x)| ≥ 2k−1. Therefore k−1χEk,1≤ g, so

k−1χEk,1∈ L1 and hence µ(Ek,1) < ∞. Since ∩n∈NEk,n ⊆ A ∩ Ac = ∅, it follows that limn→∞ µ(En,k) = 0. Now let

ε ∈ (0,∞) and for each k ∈ N choose nk ∈ N so that µ(Enk,k) < 2−kε. Define E := (∪k∈NEnk,k)∪Ac. Then µ(E) < ε,

and for each δ ∈ (0,∞) there exists k ∈ N such that 2k−1 < δ, whence |fm(x)− f(x)| < 2k−1 < δ for all x ∈ Ec and

m ∈ N with m ≥ nk. This implies that (fn)n∈N converges to f uniformly on Ec.

42. Suppose that (fn)n∈N converges to f in measure. Given ε ∈ (0,∞), there exists N ∈ N such that

µ(x ∈ N | ε ≤ |fn(x)− f(x)|) < 1

for all n ∈ N with n ≥ N. This implies that x ∈ N | ε ≤ |fn(x)− f(x)| = ∅, and hence ‖fn − f‖u ≤ ε, for all n ∈ Nwith n ≥ N. Therefore (fn)n∈N converges uniformly to f.

Now suppose that (fn)n∈N converges uniformly to f. If ε ∈ (0,∞), there exists N ∈ N such that |fn(x) − f(x)| < ε

for all x, n ∈ N with n ≥ N. In particular µ(x ∈ N | ε ≤ |fn(x)− f(x)|) = 0 for all n ∈ N with n ≥ N, which implies

that (fn)n∈N converges to f in measure.

44. For each n ∈ N define En := f−1(Bn(0)) = x ∈ [a, b] | |f(x)| < n. Then limn→∞ µ(En) = µ(∪n∈NEn) = µ([a, b]), so

there exists m ∈ N such that µ([a, b])− µ(Em) < ε3 . Define g : R→ C by

g(x) :=

f(x), x ∈ Em0, x ∈ Ecm.

Then |g| ≤ mχEm ≤ mχ[a,b], so g ∈ L1(µ). Hence for each n ∈ N there exists a compactly supported continuous

function gn : R → C such that ‖gn − g‖1 < n−1. Clearly (gn)n∈N converges to g in measure, so there exists a

subsequence (gnk)k∈N which converges to g pointwise almost everywhere. After restricting these functions to [a, b],

Egoroff’s theorem implies that there exists F ⊆ [a, b] such that µ(F ) < ε3 and (gnk

)k∈N converges to g uniformly on

[a, b] \ F. By inner regularity there exists a compact set E ⊆ Em \ F such that µ(E) > µ(Em \ F )− ε3 and hence

µ([a, b] \ E) = µ([a, b])− µ(E) < µ([a, b])− µ(Em \ F ) +ε

3≤ µ([a, b]) + µ(F )− µ(Em) +

ε

3< 3 · ε

3= ε.

Moreover (gnk)k∈N converges to g uniformly on E, so f |E = g|E is continuous.

46. Fix y ∈ Y. Then χD(x, y) = χy(x) for all x ∈ X, so∫χD(x, y) dµ(x) = µ(y) = 0. This implies that∫∫

χD(x, y) dµ(x) dν(y) =

∫0 dν(y) = 0.

Now fix x ∈ X. Clearly χD(x, y) = χx(y) for all y ∈ Y, so∫χD(x, y) dν(y) = ν(x) = 1. It follows that∫∫

χD(x, y) dν(y) dµ(x) =

∫1 dµ(x) = µ(X) = 1.

By definition∫χD d(µ× ν) = (µ× ν)(D), and hence∫

χD d(µ× ν) = inf

∞∑n=1

(µ× ν)(En)

∣∣∣∣∣ (En)∞n=1 is a sequence of finite disjoint unions of rectangles covering D

14


= inf

∞∑n=1

µ(An)ν(Bn)

∣∣∣∣∣ (An ×Bn)∞n=1 is a sequence of rectangles covering D

If (An × Bn)∞n=1 is a sequence of rectangles covering D, then (An ∩ Bn)∞n=1 covers X. Clearly this implies that

µ∗(An ∩Bn) > 0 for some n ∈ N. In particular µ(An) > 0 and ν(Bn) =∞, because the Lebesgue outer measure of a

finite set is 0. Therefore∑∞

n=1 µ(An)ν(Bn) =∞, so∫χD d(µ× ν) = inf∞ =∞.

