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 inf n→∞ 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 | lim fn (x) exists} = {x ∈ X | lim inf fn (x) = lim sup fn (x) ∈
/ {−∞, ∞}} = f −1 ({0}) = f −1 ([−∞, 0]) ∈ M.
n→∞
n→∞
n→∞
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 | lim fn (x) exists} = {x ∈ X | lim Re(fn (x)) exists} ∩ {x ∈ X | lim Im(fn (x)) exists} ∈ M
n→∞
n→∞
n→∞
4. If a ∈ R then there is a sequence (an )n∈N in Q ∩ (a, ∞) converging to a, and f −1 ((a, ∞]) = ∪n∈N f −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 {χ{a} }a∈A is a family of measurable functions, but its supremum is χA , which is not measurable
because χ−1
A ([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 := max{x0 − 21 ε, 0}
and x2 := min{x0 + 12 ε, 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 − 21 ε ≤ h(y) ≤ x0 + 12 ε,
so |h(y) − h(x)| = |h(y) − x0 | < ε. Therefore h is continuous on [0, 2].
P
(b) Note that C = { n∈N 3−n an | (an )n∈N is a sequence in {0, 2}}, which implies that
(
!
)
X
X
−n
−n
g(C) = f
3 an +
3 an (an )n∈N is a sequence in {0, 2}
n∈N
n∈N
1
Real Analysis
Chapter 2 Solutions
Jonathan Conder
)
X
a
−n n
−n
=
+
3 an (an )n∈N is a sequence in {0, 2}
2
2
n∈N
n∈N
(
)
X
=
(2−n−1 + 3−n )an (an )n∈N is a sequence in {0, 2} .
(
X
n∈N
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 21 (21−n + 31−n − 3n ) = 2−n + 3−n ). Set C 0 := ∩n∈N Cn , so that
2
m(C 0 ) = lim m(Cn ) = lim 2n (2−n + 3−n ) = 1 + lim ( )n = 1.
n→∞
n→∞
n→∞ 3
P
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
X
(2−n−1 + 3−n )an ≤
n=1
∞
X
2(2−n−1 + 3−n ) =
n=N +1
3−N −1
2−N −1
+
2
= 2−N + 3−N .
1 − 2−1
1 − 3−1
P
−n−1 + 3−n )a is the left endpoint of an interval from C ,
By induction on N it can be shown that N
n
N
n=1 (2
th
−N
−N
because the N term in the series is either 0 or 2
+ 2 · 3 , 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 0 , which shows that g(C) ⊆ C 0 .
Conversely, let x ∈ C 0 , so that x ∈ Cn for all n ∈ N. For each n ∈ N define an ∈ {0, 2} depending on whether
P
−n−1 + 3−n )a
the interval x belongs to in Cn is the left or right child of its parent in Cn−1 . Then x and N
n
n=1 (2
are from the same interval in CN , for all N ∈ N. In particular
N
X
lim x −
(2−n−1 + 3−n )an ≤ lim (2−N + 3−N ) = 0,
N →∞
N →∞ n=1
which implies that x =
P∞
−n−1
n=1 (2
+ 3−n )an ∈ g(C). Therefore C 0 ⊆ g(C), and hence m(g(C)) = m(C 0 ) = 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 (x0 , y) − f (x, y)| < ε for all x0 ∈ (x − δ, x + δ).
Moreover, there exists N ∈ N such that N1 < δ. Let n ∈ N with n ≥ N, and choose i ∈ Z so that x ∈ [ai , ai+1 ]. Since
1
n < δ it is clear that ai , ai+1 ∈ (x − δ, x + δ). Therefore
f (ai+1 , y)(x − ai ) − f (ai , y)(x − ai+1 ) − f (x, y)(ai+1 − ai ) |fn (x, y) − f (x, y)| = 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 ) (f (ai , y) − f (x, y))(x − ai+1 ) +
≤ ai+1 − ai
ai+1 − ai
x − ai
ai+1 − x
= |f (ai+1 , y) − f (x, y)| ·
+ |f (ai , y) − f (x, y)| ·
ai+1 − ai
ai+1 − ai
x − ai
ai+1 − x
≤ε·
+ε·
ai+1 − ai
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
Z
Z
Z
Z
Z
f = f χE = lim inf fn χE ≤ lim inf fn χE = lim inf
fn
n→∞
E
n→∞
n→∞
E
R
R
and similarly E c f ≤ lim inf n→∞ E c fn . But f χE + f χE c = f and fn χE + fn χE c = fn for all n ∈ N, which implies
R
R
R
R
R
R
that E c f = f − E f and (for sufficiently large n ∈ N) E c fn = fn − E fn . Therefore
Z
Z
Z
Z
Z
Z
Z
Z
f−
f=
f ≤ lim inf
fn = lim inf
fn −
fn = f − lim sup
fn ,
E
so lim supn→∞
R
E
fn ≤
R
E
Ec
n→∞
f ≤ lim inf n→∞
R
E
n→∞
Ec
fn and hence
n→∞
E
R
E
f = limn→∞
R
E
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
R
R
R
R
(χFn )n∈N converges to χF pointwise and χF = ∞ = limn→∞ χFn . However, [0,∞) χF = 0 6= 1 = limn→∞ [0,∞) χFn .
R
R
R
14. Clearly λ(E) ≥ 0 for all E ∈ M. Moreover, λ(∅) = ∅ f dµ = f χ∅ dµ = 0 dµ = 0. If {En }n∈N is a pairwise disjoint
subcollection of M then (f χ∪N En )N ∈N is a sequence of measurable functions increasing to f χ∪n∈N En , so
n=1
Z
f dµ
λ(∪n∈N En ) =
∪n∈N En
Z
=
f χ∪n∈N En dµ
Z
= lim
f χ∪N En dµ
N →∞
n=1
Z
= lim
N →∞
= lim
N →∞
3
f
N
X
χEn dµ
n=1
N Z
X
n=1
f χEn dµ
Real Analysis
Chapter 2 Solutions
= lim
N →∞
=
=
N Z
X
Jonathan Conder
f dµ
n=1 En
∞ Z
X
f dµ
n=1 En
∞
X
λ(En )
n=1
by the monotone convergence theorem. Therefore λ is a measure. Now let g ∈ L+ . If g is simple with standard
P
representation N
n=1 an χEn , then
Z
g dλ =
N
X
an λ(En ) =
n=1
N
X
Z
f dµ =
an
En
n=1
N
X
Z
an
f χEn dµ =
Z X
N
Z
an f χEn dµ =
f g dµ.
n=1
n=1
Otherwise, there exists an increasing sequence (gn )n∈N of simple functions in L+ which converges pointwise to g, so
that (f gn )n∈N increases pointwise to f g and hence
Z
Z
Z
Z
g dλ = lim
gn dλ = lim
f gn dµ = f g dµ,
n→∞
n→∞
by two applications of the monotone convergence theorem.
R
15. Since f1 < ∞, the functions {fn }n∈N and f can be adjusted on a set of measure zero (namely f1−1 ({∞})) so that
they map into [0, ∞). This does not affect their integrals. Clearly (f1 − fn )n∈N increases pointwise to f1 − f. Moreover
