Instructor's Solutions Manual to Essentials of Integration Theory for Analysis

58 Pages·2011·0.487 MB·English

by  Daniel W. Stroock

Solutions Manual
§1.1
(1.1.10): First note that, for any α∈R, αf is Riemann integrable if f is.
To prove that f ∨g is Riemann integrable if f and g are, observe that
(cid:12)(cid:12)a(cid:48)∨b(cid:48)−a∨b≤|a(cid:48)−a|∨|b(cid:48)−b|(cid:12)(cid:12)≤|a(cid:48)−a|+|b(cid:48)−b| for any a, a(cid:48), b, and b(cid:48) ∈R.
Thus, for any C,
(cid:0) (cid:1) (cid:0) (cid:1) (cid:104) (cid:105) (cid:104) (cid:105)
U f ∨g;C −L f ∨g;C ≤ U(f;C)−L(f;C) + U(g;C)−L(g;C) .
Now apply the final part of Lemma 1.1.5 and Theorem 1.1.9 to see that f∨g
is Riemann integrable if f and g are. In addition, since
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
R f ∨g;C,ξ ≥R f;C,ξ ∨R g;C,ξ
for every C and ξ ∈ Ξ(C), the corresponding inequality is obvious for the
integrals.
One can handle f ∧g by either applying an analogous line of reasoning or
simplynotingthatf∧g =−(−f)∨(−g). Finally, toprovethelastpart, first
reduce to the case when α=1=β, and then observe that
(cid:104) (cid:105) (cid:104) (cid:105)
U(f +g;C)−L(f +g;C)≤ U(f;C)−L(f;C) + U(g;C)−L(g;C) .
Hence, f +g is Riemann integrable if f and g are. The linearity assertion
follows immediately from the linearity of the approximating Riemann sums.
(1.1.11): Given C and (cid:15)>0, set
(cid:26) (cid:27)
C((cid:15))= I ∈C : supf −inff ≥(cid:15) .
I I
Assume that f is Riemann integrable, and let (cid:15) > 0 be given. Then there
exists a δ >0 such that
U(f;C)−L(f;C)<(cid:15)2
1
2 Solutions Manual
whenever (cid:107)C(cid:107)<δ. Thus, if (cid:107)C(cid:107)<δ, then
(cid:18) (cid:19)
(cid:88) (cid:88)
(cid:15) vol(I)≤ supf −inff vol(I)
I I
I∈C((cid:15)) I∈C((cid:15))
(cid:18) (cid:19)
(cid:88)
≤ supf −inff vol(I)=U(f;C)−L(f;C)<(cid:15)2
I I
I∈C
as long as (cid:107)C(cid:107)<δ.
Conversely, suppose that for each (cid:15)>0 there is a C((cid:15)) for which
(cid:88)
vol(I)<(cid:15).
I∈C((cid:15))
Then
(cid:0) (cid:1) (cid:0) (cid:1)
U f;C((cid:15)) −L f;C((cid:15))
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) (cid:88)
= supf −inff vol(I)+ supf −inff vol(I)
I I I I
I∈C((cid:15)) I∈C\C((cid:15))
(cid:88) (cid:88)
≤2(cid:107)f(cid:107) vol(I)+(cid:15) vol(I)
u
I∈C((cid:15)) I∈C\C((cid:15))
(cid:0) (cid:1)
≤2(cid:107)f(cid:107) (cid:15)+(cid:15)vol(J)=(cid:15) 2(cid:107)f(cid:107) +vol(J) −→0 as (cid:15)(cid:38)0.
u u
Hence, inf U(f;C)=sup L(f;C), and so f is Riemann integrable.
C C
(1.1.12): Suppose that f is a bounded function on J which is continuous at
all but a finite number of points, a ,...,a . If N =1, write J =[c,d]; and if
1 m
N ≥2, write J =[c,d]×J(cid:48), where J(cid:48) is a rectangle in RN−1.
Given (cid:15) > 0, choose c = α ≤ β ≤ ··· ≤ α ≤ β = d so that
0 0 m+1 m+1
max (β −α )<(cid:15)and{a ,...,a }isdisjointfrom(cid:83)m R ,where,
0≤µ≤m+1 µ µ 1 m µ=1 µ
dependingonwhetherN =1orN ≥2,R =[β ,α ]orR =[β ,α ]×
µ µ µ+1 µ µ µ+1
J(cid:48). Clearly, f (cid:22) R is continuous for each µ. Hence, for each 0 ≤ µ ≤ m, we
µ
can find a non-overlapping, finite, exact cover C of R such that U(f;C )≤
µ µ µ
L(f;C )+(cid:15).