48. Clearly∫|f | d(µ × ν) = (µ × ν)(∪∞n=1(n, n), (n + 1, n)). If (An × Bn)∞n=1 is a sequence of rectangles covering

∪∞n=1(n, n), (n+ 1, n), then (An ∩Bn)∞n=1 covers N and hence∑∞

n=1 µ(An ∩Bn) =∞. This implies that

∞∑n=1

µ(An)ν(Bn) =

∞∑n=1

µ(An)µ(Bn) ≥∞∑n=1

µ(An ∩Bn)2 ≥∞∑n=1

µ(An ∩Bn) =∞,

since µ(An ∩ Bn) ∈ 0 ∪ [1,∞] for all n ∈ N. Therefore∫|f | d(µ × ν) = inf∞ = ∞. Fix n ∈ Y. Then f(m,n) =

χn(m)− χn+1(m) for all m ∈ X, and hence∫f(m,n) dµ(m) = µ(n)− µ(n+ 1) = 0. This implies that∫∫

f(m,n) dµ(m) dν(n) =

∫0 dν(n) = 0.

Now fix m ∈ X\1. Then f(m,n) = χm(n)−χm−1(n) for all n ∈ Y, so∫f(m,n) dν(n) = ν(m)−ν(m−1) = 0.

Moreover,∫f(1, n) dν(n) =

∫χ1 dν = ν(1) = 1. It follows that∫∫

f(m,n) dν(n) dµ(m) =

∫χ1 dµ = µ(1) = 1.

49. (a) Since µ and ν are σ-finite, ∫ν(Ex) dµ(x) =

∫µ(Ey) dν(y) = (µ× ν)(E) = 0.

This implies that ν(Ex) = µ(Ey) = 0 for almost every x ∈ X and y ∈ Y.

(b) Let E ⊆ X × Y be a null set such that f(x, y) = 0 for all x ∈ X and y ∈ Y such that (x, y) /∈ E. If x ∈ X, then

fx(y) = 0 for all y ∈ Y such that y /∈ Ex. Hence fx = 0 almost everywhere, so fx is integrable with∫fx dν = 0,

for almost all x ∈ X (by the previous exercise). Similarly fy is integrable and∫fy dµ = 0 for almost every

y ∈ Y.

Now let f be L-measurable. There exists an (M ⊗ N)-measurable function g such that f = g λ-almost everywhere.

Moreover gx is N-measurable and gy is M-measurable for all x ∈ X and y ∈ Y. If f ≥ 0 then g ≥ 0 without loss of

generality, so by Tonelli’s theorem x 7→∫gx dν and y 7→

∫gy dµ are non-negative and (M⊗N)-measurable, while∫

g dλ =

∫∫g(x, y) dµ(x) dν(y) =

∫∫g(x, y) dν(y) dµ(x). (2)

Since |g| = |f | λ-almost everywhere,∫|g| d(µ× ν) =

∫|g| dλ =

∫|f | dλ and hence g ∈ L1(µ× ν) whenever f ∈ L1(λ).

By Fubini’s theorem, this implies that gx ∈ L1(ν) and gy ∈ L1(µ) for almost all x ∈ X and y ∈ Y, while x 7→∫gx dν

and y 7→∫gy dµ are in L1(µ) and L1(ν) respectively. Also (2) holds in this case. The corresponding statements about

f follow by applying part (b) of this exercise to f−g. In particular, fx−gx = 0 almost everywhere for almost all x ∈ X,so fx is N-measurable for almost all x ∈ X. Similarly fy is M-measurable for almost all y ∈ Y. Since fx − gx ∈ L1(ν)

and fy − gy ∈ L1(µ) for almost all x ∈ X and y ∈ Y, it is clear that fx ∈ L1(ν) and fy ∈ L1(µ) for almost all x ∈ X

15


and y ∈ Y, provided that f ∈ L1(λ). In either of the two cases∫gx dν =

∫(fx − gx) dν +

∫gx dν =

∫fx dν for almost

all x ∈ X, so x 7→∫fx dν is measurable and in the second case, integrable (for the first case, assume without loss of

generality that g ≤ f). The same clearly holds for y 7→∫fy dµ, so (because f = g almost everywhere)∫

f dλ =

∫∫g(x, y) dµ(x) dν(y) =

∫∫f(x, y) dµ(x) dν(y) =

∫∫g(x, y) dν(y) dµ(x) =

∫∫f(x, y) dν(y) dµ(x).