R
R
f1 − fn ∈ L+ for all n ∈ N. By the monotone convergence theorem (f1 − f ) = limn→∞ (f1 − fn ). Therefore
Z
Z
Z
Z
f = f + (f1 − f ) − (f1 − f )
Z
Z
= f1 − lim (f1 − fn )
n→∞
Z
Z
= lim
f1 − (f1 − fn )
n→∞
Z
Z
Z
= lim
fn + (f1 − fn ) − (f1 − fn )
n→∞
Z
= lim
fn ,
n→∞
since
R
(f1 − f ) ≤
R
f1 < ∞, and similarly
R
(f1 − fn ) < ∞ for all n ∈ N.
16. For each n ∈ N define En := {x ∈ X | f (x) > n−1 }. Clearly (f χEn )n∈N increases pointwise to f, so by the monotone
R
R
convergence theorem ( En f )n∈N increases to f. In particular, given ε ∈ (0, ∞) there exists n ∈ N such that
R
R
R
f − ε. Since f < ∞ it is clear that µ(En ) < ∞.
En f >
17. Let (fn )n∈N be an increasing sequence in L+ , and set f := limn→∞ f. Then f ∈ L+ , and by Fatou’s lemma
Z
Z
Z
f = lim inf fn ≤ lim inf fn .
n→∞
Since fn ≤ f and hence
R
fn ≤
R
R
R
f for all n ∈ N, it is clear that lim supn→∞ fn ≤ f. Therefore
Z
Z
Z
lim sup fn = lim inf fn = lim
fn ,
n→∞
so
R
f = limn→∞
R
n→∞
n→∞
fn .
4
n→∞
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 inf n→∞ 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 fn− ∈ L1 and g − fn− , fn + g
Z
Z
Z
h+ g =
Z
=
Z
=
Z
=
∈ L+ for all n ∈ N. Therefore, by Fatou’s lemma
Z
Z
+
−
h − h + g
Z
+
h + (g − h− )
(h + g)
lim inf (fn + g)
n→∞
Z
≤ lim inf (fn + g)
n→∞
Z
Z
= lim inf
fn+ + (g − fn− )
n→∞
Z
Z
Z
+
−
= lim inf
fn + g − fn
n→∞
Z
Z
= lim inf fn + g.
n→∞
Since
R
g < ∞, it follows that
R
lim inf n→∞ fn =
R
h ≤ lim inf n→∞
R
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
Z
Z
Z
Z
Z
Z
Z
Z
Z
−
+
−
− h = h − h = −h = lim inf −fn ≤ lim inf −fn = lim inf fn = lim inf − fn = − lim sup fn .
n→∞
Therefore lim supn→∞
R
fn ≤
R
h=
R
n→∞
n→∞
n→∞
n→∞
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 |,
R
R
R
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
Z
Z
Z
Z
lim
fn dµ = lim
fN +n dµ =
lim fN +n dµ = f dµ.
n→∞
n→∞
n→∞
(b) For each n ∈ N define fn := 2−n χ[−2n ,2n ] . Clearly (fn )n∈N converges uniformly to 0, but for each n ∈ N
Z
Z
−n
n n
fn dµ = 2 µ([−2 , 2 ]) = 2 6= 0 = 0 dµ,
R
R
which implies that fn ∈ L1 (µ) (where µ is the Lebesgue measure) and limn→∞ fn dµ 6= 0 dµ.
R
R
R
R
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
Z
∞
Z
∞
Z
Z
Z
Z
inf
gm + fm
≤ gn + fn ≤ sup
gm
+ fn ,
m=N
m=N
5
Real Analysis
Chapter 2 Solutions
which implies that
Z
∞
Z
inf
gm + fm
∞
Z
≤ inf sup
Therefore
Z
+
lim inf
gn +
n→∞
= sup
fn
n=N
Z
Z
fn
≤ lim sup
Z
gn + lim inf
n→∞
n→∞
∞
Z
m=N
Z
∞
Z
gm
m=N
Jonathan Conder
∞
Z
+ inf
gm
.
fm
m=N
Z
fn =
m=N
Z
g + lim inf
fn .
n→∞
Similarly
Z
gn −
lim inf
n→∞
fn
Z
≤
Z
g + lim inf −
n→∞
Z
fn =
Z
g − lim sup
n→∞
fn .
Since gn + fn , gn − fn ∈ L+ for all n ∈ N, Fatou’s lemma implies that
Z
Z
Z
Z
Z
Z
Z
Z
Z
g + f = (g + f ) = lim inf (gn + fn ) ≤ lim inf (gn + fn ) = lim inf
gn + fn ≤ g + lim inf fn
n→∞
and
Z
Z
g−
Z
f=
n→∞
Z
(g − f ) =
Z
Z
lim inf (gn − fn ) ≤ lim inf
n→∞
n→∞
n→∞
n→∞
(gn − fn ) = lim inf
n→∞
Z
gn −
fn
Z
≤
Z
g − lim sup
n→∞
fn .
R
R
R
R
R
R
g < ∞, it follows that lim supn→∞ fn ≤ f ≤ lim inf n→∞ fn , and hence f = limn→∞ fn .
R
21. Suppose that limn→∞ |fn − f | = 0. For every ε ∈ (0, ∞), there exists N ∈ N such that
Z
Z
Z
Z
Z
Z
|fn | − |f | = (|fn | − |f |) ≤ |fn | − |f | ≤ |fn − f | = |fn − f | − 0 < ε
Since
R
R
for all n ∈ N with n ≥ N (by the reverse triangle inequality). Therefore limn→∞ |fn | = |f |.
R
R
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∈N
converges to 2|f | ∈ L1 pointwise almost everywhere, and
Z
Z
Z
Z
Z
lim (|fn | + |f |) = lim
|fn | + |f | = 2 |f | = 2|f |.
n→∞
n→∞
Therefore, by the previous exercise, limn→∞
R
|fn − f | =
R
0 = 0.
23. (a) Let x ∈ [a, b], and suppose that H(x) = h(x). Fix ε ∈ (0, ∞). Since
lim (sup f ([a, b] ∩ [x − δ, x + δ]) − inf f ([a, b] ∩ [x − δ, x + δ])) = H(x) − h(x) = 0,
δ→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)| <
for all y ∈ [a, b] with |x − y| < η. Therefore f (x) − 3ε < f (y) < f (x) + 3ε for all y ∈ [a, b] ∩ (x − η, x + η), so
ε
ε
sup f ([a, b] ∩ [x − δ, x + δ]) − inf f ([a, b] ∩ [x − δ, x + δ]) ≤ f (x) + − f (x) + < ε
3
3
for all δ ∈ (0, η). This shows that
H(x) − h(x) = lim (sup f ([a, b] ∩ [x − δ, x + δ]) − inf f ([a, b] ∩ [x − δ, x + δ])) = 0.
δ→0+
6
ε
3
Real Analysis
Chapter 2 Solutions
Jonathan Conder
b
(b) Choose a nested sequence (Pn )n∈N of partitions of [a, b] such that (SPn f )n∈N converges to I a (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∈N En , so that
m(E) = 0. Let x ∈ [a, b] \ E, and choose δ ∈ (0, ∞) such that
ε
sup f ([a, b] ∩ [x − δ, x + δ]) < H(x) + .
2
b
There exists N ∈ N such that SPn f < I a (f ) + εδ
2 for all n ∈ N with n ≥ N. Fix n ∈ N with n ≥ N. There is an
0
0
0
0
interval [a , b ] in Pn such that x ∈ (a , b ) (because x ∈
/ En ). If [a0 , b0 ] ⊆ [x − δ, x + δ] then
GPn (x) = sup f ([a0 , b0 ]) ≤ sup f ([a, b] ∩ [x − δ, x + δ]) < H(x) +
ε
< H(x) + ε.
2
Otherwise a0 < x − δ or x + δ < b0 , so that [x − δ, x] or [x, x + δ] is contained in (a0 , b0 ). Construct a new partition