µ
Now, set I equal to [α ,β ] or [α ,β ]×J(cid:48), and observe that the union
µ µ µ µ µ
of {I : 0≤µ≤m+1} with (cid:83)m C forms a non-overlapping, finite, exact
µ µ=0 µ
cover C of J by closed rectangles. Finally,
m+1 m
(cid:88)(cid:16) (cid:17) (cid:88)(cid:0) (cid:1)
U(f;C)−L(f;C)= supf−inff vol(I )+ U(f;C )−L(f;C ) ≤C(cid:15),
µ µ µ
µ=0 Iµ Iµ µ=0
whereCisdominatedby2(cid:107)f(cid:107) (m+2)+(m+1)or2(cid:107)f(cid:107) (m+2)vol(J(cid:48))+(m+1)
u u
depending on whether N =1 or N ≥2.
Solutions Manual 3
§1.2
(1.2.17):
(i)Given(cid:15)>0,chooseδ >0sothat|ψ(cid:48)(y)−ψ(cid:48)(x)|<(cid:15)whenever|y−x|<δ.
Next, given C with (cid:107)C(cid:107)<δ, use the Mean Value Theorem to choose η(I)∈˚I
for each I ∈C so that ∆ ψ =ψ(cid:48)(cid:0)η(I)(cid:1)vol(I). Then, for any ξ ∈Ξ(C),
I
(cid:12)(cid:12)R(ϕ|ψ;C,ξ)−R(ϕψ(cid:48);C,ξ)(cid:12)(cid:12)
≤(cid:88)(cid:12)(cid:12)ϕ(cid:0)ξ(I)(cid:1)(cid:12)(cid:12)(cid:12)(cid:12)ψ(cid:48)(cid:0)η(I)(cid:1)−ψ(cid:48)(cid:0)ξ(I)(cid:1)(cid:12)(cid:12)≤(cid:15)(cid:107)ϕ(cid:107)uvol(J).
I∈C
Hence, since ϕψ(cid:48) is continuous and therefore Riemann integrable, we have
now proved that
(cid:12) (cid:90) (cid:12)
lim sup (cid:12)(cid:12)R(ϕ|ψ;C,ξ)−(R) ϕ(x)ψ(cid:48)(x)dx(cid:12)(cid:12)=0.
(cid:107)C(cid:107)→0ξ∈Ξ(C)(cid:12) J (cid:12)
(cid:12)
(ii) Given (cid:15) > 0, choose 0 < δ < min1≤m≤n(am −am−1) so that (cid:12)ϕ(y)−
(cid:12)
ϕ(x)(cid:12) < (cid:15) whenever |y−x| < δ. Next, let C with (cid:107)C(cid:107) < δ be given. Clearly,
no I ∈ C contains more than one of the a ’s and each a is in at most two
m m
I’s. In fact, for any I ∈C,
 0 if I∩{a ,...,a }=∅
 d if a ∈˚I0 n
m m
∆ ψ =
I  dd+m− ≡≡ψψ((aam+)−)−ψ(ψa(a−m)) iiff II−+ ==aam.
m m m m
Thus, if
M ={m: a ∈˚I for some I ∈C}
0 m
and
M ={m: a =I± for some I ∈C},
± m
and if ξ ∈C is chosen so that a ∈I =⇒ ξ (I)=a , then
0 m 0 m
R(cid:0)ϕ|ψ;C,ξ (cid:1)= (cid:88) ϕ(a )d + (cid:88) ϕ(a )d− + (cid:88) ϕ(a )d+
0 m m m m m m
m∈M0 m∈M+ m∈M−
n
(cid:88)
= ϕ(a )d ,
m m
m=0
since (when d− =d+ ≡0) d =d− +d+, 0≤m≤n. At the same time, for
0 n m m m
any ξ ∈Ξ(C),
n
(cid:12)(cid:12)R(cid:0)ϕ(cid:12)(cid:12)ψ;C,ξ(cid:1)−R(cid:0)ϕ(cid:12)(cid:12)ψ;C,ξ0(cid:1)(cid:12)(cid:12)≤(cid:15)(cid:88)(cid:0)|d−m|+|d+m|(cid:1).