50. Subtraction is a continuous map from [0,∞]×[0,∞)→ (−∞,∞], since it is constant on the closed set ∞×[0,∞). In

particular, the preimage of [0,∞) (or (0,∞)) under subtraction is an open subset of [0,∞]× [0,∞), so it is a countable

union of rectangles (An ×Bn)∞n=1. Hence, the preimage of [0,∞) (or (0,∞)) under the map (x, y) 7→ f(x)− y is

E := (x, y) ∈ X × [0,∞) | (f(x), y) ∈ An ×Bn for some n ∈ N

= ∪∞n=1(x, y) ∈ X × [0,∞) | x ∈ f−1(An) and y ∈ Bn

= ∪∞n=1(f−1(An)×Bn).

Clearly E is (M⊗BR)-measurable, and hence Gf = E ∪ (f−1(∞)× ∞) is also (M⊗BR)-measurable (the same

holds for the redefinition of Gf , because in that case Gf = E if we take (0,∞) instead of [0,∞) above). For the

second part, assume that m(∞) = 0 and m|[0,∞) agrees with the Lebesgue measure. By Tonelli’s theorem

(µ×m)(Gf ) = (µ×m)(E) =

∫χE =

∫∫χE(x, y) dm(y) dµ(x).

If x ∈ X then (x, y) ∈ E iff y ∈ [0,∞) and f(x) − y ≥ 0 (or > 0), so∫χE(x, y) dm(y) =

∫ f(x)0 1 dm(y) = f(x).

Therefore (µ×m)(Gf ) =∫f dµ as required.

51. (a) Define F,G : X × Y → C by F (x, y) := f(x) and G(x, y) := g(y). Then F−1(A) = f−1(A) × Y and G−1(A) =

X × g−1(A) for all A ⊆ C, so F and G are (M⊗N)-measurable. Therefore h = FG is (M⊗N)-measurable.

(b) Suppose f ≥ 0 and g ≥ 0. There exist increasing sequences (φn)∞n=1 and (ψn)∞n=1 of non-negative simple functions

which converge pointwise to f and g respectively. For each n ∈ N define Φn,Ψn : X × Y → [0,∞] as in part (a),

so that (ΦnΨn)∞n=1 converges pointwise to h. Fix n ∈ N, and write φn =∑k

i=1 aiχAi and φn =∑l

j=1 bjχBj for

some a1, a2, . . . , ak, b1, b2, . . . , bl ∈ [0,∞] and measurable sets A1, A2, . . . , Ak, B1, B2, . . . , Bl. Clearly

ΦnΨn =

(k∑i=1

aiχ(Ai×Y )

) l∑j=1

bjχ(X×Bj)

=k∑i=1

l∑j=1

aiχ(Ai×Y )bjχ(X×Bj) =k∑i=1

l∑j=1

aibjχ(Ai×Bj),

and hence∫ΦnΨn =

k∑i=1

l∑j=1

aibj(µ×ν)(Ai×Bj) =

k∑i=1

l∑j=1

aiµ(Ai)bjν(Bj) =

(k∑i=1

aiµ(Ai)

) l∑j=1

bjν(Bj)

=

∫φn·∫ψn.

By the monotone convergence theorem, it follows that∫h = lim

n→∞

∫ΦnΨn = lim

n→∞

∫φn ·

∫ψn = lim

n→∞

∫φn · lim

n→∞

∫ψn =

∫f ·∫g.

Hence, in general∫|h| =

∫|f | ·

∫|g| <∞, so h ∈ L1(µ× ν). If f(X) ⊆ R and g(Y ) ⊆ R, then∫

h =

∫h+ −

∫h−

16


=

∫F+G+ +

∫F−G− −

∫F+G− −

∫F−G+

=

∫f+ ·

∫g+ +

∫f− ·

∫g− −

∫f+ ·

∫g− −

∫f− ·

∫g+

=

∫f+

(∫g+ −

∫g−)

+

∫f−(∫

g− −∫g+

)=

(∫f+ −

∫f−)(∫

g+ −∫g−)

=

∫f ·∫g.