Pn0 of [a, b] from Pn by inserting x and one of x − δ or x + δ between a0 and b0 . In the former case
SPn0 f − SPn f = sup f ([a0 , x − δ])(x − δ − a0 ) + sup f ([x − δ, x])δ + sup f ([x, b0 ])(b0 − x) − GPn (x)(b0 − a0 )
ε
δ + GPn (x)(b0 − x) − GPn (x)(b0 − a0 )
< GPn (x)(x − δ − a0 ) + H(x) +
2
ε
= GPn (x)(x − δ − a0 + b0 − x − b0 + a0 ) + H(x) +
δ
2
ε
= GPn (x)(−δ) + H(x) +
δ
2
ε
= H(x) − GPn (x) +
δ,
2
which still holds for the latter case, by a similar calculation. It follows that
εδ
1
1
ε
ε
b
GPn (x) < (SPn f − SPn0 f ) + H(x) + <
I a (f ) +
− SPn0 f + H(x) + ≤ H(x) + ε.
δ
2
δ
2
2
Since x ∈ (a0 , b0 ), there exists η ∈ (0, ∞) such that [x − η, x + η] ⊆ (a0 , b0 ). This implies that
H(x) =
inf
ζ∈(0,∞)
sup f ([a, b] ∩ [x − ζ, x + ζ]) ≤ sup f ([a, b] ∩ [x − η, x + η]) ≤ sup f ([a0 , b0 ]) = 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
Z
Z
b
GPn dm = lim SPn f = I a (f ).
H dm = lim
n→∞
n→∞
[a,b]
A similar argument implies that (gPn )n∈N converges to h pointwise almost everywhere, for all nested sequences
R
(Pn )n∈N of partitions of [a, b] such that (sPn f )n∈N converges to I ba (f ). Therefore [a,b] h dm = I ba (f ).
25. (a) By the monotone convergence theorem and Theorem 2.28,
Z
Z
f = lim
1
n→∞ 1/n
1
x−1/2 dx = lim 2x1/2 n→∞
1/n
= lim (2 − 2n−1/2 ) = 2.
n→∞
(1)
R
P∞ −n R
P∞ −n R
Therefore |g| =
f (x − rn ) dx =
f = 2, by the monotone convergence theorem. It
n=1 2
n=1 2
1
follows that g ∈ L (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 E c . 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 ) ∩ I
has positive measure, so there exists xk ∈ ((rn , rn + k −1 ) ∩ I) \ E. Clearly limk→∞ xk = rn , in which case
limk→∞ 2−n f (xk − rn ) = 2−n limk→∞ (xk − rn )−1/2 = ∞. But 2−n f (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) g 2 < ∞ 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
Z δ
Z δ
Z
Z rn +δ
Z rn +δ
x−1 dx = ∞,
f 2 = 2−2n
g2 ≥
g2 ≥
2−2n f (x − rn )2 dx = 2−2n
I
rn
0
rn
0
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
Z
Z
lim
fn = 0 = 0,
n→∞
R by the dominated convergence theorem. Choose n ∈ N such that fn < ε, and let y ∈ (x − 2−n , x + 2−n ). Then
Z x
Z y
|F (x) − F (y)| = f (t) dt −
f (t) dt
−∞
Z−∞
Z
= f χ(−∞,x] − f χ(−∞,y] Z
= f · (χ(−∞,x] − χ(−∞,y] )
Z
≤ |f | · |χ(−∞,x] − χ(−∞,y] |
Z
= |f | · χ[min{x,y},max{x,y}]
Z
≤ |f | · χ[x−2−n ,x+2−n ]
Z
= fn
Z
= 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 + nx )−n . Then
x n(−xn−2 )
x
x
(log ◦f )0 (n) = − log 1 +
−
=
−
log
1
+
n
1 + nx
n+x
n
for all n ∈ [1, ∞). Note that
X
k
k
∞ ∞
x
X
x
1
x
x
exp
=
≤1+
= 1 + n+xx = 1 +
n+x
k! n + x
n+x
1 − n+x
k=0
k=1
8
x
n+x
n
n+x
=1+
x
n
Real Analysis
Chapter 2 Solutions
Jonathan Conder
x
and hence n+x
≤ log 1 + nx 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 + nx )−n . In particular fn ≤ f2 for all n ∈ N with n ≥ 2. By the monotone convergence theorem
Z n
Z ∞
Z ∞
Z n
x −2
d
x −1
f2 χ[0,n] = lim
1+
f2 = lim
−2 1 +
dx = lim
dx,
n→∞ 0
n→∞ 0
n→∞ 0 dx
2
2
0
so by the fundamental theorem of calculus
Z ∞
n −1
2
−1
f2 = lim 2(1 + 0) − 2 1 +
= 2 − lim
n→∞
n→∞
2
1+
0
n
2
= 2.
Therefore f2 ∈ L1 , so (since | sin | ≤ 1) by the dominated convergence theorem
Z ∞
Z ∞
Z ∞
x
x
x −n
x −n
lim
1+
sin
lim 1 +
sin
0 dx = 0.
dx =
dx =
n→∞ 0
n
n
n
n
0 n→∞
0
Indeed, if x ∈ [0, ∞) then limn→∞ sin( nx ) = sin(limn→∞ nx ) = sin(0) = 0, and
lim
n→∞
1+
x −n
x = exp lim −n log 1 +
n→∞
n
n !!
x
log 1 + n
= exp − lim
1
n→∞
= exp −x lim
n
log 1 +
n→∞
x
n
x
n
− log(1)
!!
= exp −x log0 (1)
= e−x .
(b) If x ∈ [0, 1] and n ∈ N with n ≥ 1 then 0 ≤ (1 + nx2 )(1 + x2 )−n ≤ 1 because
n n X
X
n
n
2 k
2
(x2 )k ≥ 1 + nx2 .
(x ) = 1 + nx +
(1 + x ) =
k
k
2 n
k=2
k=0
Moreover
R1
0
1 dx = 1. Hence, by the dominated convergence theorem
Z
lim
n→∞ 0
1
2
2 −n
(1 + nx )(1 + x )
Z
dx =
1
2
2 −n
lim (1 + nx )(1 + x )
0 n→∞
Indeed, (1 + nx2 )(1 + x2 )−n ≤ (1 + nx2 )(1 + nx2 +
n
2
Z
dx =
1
0 dx = 0.
0
x4 )−1 for all x ∈ [0, 1] and n ∈ N with n ≥ 2, and
1 + nx2
1 + nx2
n−2 + n−1 x2
0+0
=
lim
=
lim
=
=0
n
n−1
n(n−1)
2
4
−2
−1
2
4
n→∞ 1 + nx +
n→∞ 1 + nx2 +
0 + 0 + 12 x4
x4 n→∞ n + n x + 2n x
2 x
lim
2
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 0 (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,
Z ∞
Z ∞
Z n
d
π
2 −1
2 −1
(1+x ) dx = lim
(1+x ) χ[0,n] (x) dx = lim
tan−1 (x) dx = lim (tan−1 (n)−tan−1 (0)) = .
n→∞ 0
n→∞ 0 dx
n→∞
2
0
9
Real Analysis
Chapter 2 Solutions
Jonathan Conder
Since | sin( nx )/ nx | ≤ 1 for all x ∈ [0, ∞) and n ∈ N, the dominated convergence theorem implies that
Z
∞
lim
n→∞ 0
n sin
x
n
2
−1
(x(1 + x ))
Z
∞
dx = lim
n→∞ 0
Z ∞
=
Z
Z
lim
sin( nx ) − sin(0)
n→∞
0
∞
=
Z0 ∞
=
Z0 ∞
=
(1 + x2 )−1 dx
sin( nx )
∞
=
x
n
lim
n→∞
0
sin( nx )
x
n
(1 + x2 )−1 dx
x
n
(1 + x2 )−1 dx
sin0 (0)(1 + x2 )−1 dx
cos(0)(1 + x2 )−1 dx
(1 + x2 )−1 dx
0
=
π
.
2
(d) Fix n ∈ N. By the monotone convergence theorem and the fundamental theorem of calculus,
Z ∞
Z m
π
d
2 2 −1
tan−1 (nx) dx = lim tan−1 (nm) − tan−1 (na) = − tan−1 (na).
n(1 + n x ) dx = lim
m→∞
m→∞
2
a
a dx
Therefore



