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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
Real Analysis Chapter 2 Solutions Jonathan Conder
=
∑n∈N
2−nan2
+∑n∈N
3−nan
∣∣∣∣∣ (an)n∈N is a sequence in 0, 2
=
∑n∈N
(2−n−1 + 3−n)an
∣∣∣∣∣ (an)n∈N is a sequence in 0, 2
.
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
Real Analysis Chapter 2 Solutions Jonathan Conder
=
∣∣∣∣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
Real Analysis Chapter 2 Solutions Jonathan Conder
= 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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
(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
Real Analysis Chapter 2 Solutions Jonathan Conder
(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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
(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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
= 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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
=
∫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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
=
∫ 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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
(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
,
22
Real Analysis Chapter 2 Solutions Jonathan Conder
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
Real Analysis Chapter 2 Solutions Jonathan Conder
(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.
24