Since FG = (Re(F )+i Im(F ))(Re(G)+i Im(G)) = Re(F ) Re(G)−Im(F ) Im(G)+i(Re(F ) Im(G)+Im(F ) Re(G)),∫h =

∫Re(h) + i

∫Im(h)

=

∫Re(F ) Re(G)−

∫Im(F ) Im(G) + i

∫Re(F ) Im(G) + i

∫Im(F ) Re(G)

=

∫Re(f) ·

∫Re(g)−

∫Im(f) ·

∫Im(g) + i

∫Re(f) ·

∫Im(g) + i

∫Im(f) ·

∫Re(g)

=

∫Re(f)

(∫Re(g) + i

∫Im(g)

)−∫

Im(f)

(∫Im(g)− i

∫Re(g)

)=

∫Re(f)

(∫Re(g) + i

∫Im(g)

)+ i

∫Im(f)

(∫Re(g) + i

∫Im(g)

)=

(∫Re(f) + i

∫Im(f)

)(∫Re(g) + i

∫Im(g)

)=

∫f ·∫g.

55. (a) Fix y ∈ (0, 1], and define F : [0, 1]→ R by F (x) := x(x2 + y2)−1. By the quotient rule

F ′(x) =(x2 + y2)− x(2x)

(x2 + y2)2=

y2 − x2

(x2 + y2)2= −fy(x)

for all x ∈ [0, 1]. This implies that∫ 1

0(fy)− =

∫ y

0−fy = F (y)− F (0) = F (y) =

y

2y2=

1

2y

and similarly∫ 1

0 (fy)+ =∫ 1y f

y = −F (1)+F (y) = 12y −

11+y2

. By the Tonelli and monotone convergence theorems∫Ef− =

∫ 1

0

∫ 1

0f−(x, y) dx dy =

∫ 1

0

1

2ydy = lim

n→∞

∫ 1

1n

1

2ydy = lim

n→∞

− log( 1n)

2=∞,

which implies that ∫Ef+ =

∫ 1

0

∫ 1

0f+(x, y) dx dy =

∫ 1

0

(1

2y− 1

1 + y2

)dy =∞

because∫ 1

01

1+y2dy ≤

∫ 10 1 dy <∞ and∫ 1

0

(1

2y− 1

1 + y2

)dy +

∫ 1

0

1

1 + y2dy =

∫ 1

0

1

2ydy =∞

17


This shows that∫E f is not defined. However,∫ 1

0

∫ 1

0f(x, y) dx dy =

∫ 1

0(−F (1) + F (0)) dy = −

∫ 1

0

1

1 + y2dy = −(tan−1(1)− tan−1(0)) = −π

4.

Since f(x, y) = −f(y, x) for all x, y ∈ (0, 1], it follows that∫ 1

0

∫ 1

0f(x, y) dy dx =

π

4.

(b) Since f(x, y) ≥ 0 for all (x, y) ∈ E \ (1, 1), all three integrals exist and Tonelli’s theorem implies that they are

equal.

(c) By Tonelli’s theorem, the fundamental theorem of calculus and the monotone convergence theorem,∫Ef+ =

∫ 1

0

∫ 1

0f+(x, y) dx dy

=

∫ 12

0

∫ 1

12

+y

(x− 1

2

)−3

dx dy

=

∫ 12

0(−2)−1

((1

2

)−2

− y−2

)dy

=1

2

∫ 12

0(y−2 − 4) dy

= limn→∞

1

2

∫ 12

1n+4

(y−2 − 4) dy

= limn→∞

1

2

(−(

1

2

)−1

+

(1

n+ 4

)−1

− 4

2+

4

n+ 4

)= lim

n→∞

n

2

=∞.

Similarly ∫Ef− =

∫ 1

0

∫ 1

0f−(x, y) dx dy

=

∫ 12

0

∫ 12−y

0

(1

2− x)−3

dx dy

=

∫ 12

02−1

(y−2 −

(1

2

)−2)dy

=1

2

∫ 12

0(y−2 − 4) dy

=∞.

Therefore∫E f does not exist. However, the above working implies that∫ 1

0

∫ 1

0f(x, y) dx dy =

∫ 12

0

∫ 1

0f(x, y) dx dy

18


=

∫ 12

0

(∫ 1

0f+(x, y) dx−

∫ 1

0f−(x, y) dx

)dy

=

∫ 12

0

(1

2(y−2 − 4)− 1

2(y−2 − 4)

)dy

= 0.

If x ∈ [0, 12 ], then fx ≤ 0 and hence∫ 1

0f(x, y) dy = −

∫ 1

0f−(x, y) dy = −

∫ 12−x

0

(1

2− x)−3

dy = −(

1

2− x)−2

.

Similarly, if x ∈ [12 , 1] then∫ 1

0f(x, y) dy =

∫ 1

0f+(x, y) dy =

∫ x− 12

0

(x− 1

2

)−3

dy =

(x− 1

2

)−2

.