0, a > 0

Z ∞

2 2 −1
lim
n(1 + n x ) dx = π2 , a = 0
n→∞ a



π, a < 0.
R
R∞
This implies that there is no f ∈ L1 such that limn→∞ a n(1 + n2 x2 )−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)
k
−tk
e
e0
1
e−tx = lim −
+
= .
e
dx = lim
e
dx = lim
k→∞
k→∞
k→∞
−t 0
t
t
t
0
0
R∞
Given n ∈ N, define fn : [0, ∞) × [ 12 , 2] → [0, ∞) by fn (x, t) := xn e−tx . We claim that 0 fn (x, t) dx = n! t−(n+1) for
R∞
each t ∈ [ 21 , 2]. By induction, we may assume that 0 fn−1 (x, t) dx = (n − 1)! t−n . Note that
Z
∞
−tx
Z
k
−tx
fn−1 (x, t) = xn−1 e−tx ≤ 3n−1 (n − 1)! ex/3 e−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
∂
fn−1 (x, t) = |−xn e−tx | = fn (x, t) ≤ 3n n! e−(1/6)x
∂t
for all x ∈ [0, ∞), so by Theorem 2.27
Z ∞
Z ∞
∂
∂
fn (x, t) dx = −
fn−1 (x, t) dx = − (n − 1)! t−n = n(n − 1)! t−(n+1) = n! t−(n+1) ,
∂t
∂t
0
0
R ∞ n −x
as claimed. In particular 0 x e dx = n!.
10
Real Analysis
Chapter 2 Solutions
Jonathan Conder
p
R∞
2
If t ∈ (0, ∞), then −∞ e−tx 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 × [ 21 , 2] → [0, ∞)
R∞
√ −(2n+1)/2
2
by fn (x, t) := x2n e−tx . We claim that −∞ fn (x, t) dx = (2n)!
πt
for each t ∈ [ 12 , 2]. By induction, we may
n n!
4
R∞
√
(2n−2)!
assume that −∞ fn−1 (x, t) dx = 4n−1
πt−(2n−1)/2 . Note that
(n−1)!
2
2 /3
fn−1 (x, t) = x2n−2 e−tx ≤ 3n−1 (n − 1)! ex
2
2
e−tx = 3n−1 (n − 1)! e−(t−1/3)x ≤ 3n−1 (n − 1)! e−(1/6)x
2
for all x ∈ R, so fn−1 (−, t) ∈ L1 (R). Moreover
∂
fn−1 (x, t) = |−x2n e−tx2 | = fn (x, t) ≤ 3n n! e−(1/6)x2
∂t
for all x ∈ R, so by Theorem 2.27
Z ∞
∂
fn (x, t) dx = −
fn−1 (x, t) dx
∂t −∞
−∞
∂ (2n − 2)! √ −(2n−1)/2
=−
πt
∂t 4n−1 (n − 1)!
(2n − 2)!(2n − 1) √ −(2n+1)/2
=
πt
2 · 4n−1 (n − 1)!
√ −(2n+1)/2
(2n)!
πt
=
4n · 4n−1 (n − 1)!
(2n)! √ −(2n+1)/2
= n
πt
,
4 n!
R∞
√
2
as claimed. In particular −∞ x2n e−x dx = (2n)!
4n n! π.
Z
∞
30. For each k ∈ N, we claim that (1 − k −1 x)k ≤ e−x for all x ∈ (0, k). If so, xn (1 − k −1 x)k χ(0,k) (x) ≤ xn e−x for all k ∈ N
and x ∈ [0, ∞), and by the previous exercise we may apply the dominated convergence theorem to show that
Z k
Z ∞
Z k
lim
xn (1 − k −1 x)k dx =
lim xn (1 − k −1 x)k χ(0,k) dx =
xn e−x dx = n!
k→∞ 0
0
k −1 x)k
k→∞
0
e−x ,
(to prove that limk→∞ (1 −
=
take logarithms and apply l’Hôpital’s rule). Now we prove the claim. It
−1
suffices to show that k log(1 − k x) + x ≤ 0 for all x ∈ (0, k). This is certainly true for x = 0. Moreover,
k(−k −1 )
1
−k −1 x
d
(k log(1 − k −1 x) + x) =
+
1
=
1
−
=
<0
dx
1 − k −1 x
(1 − k −1 x)
1 − k −1 x
for all x ∈ (0, k). By the mean value theorem k log(1 − k −1 x) + x = (k log(1 − k −1 x) + x) − (k log(1 − 0) + 0) < 0 for
all x ∈ (0, k), which proves the claim.
R
R
R
R
33. There clearly exists a subsequence ( fnk )k∈N of ( fn )n∈N such that limk→∞ fnk = lim inf n→∞ fn . Moreover
(fnk )k∈N converges to f in measure, because for every ε ∈ (0, ∞)
lim µ({x ∈ X | |fnk (x) − f (x)| ≥ ε}) = lim µ({x ∈ X | |fn (x) − f (x)| ≥ ε}) = 0.
n→∞
k→∞
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 , δ ∈ (0, ∞)). Therefore, by Fatou’s lemma
Z
Z
Z
Z
Z
Z
+
f = g ≤ lim inf fnki = lim
fnki = lim
fnk = lim inf fn .
i→∞
i→∞
11
k→∞
n→∞
Real Analysis
Chapter 2 Solutions
34. (a) It suffices to show that limn→∞
R
Re(fn ) =
R
Jonathan Conder
Re(f ) and limn→∞
R
Im(fn ) =
R
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
Z
Z
Z
Z
Z
Z
Z
Z
g + f = (g + f ) ≤ lim inf (g + fn ) = lim inf
g + fn = g + lim inf fn
n→∞
and
Z
Z
g−
Since
R
Z
f=
Z
Z
(g − f ) ≤ lim inf
n→∞
g < ∞, it follows that lim supn→∞
n→∞
n→∞
R
(g − fn ) = lim inf
n→∞
fn ≤
R
f ≤ lim inf n→∞
Z
g−
R
fn
Z
=
fn , and hence
Z
g − lim sup
n→∞
R
f = limn→∞
fn
R
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),
Z
Z
lim kfn − f k1 = lim
|fn − f | = 0 = 0.
n→∞
n→∞
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 := min{ε, 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
lim φ(fn (x)) = φ( lim fn (x)) = φ(f (x)),
n→∞
n→∞
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 E c , and hence (φ ◦ fn )n∈N converges to φ ◦ f uniformly on
E c 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
lim µ({x ∈ X | |fn (x) − f (x)| ≥ δ}) = 0.
n→∞
Clearly {x ∈ X | |φ(fn (x)) − φ(f (x))| ≥ ε} ⊆ {x ∈ X | |fn (x) − f (x)| ≥ δ} for all n ∈ N, so
lim µ({x ∈ X | |φ(fn (x)) − φ(f (x))| ≥ ε}) = 0
n→∞
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