0
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Hence, we have now shown that
(cid:12) (cid:12)
(cid:12) (cid:88)n (cid:12)
lim sup (cid:12)R(ϕ|ψ;C,ξ)− ϕ(a )d (cid:12)=0.
(cid:12) m m(cid:12)
(cid:107)C(cid:107)→0ξ∈Ξ(C)(cid:12) m=0 (cid:12)
(iii)Withoutlossofgenerality,assumethatα=β =1. Given(cid:15)>0,choose
δ >0 so that
(cid:12) (cid:90) (cid:12)
(cid:12) (cid:12) (cid:12)
(cid:107)C(cid:107)<δ =⇒ max sup (cid:12)R(ϕi (cid:12)ψ;C,ξ)−(R) ϕi(x)dψ(x)(cid:12)<(cid:15).
i∈{1,2}ξ∈Ξ(C)(cid:12) J (cid:12)
Because R(ϕ +ϕ |ψ;C,ξ)=R(ϕ |ψ;C,ξ)+R(ϕ |ψ;C,ξ), it is then clear
1 2 1 2
that
(cid:12) (cid:90) (cid:90) (cid:12)
(cid:12) (cid:0) (cid:12) (cid:1) (cid:12)
sup (cid:12)R ϕ1+ϕ2 (cid:12)ψ;C,ξ −(R) ϕ1(x)dψ(x)−(R) ϕ2(x)dψ(x)(cid:12)<2(cid:15)
(cid:12) (cid:12)
ξ∈Ξ(C) J J
whenever (cid:107)C(cid:107)<δ.
(iv) We begin by showing that ϕ is ψ-Riemann integrable on both J and
1
J . To this end, let (cid:15)>0 be given, and choose δ >0 so that
2
(cid:12) (cid:90) (cid:12)
(cid:12) (cid:12)
(*) (cid:107)C(cid:107)<δ =⇒ sup (cid:12)R(ϕ|ψ;C,ξ)−(R) ϕ(x)dψ(x)(cid:12)<(cid:15).
(cid:12) (cid:12)
ξ∈Ξ(C) J
Now, suppose that C and C(cid:48) are finite, non-overlapping covers of J with
1 1 1
mesh size less than δ. Let C be any finite, non-overlapping cover of J with
2 2
meshsizelessthanδ,andsetC =C ∪C andC(cid:48) =C(cid:48)∪C . Next,letξ ∈Ξ(C )
1 2 1 2 1 1
and ξ(cid:48) ∈Ξ(C(cid:48)) be given, and define ξ ∈Ξ(C) and ξ(cid:48) ∈Ξ(C(cid:48)) so that
1 1
(cid:26) ξ (I) if I ∈C (cid:26) ξ(cid:48)(I) if I ∈C(cid:48)
ξ(I)= 1 1 and ξ(cid:48)(I)= 1 1
ξ (I) if I ∈C ξ (I) if I ∈C ,
2 2 2 2
where ξ ∈Ξ(C ). Then, by (*),
2 2
(cid:12)(cid:12)R(ϕ|ψ;C,ξ)−R(ϕ|ψ;C(cid:48),ξ(cid:48))(cid:12)(cid:12)<2(cid:15).
At the same time,
R(ϕ|ψ;C ,ξ )−R(ϕ|ψ;C(cid:48),ξ(cid:48))=R(ϕ|ψ;C,ξ)−R(ϕ|ψ;C(cid:48),ξ(cid:48)).
1 1 1 1
Thus, by Cauchy’s convergence criterion, we have now shown that ϕ is ψ-
Riemann integrable on J . The same argument shows that ϕ is ψ-Riemann
1
integrable on J .
2
Solutions Manual 5
To prove the asserted equality, let (cid:15)>0 be given, and choose δ >0 so that
(*) holds. Next, choose C for J and ξ ∈Ξ(C ) so that (cid:107)C (cid:107)<δ and
i i i i i
(cid:12) (cid:90) (cid:12)
(cid:12) (cid:0) (cid:1) (cid:12)
(**) max (cid:12)R ϕ|ψ;Ci,ξi −(R) ϕ(x)dψ(x)(cid:12)<(cid:15).
i∈{1,2}(cid:12) Ji (cid:12)
Finally, set C =C ∪C , and define ξ ∈Ξ(C) by
1 2
(cid:26) ξ (I) if I ∈C
1 1
ξ(I)=
ξ (I) if I ∈C .