By the monotone convergence theorem and the fundamental theorem of calculus, this implies that∫ 1

0

(∫ 1

0f(x, y) dy

)+

dx =

∫ 1

12

(x− 1

2

)−2

dx

= limn→∞

∫ 1

12

+ 1n+2

(x− 1

2

)−2

dx

= limn→∞

(−(

1

2

)−1

+

(1

n+ 2

)−1)

= limn→∞

n

=∞.

Similarly ∫ 1

0

(∫ 1

0f(x, y) dy

)−dx =

∫ 12

0

(1

2− x)−2

dx

= limn→∞

∫ 12− 1

n+2

0

(1

2− x)−2

dx

= limn→∞

((1

n+ 2

)−1

−(

1

2

)−1)

= limn→∞

n

=∞.

This shows that∫ 1

0

∫ 10 f(x, y) dy dx does not exist.

56. Define a measurable function h : (0, a)2 → C by h(t, x) := t−1f(t)χE(t, x), where E := (t, x) ∈ (0, a)2 | x < t is

open, hence measurable. Then g(x) =∫ a

0 t−1f(t)χ(x,a)(t) dt =

∫hx for all x ∈ (0, a). By Tonelli’s theorem∫

|h| =∫ a

0

∫ a

0t−1|f(t)|χE(t, x) dx dt =

∫ a

0

∫ t

0t−1|f(t)| dx dt =

∫ a

0|f(t)| dt <∞.

19


Hence h is integrable, so g is integrable by Fubini’s theorem, and∫ a

0g =

∫ a

0

∫ a

0hx(t) dt dx =

∫h =

∫ a

0

∫ a

0t−1f(t)χE(t, x) dx dt =

∫ a

0

∫ t

0t−1f(t) dx dt =

∫ a

0f(t) dt.

58. Let s ∈ (0,∞) and define f : [0,∞) × [0, 1] → [0, 1] by f(x, y) := e−sx sin(2xy). Clearly |f(x, y)| ≤ e−sx for all

x, y ∈ [0,∞)× [0, 1], so f ∈ L1. Since∫ ∞0

∫ 1

0e−sx sin(2xy) dy dx =

∫ ∞0

(−e−sx cos(2xy)

2x

)∣∣∣∣10

dx =

∫ ∞0

e−sx1− cos(2x)

2xdx =

∫ ∞0

e−sxx−1 sin2(x) dx,

Fubini’s theorem implies that∫ ∞0

e−sxx−1 sin2(x) dx =

∫ 1

0

∫ ∞0

e−sx sin(2xy) dx dy =

∫ 1

0

2y

s2 + 4y2dy =

1

4log(4−1s2 + y2)

∣∣∣∣10

=1

4log(1 + 4s−2)

(the middle step can be done using integration by parts or by expressing sin as a difference of complex exponentials).

59. (a) Let n ∈ N. If x ∈ [(n+ 16)π, (n+ 5

6)π] then | sin(x)| ≥ 12 , and x−1 ≥ (n+ 1)−1π−1. Therefore∫ ∞

0|f | ≥

∫ ∞∑n=1

1

2(n+ 1)πχ[(n+ 1

6)π,(n+ 5

6)π] =

∞∑n=1

∫1

2(n+ 1)πχ[(n+ 1

6)π,(n+ 5

6)π] =

∞∑n=1

46π

2(n+ 1)π=∞,

by the monotone convergence theorem.

(b) Fix b ∈ (0,∞), and define f : (0, b)2 → R by f(x, y) := e−xy sin(x). Clearly |f | ≤ 1, so∫|f | ≤ b2 < ∞. Hence,

by Fubini’s theorem and the fundamental theorem of calculus∫ b

0

∫ b

0f(x, y) dx dy =

∫ b

0

∫ b

0e−xy sin(x) dy dx

=

∫ b

0

(e−xb sin(x)

−x− e0 sin(x)

−x

)dx

=

∫ b

0

(sin(x)

x− e−xb sin(x)

x

)dx (3)

Since | sin(x)| ≤ x for all x ∈ (0,∞) it is clear that∣∣∣∣∫ b

0

e−xb sin(x)

xdx

∣∣∣∣ ≤ ∫ b

0

∣∣∣∣e−xb sin(x)

x

∣∣∣∣ dx ≤ ∫ b

0e−xb dx =

e−b2

−b− e0

−b=

1− e−b2

b. (4)