1,
Re(z) > 0
φ(z) :=
−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) := z 2 . Let E ⊆ R and suppose that (φ ◦ fn )n∈N converges to φ ◦ f 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−N x + 2−2N |
for all x ∈ E, it follows that E ⊆ (−2N −1 − 2−N −1 , 2N −1 − 2−N −1 ). In particular µ(E c ) = ∞, so (φ ◦ fn )n∈N
does not converge to φ ◦ f uniformly or almost uniformly. If ε ∈ (0, ∞) and n ∈ N, then
[2n−1 ε, ∞) ⊆ {x ∈ R | |φ(fn (x)) − φ(f (x))| ≥ ε}
because |φ(fn (x)) − φ(f (x))| = 21−n x + 2−2n ≥ ε + 2−2n for all x ∈ [2n−1 ε, ∞). Therefore
lim µ({x ∈ R | |φ(fn (x)) − φ(f (x))| ≥ ε}) = ∞,
n→∞
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 Enc . Define E := ∩n∈N ∪∞
k=n Ek .
∞
∞
1−n
c
Then µ(E) = limn→∞ µ(∪k=n Ek ) = 0, since µ(∪k=n Ek ) ≤ 2
for all n ∈ N. Moreover, if x ∈ E there exists n ∈ N
c
such that x ∈ En 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 E c . There exists N ∈ N
such that |fn (x) − f (x)| < ε for all x ∈ E c 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
−1
all x ∈ A. Fix k ∈ N, and for each n ∈ N define Ek,n := ∪∞
m=n {x ∈ A | |fm (x) − f (x)| ≥ 2k }. 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 −1 χEk,1 ≤ g, so
k −1 χEk,1 ∈ L1 and hence µ(Ek,1 ) < ∞. Since ∩n∈N Ek,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−k ε. Define E := (∪k∈N Enk ,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 ∈ E c and
m ∈ N with m ≥ nk . This implies that (fn )n∈N converges to f uniformly on E c .
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 kfn − f ku ≤ ε, for all n ∈ N
with 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∈N En ) = µ([a, b]), so
there exists m ∈ N such that µ([a, b]) − µ(Em ) < 3ε . Define g : R → C by

f (x), x ∈ E
m
g(x) :=
0,
x ∈ Ec .
m
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 kgn − gk1 < 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 ) < 3ε 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 ) − 3ε and hence
µ([a, b] \ E) = µ([a, b]) − µ(E) < µ([a, b]) − µ(Em \ F ) +
ε
ε
ε
≤ µ([a, b]) + µ(F ) − µ(Em ) + < 3 · = ε.
3
3
3
Moreover (gnk )k∈N converges to g uniformly on E, so f |E = g|E is continuous.
R
46. Fix y ∈ Y. Then χD (x, y) = χ{y} (x) for all x ∈ X, so χD (x, y) dµ(x) = µ({y}) = 0. This implies that
ZZ
Z
χD (x, y) dµ(x) dν(y) = 0 dν(y) = 0.
R
Now fix x ∈ X. Clearly χD (x, y) = χ{x} (y) for all y ∈ Y, so χD (x, y) dν(y) = ν({x}) = 1. It follows that
ZZ
Z
χD (x, y) dν(y) dµ(x) = 1 dµ(x) = µ(X) = 1.
R
χD d(µ × ν) = (µ × ν)(D), and hence
(∞
)
Z
X
χD d(µ × ν) = inf
(µ × ν)(En ) (En )∞
n=1 is a sequence of finite disjoint unions of rectangles covering D
By definition
n=1
14
Real Analysis
Chapter 2 Solutions
(
= inf
∞
X
n=1
Jonathan Conder
)
∞
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
R
P
finite set is 0. Therefore ∞
χD d(µ × ν) = inf{∞} = ∞.
n=1 µ(An )ν(Bn ) = ∞, so
R
∞
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
P
∞
∞
∪∞
n=1 {(n, n), (n + 1, n)}, then (An ∩ Bn )n=1 covers N and hence
n=1 µ(An ∩ Bn ) = ∞. This implies that
∞
X
µ(An )ν(Bn ) =
n=1
∞
X
µ(An )µ(Bn ) ≥
n=1
∞
X
n=1
2
µ(An ∩ Bn ) ≥
∞
X
µ(An ∩ Bn ) = ∞,
n=1
R
since µ(An ∩ Bn ) ∈ {0} ∪ [1, ∞] for all n ∈ N. Therefore |f | d(µ × ν) = inf{∞} = ∞. Fix n ∈ Y. Then f (m, n) =
R
χ{n} (m) − χ{n+1} (m) for all m ∈ X, and hence f (m, n) dµ(m) = µ({n}) − µ({n + 1}) = 0. This implies that
ZZ
Z
f (m, n) dµ(m) dν(n) = 0 dν(n) = 0.
R
Now fix m ∈ X \{1}. Then f (m, n) = χ{m} (n)−χ{m−1} (n) for all n ∈ Y, so f (m, n) dν(n) = ν({m})−ν({m−1}) = 0.
R
R
Moreover, f (1, n) dν(n) = χ{1} dν = ν({1}) = 1. It follows that
ZZ
Z
f (m, n) dν(n) dµ(m) = χ{1} dµ = µ({1}) = 1.
49. (a) Since µ and ν are σ-finite,
Z
Z
ν(Ex ) dµ(x) =
µ(E y ) dν(y) = (µ × ν)(E) = 0.
This implies that ν(Ex ) = µ(E y ) = 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
R
fx (y) = 0 for all y ∈ Y such that y ∈
/ Ex . Hence fx = 0 almost everywhere, so fx is integrable with fx dν = 0,
R
for almost all x ∈ X (by the previous exercise). Similarly f y is integrable and f y 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 g y is M-measurable for all x ∈ X and y ∈ Y. If f ≥ 0 then g ≥ 0 without loss of
R
R
generality, so by Tonelli’s theorem x 7→ gx dν and y 7→ g y dµ are non-negative and (M ⊗ N)-measurable, while
Z
ZZ
ZZ
g dλ =
g(x, y) dµ(x) dν(y) =
g(x, y) dν(y) dµ(x).
(2)
R
R
R
Since |g| = |f | λ-almost everywhere, |g| d(µ × ν) = |g| dλ = |f | dλ and hence g ∈ L1 (µ × ν) whenever f ∈ L1 (λ).
R
By Fubini’s theorem, this implies that gx ∈ L1 (ν) and g y ∈ L1 (µ) for almost all x ∈ X and y ∈ Y, while x 7→ gx dν
R
and y 7→ g y 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 f y is M-measurable for almost all y ∈ Y. Since fx − gx ∈ L1 (ν)
and f y − g y ∈ L1 (µ) for almost all x ∈ X and y ∈ Y, it is clear that fx ∈ L1 (ν) and f y ∈ L1 (µ) for almost all x ∈ X
15
Real Analysis
Chapter 2 Solutions
Jonathan Conder
R
R
R
R
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
R
all x ∈ X, so x 7→ fx dν is measurable and in the second case, integrable (for the first case, assume without loss of
R
generality that g ≤ f ). The same clearly holds for y 7→ f y dµ, so (because f = g almost everywhere)
Z
ZZ
ZZ
ZZ
ZZ
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}
−1
= ∪∞
(An ) and y ∈ Bn }
n=1 {(x, y) ∈ X × [0, ∞) | x ∈ f
−1
= ∪∞
(An ) × Bn ).
n=1 (f
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
Z
ZZ
(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
R
Therefore (µ × m)(Gf ) = f dµ as required.
R
χE (x, y) dm(y) =
R f (x)
0
1 dm(y) = f (x).
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 = F G 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),
Pl
Pk
so that (Φn Ψn )∞
n=1 converges pointwise to h. Fix n ∈ N, and write φn =
j=1 bj χBj for
i=1 ai χAi and φn =
some a1 , a2 , . . . , ak , b1 , b2 , . . . , bl ∈ [0, ∞] and measurable sets A1 , A2 , . . . , Ak , B1 , B2 , . . . , Bl . Clearly