2 2
Then,
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
R ϕ|ψ,C,ξ =R ϕ|ψ;C ,ξ +R ϕ|ψ;C ,ξ ,
1 1 2 2
and so (*) and (**) imply that
(cid:12) (cid:90) (cid:90) (cid:90) (cid:12)
(cid:12) (cid:12)
(cid:12)(R) ϕ(x)dψ(x)−(R) ϕ(x)dψ(x)−(R) ϕ(x)dψ(x)(cid:12)<3(cid:15).
(cid:12) (cid:12)
J J1 J2
(1.2.18): It is obvious that ψ is continuous. To see that it has unbounded
variation, set t = 1 for n≥1. Then, for each n≥1,
n nπ
n (cid:18) (cid:19)
(cid:0) (cid:1) 1 (cid:88) 1 1
Var ψ;[0,1] ≥ + ,
π 2m 2m−1
m=1
and the right hand side tends to ∞ as n→∞.
To handle the second part of the exercise, consider the function
(cid:26) 0 if t∈[0,1]\{1}
ψ(t)= 2
1 if t= 1.
2
Clearly Var(cid:0)ψ;[0,1](cid:1) = 2. On the other hand, for any ϕ ∈ C(cid:0)[0,1];R(cid:1), one
has that R(cid:0)ϕ|ψ;C,ξ(cid:1)=0 as long as 1 is not an endpoint of any I ∈C. Thus,
2
(R)(cid:82) ϕ(x)dψ(x)=0 for every ϕ∈C(cid:0)[0,1];R(cid:1).
[0,1]
(1.2.19): Suppose that ϕ is ψ-Riemann integrable on J. Given (cid:15)>0, choose
δ >0 so that
(cid:12) (cid:12)
(cid:12)(cid:88)(cid:18) (cid:19) (cid:12)
(cid:107)C(cid:107)<δ =⇒ (cid:12) supϕ−infϕ ∆ ψ(cid:12)<(cid:15)2.
(cid:12) I (cid:12)
(cid:12) I I (cid:12)
I∈C
Then, since
(cid:18) (cid:19)
(cid:88) (cid:88)
(cid:15) ∆ ψ ≤ supϕ−infϕ ∆ ψ,
I I
I I
{I∈C:supIϕ−infIϕ≥(cid:15)} I∈C
6 Solutions Manual
it is clear that (1.2.20) holds whenever (cid:107)C(cid:107)<δ.
Conversely,supposethat,forevery(cid:15)>0,thereisaδ >0forwhich(1.2.20)
holds. Since
(cid:18) (cid:19)
(cid:88) (cid:88)
supϕ−infϕ ∆ ψ ≤(cid:15)∆ ψ+2(cid:107)ϕ(cid:107) ∆ ψ,
I J u I
I I
I∈C {I∈C:supIϕ−infIϕ≥(cid:15)}
it follows that ϕ is ψ-Riemann integrable.
Finally, the argument just given shows that the existence for every (cid:15)>0 of
a C for which (1.2.20) holds is equivalent to the existence for every (cid:15)>0 of a
C for which
(cid:18) (cid:19)
(cid:88)
supϕ−infϕ ∆ ψ <(cid:15)
I
I I
I∈C
holds. Thus, what we must show when ψ ∈C(J;R) is that, for every C,
(cid:88) (cid:88)
lim ∆ ψ ≤ ∆ ψ
I I
(cid:107)C(cid:48)(cid:107)→0
{I∈C(cid:48):supIϕ−infIϕ≥(cid:15)} {I∈C:supIϕ−infIϕ≥(cid:15)}
and
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) (cid:88)
lim supϕ−infϕ ∆ ψ ≤ supϕ−infϕ ∆ ψ.
I I
(cid:107)C(cid:48)(cid:107)→0 I I I I
I∈C(cid:48) I∈C
To this end, let C be given and set S = {I+ : I ∈ C and I+ (cid:54)= J+}. Given a
C(cid:48), set
(cid:26) (cid:27)
C(cid:48)((cid:15))= I(cid:48) ∈C(cid:48) : supϕ−infϕ≥(cid:15) .