Integrating f by parts twice with respect to x suggests we define a function F : (0, b)2 → R by

F (x, y) := −e−xy y sin(x) + cos(x)

y2 + 1,

so that

∂

∂xF (x, y) = ye−xy

y sin(x) + cos(x)

y2 + 1− e−xy y cos(x)− sin(x)

y2 + 1

= e−xyy2 sin(x) + y cos(x)− y cos(x) + sin(x)

y2 + 1

= f(x, y)

20


and hence ∫ b

0

∫ b

0f(x, y) dx dy =

∫ b

0(F (b, y)− F (0, y)) dy

=

∫ b

0

(e0 y sin(0) + cos(0)

y2 + 1− e−by y sin(b) + cos(b)

y2 + 1

)dy

=

∫ b

0

(1

y2 + 1− e−by y sin(b) + cos(b)

y2 + 1

)dy. (5)

Either y ≤ 1 or y ≤ y2 for all y ∈ (0,∞), so∣∣∣∣∫ b

0e−by

y sin(b) + cos(b)

y2 + 1dy

∣∣∣∣ ≤ ∫ b

0e−by

(y| sin(b)|y2 + 1

+| cos(b)|y2 + 1

)dy ≤

∫ b

02e−by dy = 2

1− e−b2

b. (6)

Together (3), (4), (5) and (6) imply that

limb→∞

∫ b

0

sin(x)

xdx+ 0 = lim

b→∞

∫ b

0

∫ b

0f(x, y) dx dy = lim

b→∞

∫ b

0

1

y2 + 1dy + 0 = lim

b→∞(tan−1(b)− tan−1(0)) =

π

2.

60. If x, y ∈ (0,∞) then, by Exercise 51

Γ(x)Γ(y) =

∫ ∞0

sx−1e−s ds

∫ ∞0

ty−1e−t dt =

∫ ∞0

∫ ∞0

sx−1ty−1e−(s+t) ds dt.

Define G : (0,∞) × (0, 1) → (0,∞)2 by G(u, v) := (uv, u(1 − v)), and check that G is a C1-diffeomorphism with

Jacobian determinant −u at the point (u, v). By Theorem 2.47, Tonelli’s theorem and Exercise 51, Γ(x)Γ(y) is∫ 1

0

∫ ∞0

(uv)x−1(u(1− v))y−1e−uu du dv =

∫ 1

0vx−1(1− v)y−1 dv

∫ ∞0

ux+y−1e−u du = Γ(x+ y)

∫ 1

0tx−1(1− t)y−1 dt.

61. (a) Let α, β ∈ (0,∞) and x ∈ [0,∞). Note that

Iα(Iβf)(x) =1

Γ(α)

∫ x

0(x− t)α−1Iβf(t) dt

=1

Γ(α)

∫ x

0(x− t)α−1 1

Γ(β)

∫ t

0(t− s)β−1f(s) ds dt

=1

Γ(α)Γ(β)

∫ x

0

∫ t

0(x− t)α−1(t− s)β−1f(s) ds dt.

Since f is bounded on [0, x] and∫ 1

0 tγ dt <∞ for all γ ∈ (−1,∞), the Fubini-Tonelli theorem implies that

Iα(Iβf)(x) =1

Γ(α)Γ(β)

∫ x

0

∫ x

s(x− t)α−1(t− s)β−1f(s) dt ds.

For each s we apply the substitution u := (t− s)/(x− s) to the inner integral, and obtain

Iα(Iβf)(x) =1

Γ(α)Γ(β)

∫ x

0

∫ 1

0(x− s− u(x− s))α−1(u(x− s))β−1f(s)(x− s) du ds

=1

Γ(α)Γ(β)

∫ x

0

∫ 1

0(x− s)α−1(1− u)α−1uβ−1(x− s)β−1f(s)(x− s) du ds

=1

Γ(α)Γ(β)

∫ x

0(x− s)α+β−1f(s) ds

∫ 1

0(1− u)α−1uβ−1 du

=1

Γ(α+ β)

∫ x

0(x− s)α+β−1f(s) ds

= Iα+βf(x).

21


(b) Clearly I1f is an antiderivative of f. Given n ∈ N, we aim to show that Inf is an nth-order antiderivative of f.

By induction, we may assume that n > 1 and that In−1(I1f) is an (n− 1)th-order antiderivative of I1f. Hence

(Inf)(n) = ((In−1(I1f))(n−1))′ = (I1f)′ = f.