! l
k
k X
l
k X
l
X
X
X
X


Φn Ψn =
ai χ(Ai ×Y )
bj χ(X×Bj ) =
ai χ(Ai ×Y ) bj χ(X×Bj ) =
ai bj χ(Ai ×Bj ) ,
i=1
j=1
i=1 j=1
i=1 j=1
and hence
Z
Φn Ψn =
l
k X
X
ai bj (µ×ν)(Ai ×Bj ) =
i=1 j=1
k X
l
X
ai µ(Ai )bj ν(Bj ) =
i=1 j=1
k
X
!
ai µ(Ai ) 
i=1
l
X

Z
bj ν(Bj ) =
j=1
By the monotone convergence theorem, it follows that
Z
Z
Z
Z
Z
Z
Z
Z
h = lim
Φn Ψn = lim
φn · ψn = lim
φn · lim
ψn = f · g.
n→∞
Hence, in general
R
|h| =
Z
n→∞
n→∞
n→∞
R
|f | · |g| < ∞, so h ∈ L1 (µ × ν). If f (X) ⊆ R and g(Y ) ⊆ R, then
Z
Z
+
h = h − h−
R
16
Z
φn · ψn .
Real Analysis
Chapter 2 Solutions
Z
Z
Z
=
+
−
Z
F G + F G − F G − F − G+
Z
Z
Z
Z
Z
Z
Z
Z
+
+
−
−
+
−
−
= f · g + f · g − f · g − f · g+
Z
Z
Z
Z
Z
Z
g− − g+
g+ − g− + f −
= f+
Z
Z
Z
Z
+
−
+
−
=
f − f
g − g
Z
Z
= f · g.
+
−
Jonathan Conder
+
−
Since F G = (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)),
Z
Z
Z
h = Re(h) + i Im(h)
Z
Z
Z
Z
= Re(F ) Re(G) − Im(F ) Im(G) + i Re(F ) Im(G) + i Im(F ) Re(G)
Z
Z
Z
Z
Z
Z
Z
Z
= Re(f ) · Re(g) − Im(f ) · Im(g) + i Re(f ) · Im(g) + i Im(f ) · Re(g)
Z
Z
Z
Z
Z
Z
= Re(f )
Re(g) + i Im(g) − Im(f )
Im(g) − i Re(g)
Z
Z
Z
Z
Z
Z
= Re(f )
Re(g) + i Im(g) + i Im(f )
Re(g) + i Im(g)
Z
Z
Z
Z
=
Re(f ) + i Im(f )
Re(g) + i Im(g)
Z
Z
= f · g.
55. (a) Fix y ∈ (0, 1], and define F : [0, 1] → R by F (x) := x(x2 + y 2 )−1 . By the quotient rule
F 0 (x) =
(x2 + y 2 ) − x(2x)
y 2 − x2
=
= −f y (x)
(x2 + y 2 )2
(x2 + y 2 )2
for all x ∈ [0, 1]. This implies that
1
Z
(f y )− =
0
and similarly
R1
0
R1
(f y )+ =
Z
f
−
y
−f y = F (y) − F (0) = F (y) =
0
f y = −F (1) + F (y) =
1Z 1
Z
y
Z
=
Z
−
f (x, y) dx dy =
E
0
0
0
1
2y
1
1
y
=
2y 2
2y
1
− 1+y
2 . By the Tonelli and monotone convergence theorems
1
dy = lim
n→∞
2y
Z
1
1
n
− log( n1 )
1
dy = lim
= ∞,
n→∞
2y
2
which implies that
Z
f
+
Z
=
E
because
R1
1
0 1+y 2
dy ≤
R1
0
1Z 1
Z
+
f (x, y) dx dy =
0
0
0
1
1
1
−
2y 1 + y 2
dy = ∞
1 dy < ∞ and
Z
0
1
1
1
−
2y 1 + y 2
Z
dy +
0
17
1
1
dy =
1 + y2
Z
0
1
1
dy = ∞
2y
Real Analysis
Chapter 2 Solutions
This shows that
Z
R
E
f is not defined. However,
1Z 1
Z
1
Z
(−F (1) + F (0)) dy = −
f (x, y) dx dy =
0
Jonathan Conder
0
0
0
1
1
π
dy = −(tan−1 (1) − tan−1 (0)) = − .
2
1+y
4
Since f (x, y) = −f (y, x) for all x, y ∈ (0, 1], it follows that
Z 1Z 1
π
f (x, y) dy dx = .
4
0
0
(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,
Z
Z 1Z 1
+
f =
f + (x, y) dx dy
E
0
0
1
2
1 −3
=
x−
dx dy
1
2
0
+y
2
!
−2
Z 1
2
1
(−2)−1
=
− y −2 dy
2
0
Z 1
1 2 −2
(y − 4) dy
=
2 0
Z 1
1 2
= lim
(y −2 − 4) dy
n→∞ 2
1
n+4
!
−1
−1 1
1
4
4
1
= lim
+
− +
−
n→∞ 2
2
n+4
2 n+4
n
= lim
n→∞ 2
= ∞.
Z
Z
1
Similarly
Z
f
−
1Z 1
Z
=
E
0
1
2
Z
=
0
1
2
=
1
−y
2
Z
0
Z
f − (x, y) dx dy
0
2−1
0
1
=
2
1
2
Z
−3
1
−x
dx dy
2
−2 !
1
−2
dy
y −
2
(y −2 − 4) dy
0
= ∞.
Therefore
R
E
f does not exist. However, the above working implies that
Z
1Z 1
Z
f (x, y) dx dy =
0
0
0
1
2
Z
1
f (x, y) dx dy
0
18
Real Analysis
Chapter 2 Solutions
1
2
Z
=
Jonathan Conder
1
Z
Z
+
f (x, y) dx −
0
1
2
Z
0
−
f (x, y) dx
dy
0
0
=
1
1 −2
1
(y − 4) − (y −2 − 4) dy
2
2
= 0.
If x ∈ [0, 21 ], then fx ≤ 0 and hence
Z
1
Z
1
1
−x
2
Z
−
f (x, y) dy = −
f (x, y) dy = −
0
0
0
1
−x
2
−3
dy = −
1
−x
2
−2
.
Similarly, if x ∈ [ 12 , 1] then
1
Z
Z
1
f (x, y) dy =
f (x, y) dy =
0
0
x− 12
Z
+
0
1 −3
1 −2
x−
dy = x −
.
2
2
By the monotone convergence theorem and the fundamental theorem of calculus, this implies that
Z
+
1 Z 1
f (x, y) dy
0
Z
1
x−
dx =
1
2
0
Z
= lim
n→∞
= lim
n→∞
1
2
−2
dx
1 −2
dx
x−
1
1
2
+ n+2
2
−1 −1 !
1
1
−
+
2
n+2
1
= lim n
n→∞
= ∞.
Similarly
−
1 Z 1
Z
f (x, y) dy
0
0
1
2
−2
1
−x
dx
2
0
−2
Z 1− 1 2
n+2
1
= lim
−x
dx
n→∞ 0
2
−1 −1 !
1
1
= lim
−
n→∞
n+2
2
Z
dx =
= lim n
n→∞
= ∞.
This shows that
R1R1
0
0
f (x, y) dy dx does not exist.
56. Define a measurable function h : (0, a)2 → C by h(t, x) := t−1 f (t)χE (t, x), where E := {(t, x) ∈ (0, a)2 | x < t} is
Ra
R
open, hence measurable. Then g(x) = 0 t−1 f (t)χ(x,a) (t) dt = hx for all x ∈ (0, a). By Tonelli’s theorem
Z
Z
aZ a
|h| =
0
0
t−1 |f (t)|χE (t, x) dx dt =
Z
0
19
aZ t
0
t−1 |f (t)| dx dt =
Z
a
|f (t)| dt < ∞.
0
Real Analysis
Chapter 2 Solutions
Jonathan Conder
Hence h is integrable, so g is integrable by Fubini’s theorem, and
Z a
Z aZ a
Z
Z aZ a
Z
−1
g=
hx (t) dt dx = h =
t f (t)χE (t, x) dx dt =
0
0
0
0
0
aZ t
0
−1
t
Z
f (t) dx dt =
0
a
f (t) dt.
0
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
∞Z 1
Z
e
0
−sx
Z
∞
−sx cos(2xy)
−e
sin(2xy) dy dx =
2x
0
0
1
Z
dx =
0
0
∞
e
−sx 1
− cos(2x)
dx =
2x
Z
∞
e−sx x−1 sin2 (x) dx,
0
Fubini’s theorem implies that
Z
∞
−sx −1
e
x
Z
2
1Z ∞
sin (x) dx =
0
e
0
−sx
1
Z
sin(2xy) dx dy =
0
0
1
2y
1
1
−1 2
2 dy = log(4 s + y ) = log(1 + 4s−2 )
2
2
s + 4y
4
4
0
(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 + 61 )π, (n + 56 )π] then | sin(x)| ≥ 21 , and x−1 ≥ (n + 1)−1 π −1 . Therefore
Z
∞
|f | ≥
Z X
∞
0
n=1
∞
X
1
χ[(n+ 1 )π,(n+ 5 )π] =
6
6
2(n + 1)π
Z
n=1
∞
4
X
1
6π
χ[(n+ 1 )π,(n+ 5 )π] =
= ∞,
6
6
2(n + 1)π
2(n + 1)π
n=1
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
by Fubini’s theorem and the fundamental theorem of calculus
Z bZ b
Z bZ b
f (x, y) dx dy =
e−xy sin(x) dy dx
0
0
0
0
Z b −xb
e
sin(x) e0 sin(x)
=
−
dx
−x
−x
0
Z b
sin(x) e−xb sin(x)
−
dx
=
x
x
0
R
|f | ≤ b2 < ∞. Hence,
(3)
Since | sin(x)| ≤ x for all x ∈ (0, ∞) it is clear that
Z
0
b
Z b −xb
Z b
2
2
e
sin(x) e−xb sin(x) e−b
e0
1 − e−b
−xb
dx ≤
e
dx =
−
=
.
dx ≤
x
x
−b
−b
b
0
0
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
∂
y sin(x) + cos(x)
y cos(x) − sin(x)
F (x, y) = ye−xy
− e−xy
2
∂x
y +1
y2 + 1
y 2 sin(x) + y cos(x) − y cos(x) + sin(x)
= e−xy
y2 + 1
= f (x, y)
20
(4)
Real Analysis
Chapter 2 Solutions
Jonathan Conder
and hence
Z bZ
b
Z
b
(F (b, y) − F (0, y)) dy
Z b
0 y sin(0) + cos(0)
−by y sin(b) + cos(b)
e
=
dy
−e
y2 + 1
y2 + 1
0
Z b
1
−by y sin(b) + cos(b)
=
dy.
−e
y2 + 1
y2 + 1
0
f (x, y) dx dy =
0
0
0
Either y ≤ 1 or y ≤ y 2 for all y ∈ (0, ∞), so
Z b
Z b
Z b
2
1 − e−b
| cos(b)|
−by y sin(b) + cos(b)
−by
−by y| sin(b)|
e
dy ≤
2e
dy = 2
dy ≤
+ 2
.
e
y2 + 1
y2 + 1
y +1
b
0
0
0
(5)
(6)
Together (3), (4), (5) and (6) imply that
Z b
Z bZ b
Z b
1
sin(x)
π
f (x, y) dx dy = lim
dx + 0 = lim
dy + 0 = lim (tan−1 (b) − tan−1 (0)) = .
lim
2
b→∞ 0 y + 1
b→∞ 0
b→∞
b→∞ 0
x
2
0
60. If x, y ∈ (0, ∞) then, by Exercise 51
Z ∞
Z
x−1 −s
Γ(x)Γ(y) =
s e ds
0
∞
t
y−1 −t
e
0
Z
∞Z ∞
dt =
0
sx−1 ty−1 e−(s+t) ds dt.
0
Define G : (0, ∞) × (0, 1) → (0, ∞)2 by G(u, v) := (uv, u(1 − v)), and check that G is a C 1 -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
Z 1Z ∞
Z 1
Z ∞
Z 1
(uv)x−1 (u(1 − v))y−1 e−u u du dv =
v x−1 (1 − v)y−1 dv
ux+y−1 e−u du = Γ(x + y)
tx−1 (1 − t)y−1 dt.
0
0
0
0
0
61. (a) Let α, β ∈ (0, ∞) and x ∈ [0, ∞). Note that
Z x
1
Iα (Iβ f )(x) =
(x − t)α−1 Iβ f (t) dt
Γ(α) 0
Z x
Z t
1
α−1 1
=
(x − t)
(t − s)β−1 f (s) ds dt
Γ(α) 0
Γ(β) 0
Z xZ t
1
=
(x − t)α−1 (t − s)β−1 f (s) ds dt.
Γ(α)Γ(β) 0 0
R1
Since f is bounded on [0, x] and 0 tγ dt < ∞ for all γ ∈ (−1, ∞), the Fubini-Tonelli theorem implies that
Z xZ x
1
Iα (Iβ f )(x) =
(x − t)α−1 (t − s)β−1 f (s) dt ds.
Γ(α)Γ(β) 0 s
For each s we apply the substitution u := (t − s)/(x − s) to the inner integral, and obtain
Z xZ 1
1
Iα (Iβ f )(x) =
(x − s − u(x − s))α−1 (u(x − s))β−1 f (s)(x − s) du ds
Γ(α)Γ(β) 0 0
Z xZ 1
1
=
(x − s)α−1 (1 − u)α−1 uβ−1 (x − s)β−1 f (s)(x − s) du ds
Γ(α)Γ(β) 0 0
Z x
Z 1
1
α+β−1
=
(x − s)
f (s) ds
(1 − u)α−1 uβ−1 du
Γ(α)Γ(β) 0
0
Z x
1
α+β−1
=
(x − s)
f (s) ds
Γ(α + β) 0
= Iα+β f (x).
21
Real Analysis
Chapter 2 Solutions
Jonathan Conder
(b) Clearly I1 f is an antiderivative of f. Given n ∈ N, we aim to show that In f is an nth-order antiderivative of f.
By induction, we may assume that n > 1 and that In−1 (I1 f ) is an (n − 1)th-order antiderivative of I1 f. Hence
(In f )(n) = ((In−1 (I1 f ))(n−1) )0 = (I1 f )0 = f.
62. Let E ⊆ S n−1 be measurable and T ∈ SO(n) a rotation. We
−1
Φ ((0, 1) × E) = x ∈ Rn \ {0}
want to show that σ(T (E)) = σ(E). Note that
|x| ∈ (0, 1) and x ∈ E ,
|x|
while
Φ
−1
x
((0, 1) × T (E)) = x ∈ R \ {0} |x| ∈ (0, 1) and
∈ T (E)
|x|
−1
T −1 x
n
= x ∈ R \ {0} |T x| ∈ (0, 1) and −1 ∈ E
|T x|
n
= {x ∈ Rn \ {0} | T −1 x ∈ 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)) =
R1
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
Z 1/2
Z ∞
α
b
r | log(r)| dr and
rα | log(r)|b dr
0
2
are finite. We claim that, given ε ∈ (0, ∞), there exists δ ∈ (0, 21 ) 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
as claimed. If α > −1 and b > 0, set ε :=
1
2b (α
δ
Z
α
) = log(n!) + r−1/n = δ −1/n + r−1/n ≤ 2r−1/n ≤ 2r−ε ,
+ 1), take the corresponding δ and note that
δ
Z
b
−1/n
r | log(r)| dr ≤
α b −(α+1)/2
r 2r
0
b
Z
dr = 2
0
δ
r(α−1)/2 dr < ∞,
0
as 21 (α − 1) > −1. Since rα | log(r)|b is bounded for r ∈ (δ, 12 ), 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
1
infinite if α < −1 and b > 0. If α < −1 and b < 0, set ε := − 2b
(α + 1), take the corresponding δ and note that
Z
δ
rα | log(r)|b dr ≥
δ
Z
0
rα 2b r(α+1)/2 dr = 2b
0
Z
δ
r(3α+1)/2 dr = ∞,
0
as 12 (3α + 1) < −1. Thus, the first integral is finite. Finally, set α := −1. By the monotone convergence theorem
Z
1/2
α
b
Z
r | log(r)| dr = − lim
0
t→0 t
1/2
1/2
(− log(r))b+1 (− log(r)) d(− log(r)) = − lim
,
t→0
b+1
t
b
22
Real Analysis
Chapter 2 Solutions
Jonathan Conder
R 1/2
which is finite iff b < −1 (the case b = −1 should really be handled separately). In summary, 0 rα | log(r)|b dr
is 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
Z ∞
Z 1/2
rα | log(r)|b dr =
s−α−2 | log(s)|b ds,
2
0
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|)
q
|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)
q
= |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 , . . . , F n−1 . The Jacobian of G at a point (r, φ1 , . . . , φn−2 , θ) ∈ Rn is