62. Let E ⊆ Sn−1 be measurable and T ∈ SO(n) a rotation. We want to show that σ(T (E)) = σ(E). Note that

Φ−1((0, 1)× E) =

x ∈ Rn \ 0

∣∣∣∣ |x| ∈ (0, 1) andx

|x|∈ E

,

while

Φ−1((0, 1)× T (E)) =

x ∈ Rn \ 0

∣∣∣∣ |x| ∈ (0, 1) andx

|x|∈ T (E)

=

x ∈ Rn \ 0

∣∣∣∣ |T−1x| ∈ (0, 1) andT−1x

|T−1x|∈ E

= x ∈ Rn \ 0 | T−1x ∈ Φ−1((0, 1)× E)

= T (Φ−1((0, 1)× E)).

Therefore

ρ((0, 1))σ(E) = m∗((0, 1)×E) = m(Φ−1((0, 1)×E)) = m(T (Φ−1((0, 1)×E))) = m∗((0, 1)×T (E)) = ρ((0, 1))σ(T (E)).

Since ρ((0, 1)) =∫ 1

0 rn−1 dr = n−1 > 0, it follows that σ(T (E)) = σ(E).

64. Let a, b ∈ R and set α := a+ n− 1. By Corollary 2.51 it suffices to determine when∫ 1/2

0rα| log(r)|b dr and

∫ ∞2

rα| log(r)|b dr

are finite. We claim that, given ε ∈ (0,∞), there exists δ ∈ (0, 12) such that | log(r)| ≤ 2r−ε for all r ∈ (0, δ). To prove

this, choose n ∈ N such that n−1 < ε and define δ := log(n!)−n. If r ∈ (0, δ) then

| log(r)| = log(r−1) = log((r−1/n)n) ≤ log(n! er−1/n

) = log(n!) + r−1/n = δ−1/n + r−1/n ≤ 2r−1/n ≤ 2r−ε,

as claimed. If α > −1 and b > 0, set ε := 12b(α+ 1), take the corresponding δ and note that∫ δ

0rα| log(r)|b dr ≤

∫ δ

0rα2br−(α+1)/2 dr = 2b

∫ δ

0r(α−1)/2 dr <∞,

as 12(α− 1) > −1. Since rα| log(r)|b is bounded for r ∈ (δ, 1

2), it follows that the first integral is finite. If α > −1 and

b ≤ 0 then | log(r)|b = log(r−1)b ≤ log(2)b for all r ∈ (0, 12), in which case the first integral is finite. Similarly, it is

infinite if α < −1 and b > 0. If α < −1 and b < 0, set ε := − 12b(α+ 1), take the corresponding δ and note that∫ δ

0rα| log(r)|b dr ≥

∫ δ

0rα2br(α+1)/2 dr = 2b

∫ δ

0r(3α+1)/2 dr =∞,

as 12(3α+ 1) < −1. Thus, the first integral is finite. Finally, set α := −1. By the monotone convergence theorem∫ 1/2

0rα| log(r)|b dr = − lim

t→0

∫ 1/2

t(− log(r))b d(− log(r)) = − lim

t→0

(− log(r))b+1

b+ 1

∣∣∣∣1/2t

,

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which is finite iff b < −1 (the case b = −1 should really be handled separately). In summary,∫ 1/2

0 rα| log(r)|b dris finite iff a > −n or (a = −n and b < −1). The second integral can be treated similarly; alternatively we can

substitute s := r−1 and note that ∫ ∞2

rα| log(r)|b dr =

∫ 1/2

0s−α−2| log(s)|b ds,

which is finite iff −α− 2 > −1 or (−α− 2 = −1 and b < −1); in other words a < −n or (a = −n and b < −1).

65. (a) This is clear if n = 2. If n ≥ 3, let F : Rn−1 → Rn−1 be the map corresponding to G. Also let π : Rn−1 → Rn−2

and ρ : Rn−1 → R be projections on to the first n− 2 and last coordinate(s), respectively. Clearly

G(r, φ1, . . . , φn−2, θ) = (π(F (r, φ1, . . . , φn−2)), ρ(F (r, φ1, . . . , φn−2)) cos(θ), ρ(F (r, φ1, . . . , φn−2)) sin(θ)).

If x ∈ Rn then (by induction) we may assume that (x1, . . . , xn−2, |(xn−1, xn)|) = F (r, φ1, . . . , φn−2) for some

r, φ1, . . . , φn−2 ∈ R, in which case it is clear that x = G(r, φ1, . . . , φn−2, θ) for some θ ∈ R. Moreover, if

r, φ1, . . . , φn−2, θ ∈ R then (assuming, by induction, that |F (r, φ1, . . . , φn−2)| = |r|)

|G(r, φ1, . . . , φn−2, θ)| =√|π(F (r, φ1, . . . , φn−2))|2 + ρ(F (r, φ1, . . . , φn−2))2 cos2(θ) + ρ(F (r, φ1, . . . , φn−2))2 sin2(θ)

=

√|r|2 − ρ(F (r, φ1, . . . , φn−2))2 + ρ(F (r, φ1, . . . , φn−2))2(cos2(θ) + sin2(θ))

=√r2

= |r|.

(b) Denote the component functions of F by F 1, . . . , Fn−1. The Jacobian of G at a point (r, φ1, . . . , φn−2, θ) ∈ Rn is

F 1r F 1

φ1· · · F 1

φn−20

......

. . ....

...

Fn−2r Fn−2

φ1· · · Fn−2

φn−20

Fn−1r cos(θ) Fn−1

φ1cos(θ) · · · Fn−1

φn−2cos(θ) −Fn−1 sin(θ)

Fn−1r sin(θ) Fn−1

φ1sin(θ) · · · Fn−1

φn−2sin(θ) Fn−1 cos(θ)

,

so its determinant is Fn−1 sin2(θ) det(DF ) + Fn−1 cos2(θ) det(DF ) = r sin(φ1) . . . sin(φn−2) det(DF ). It easily

follows by induction that det(DG) has the required form (the case n = 2 is trivial).

(c) This is well-known for n = 2, so we may assume that n ≥ 3 and F |(0,∞)×(0,π)n−3×(0,2π) is injective. We may refine

our argument from part (a) to show that G(Ω) contains the points of Rn whose coordinates are all nonzero, in

which case Rn \ G(Ω) is clearly a null set. If (r, φ1, . . . , φn−2, θ) ∈ Ω and (R,Φ1, . . . ,Φn−2,Θ) ∈ Ω map to the

same point under G, then

π(F (r, φ1, . . . , φn−2)) = π(F (R,Φ1, . . . ,Φn−2))

and ρ(F (r, φ1, . . . , φn−2))2 = ρ(F (R,Φ1, . . . ,Φn−2))2 (because cos2(θ) + sin2(θ) = cos2(Θ) + sin2(Θ)). By defini-

tion ρ(F (r, φ1, . . . , φn−2)) = r sin(φ1) . . . sin(φn−2), which is positive by the definition of Ω. Therefore

ρ(F (r, φ1, . . . , φn−2)) = ρ(F (R,Φ1, . . . ,Φn−2)),

and hence (r, φ1, . . . , φn−2) = (R,Φ1, . . . ,Φn−2). It clearly follows that θ = Θ. This shows that G|Ω is injective,

so it has an inverse defined on G(Ω); the inverse function theorem (cf. part (b)) implies that the inverse is smooth

and G(Ω) is open, in which case G|Ω is a diffeomorphism.

23


(d) If we view Sn−1 as a smooth manifold it is straightforward to show that (F |Ω′)−1 is a diffeomorphism, but

that’s outside the scope of this course. Given an integrable function f : S1 → C, define g : Rn → C by

g(x) := f( x|x|)χB1(0)(x). By Theorem 2.49 and Exercise 51∫

Rn

g =

∫ ∞0

∫Sn−1

g(rx)rn−1 dσ(x) dr =

∫ 1

0

∫Sn−1

f(x)rn−1 dσ(x) dr =1

n

∫Sn−1

f.

On the other hand,∫Ω′f(F (φ1, . . . , φn−2, θ)) sinn−2(φ1) . . . sin(φn−2) dφ1 · · · dφn−2 dθ

= n

∫ 1

0rn−1 dr

∫Ω′f(F (φ1, . . . , φn−2, θ))| sinn−2(φ1) . . . sin(φn−2)| dφ1 · · · dφn−2 dθ

= n

∫ ∞0

∫Ω′g(rF (φ1, . . . , φn−2, θ))r

n−1| sinn−2(φ1) . . . sin(φn−2)| dφ1 · · · dφn−2 dθ dr

= n

∫Ωg(G(r, φ1, . . . , φn−2, θ))|det(D(r,φ1,...,φn−2)G)| dφ1 · · · dφn−2 dθ dr

= n

∫G(Ω)

g

=

∫Sn−1

f.

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