Fr1
Fφ11
···
Fφ1n−2
0


..
..
..
..
..


.
.
.
.
.




n−2
n−2
 Frn−2
,
F
·
·
·
F
0
φ
φ
1
n−2


F n−1 cos(θ) F n−1 cos(θ) · · · F n−1 cos(θ) −F n−1 sin(θ)
 r

φ1
φn−2
Frn−1 sin(θ)
Fφn−1
sin(θ) · · ·
1
Fφn−1
sin(θ)
n−2
F n−1 cos(θ)
so its determinant is F n−1 sin2 (θ) det(DF ) + F n−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 definition 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 S n−1 as a smooth manifold it is straightforward to show that (F |Ω0 )−1 is a diffeomorphism, but
that’s outside the scope of this course. Given an integrable function f : S 1 → C, define g : Rn → C by
x
g(x) := f ( |x|
)χB1 (0) (x). By Theorem 2.49 and Exercise 51
Z
Z
∞Z
g=
Rn
g(rx)r
0
n−1
Z
1Z
dσ(x) dr =
S n−1
f (x)r
0
On the other hand,
Z
S n−1
n−1
1
dσ(x) dr =
n
Z
f.
S n−1
f (F (φ1 , . . . , φn−2 , θ)) sinn−2 (φ1 ) . . . sin(φn−2 ) dφ1 · · · dφn−2 dθ
Z
Z 1
f (F (φ1 , . . . , φn−2 , θ))| sinn−2 (φ1 ) . . . sin(φn−2 )| dφ1 · · · dφn−2 dθ
rn−1 dr
=n
0
Ω
Z0 ∞ Z
=n
g(rF (φ1 , . . . , φn−2 , θ))rn−1 | sinn−2 (φ1 ) . . . sin(φn−2 )| dφ1 · · · dφn−2 dθ dr
0
Ω
Z0
= n g(G(r, φ1 , . . . , φn−2 , θ))| det(D(r,φ1 ,...,φn−2 ) G)| dφ1 · · · dφn−2 dθ dr
ZΩ
=n
g
G(Ω)
Z
=
f.
Ω0
S n−1
24