I(cid:48) I(cid:48)
Then
(cid:88) (cid:88) (cid:88) (cid:88) (cid:88)
∆ ψ ≤ ∆ ψ+ ∆ ψ
I(cid:48) I(cid:48) I(cid:48)
I(cid:48)∈C(cid:48)((cid:15)) I∈C{I(cid:48)∈C(cid:48)((cid:15)):I(cid:48)⊆I} t∈S{I(cid:48)∈C(cid:48)((cid:15)):t∈˚I(cid:48)}
(cid:88)
≤ ∆ ψ+card(S)max∆ ψ.
I I(cid:48)
I(cid:48)∈C(cid:48)
I∈C((cid:15))
But, by continuity, the last term tends to 0 as (cid:107)C(cid:48)(cid:107) −→ 0, and so we have
proved the first of the above. The proof of the second is essentially the same,
only this time one uses
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) (cid:88) (cid:88)
supϕ−infϕ ∆ ψ ≤ supϕ−infϕ ∆ ψ.
I(cid:48) I
I(cid:48) I(cid:48) I I
I∈C{I(cid:48)∈C(cid:48):I(cid:48)⊆I} I∈C
(1.2.21): Let ψ ∈C(J;R). What we must show is that, for each C and (cid:15)>0,
there is a δ >0 such that
S(cid:0)ψ;C(cid:1)−S(ψ;C(cid:48))<(cid:15)
Solutions Manual 7
whenever (cid:107)C(cid:48)(cid:107)<δ. To this end, suppose that
(cid:110)(cid:2) (cid:3) (cid:2) (cid:3)(cid:111)
C = a ,a ,..., a ,a
0 1 n−1 n
where J− = a < ··· < a = J+. Next, given (cid:15) > 0, choose 0 < δ <
0 n
(cid:0) (cid:1)
min a −a so that
1≤m≤n m m−1
(cid:110) (cid:111) (cid:15)
ω (δ)≡sup |ψ(t)−ψ(s)|: s, t∈J and |t−s|<δ < .
ψ 2n
If (cid:107)C(cid:48)(cid:107)<δ and A is the set of those I(cid:48) ∈C(cid:48) for which there is an I ∈C with
I(cid:48) ⊆I, thenforeachI(cid:48) ∈B ≡C(cid:48)\A, thereispreciselyonem∈{1,...,n−1}
forwhicha ∈˚I(cid:48). Inparticular,becauseC(cid:48) isnon-overlapping,B hasatmost
m
n elements. Moreover, if I(cid:48) ∈ B, a ∈ ˚I(cid:48), and we use L(I(cid:48)) and R(I(cid:48)) to
m
denote I(cid:48)∩(cid:2)a ,a (cid:3) and I(cid:48)∩(cid:2)a ,a (cid:3), respectively, then
m−1 m m m+1
C∨C(cid:48) =A∪ (cid:91) (cid:8)[L(I(cid:48)),R(I(cid:48))](cid:9).
I(cid:48)∈B
Hence,
S(cid:0)ψ;C(cid:1)−S(ψ;C(cid:48))≤S(cid:0)ψ;C(cid:48)∨C(cid:1)−S(ψ;C)
(cid:88) (cid:0)(cid:12) (cid:12) (cid:12) (cid:12)(cid:1)
≤ (cid:12)∆L(I(cid:48))ψ(cid:12)+(cid:12)∆R(I(cid:48))ψ(cid:12) ≤2nωψ(δ)<(cid:15).
I(cid:48)∈B
Finally, by (1.2.10) and (1.2.11), the analogous result for Var and Var
− +
follow immediately.
Now, suppose that ψ ∈ C1(J;R). For n ∈ Z+ and 1 ≤ m < n, set a =
m,n
J−+m∆J,andusetheMeanValueTheoremtochooseξ ∈(a ,a )
n m,n m,n m+1,n
so that ϕ(a )−ϕ(a )=ϕ(cid:48)(ξ )∆J. Then
m+1,n m,n m,n n
∆J n(cid:88)−1 (cid:90)
Var (ψ)= lim ψ(cid:48)(ξ )+ =(R) ϕ(cid:48)(t)+dt,
+ n→∞ n m,n
m=0 J
and similarly for Var (ψ) and Var(ψ).
−
(1.2.22): Let ψ and ψ be given. Then, for any a≤s<t≤b,
1 2
(cid:16) (cid:17) (cid:16) (cid:17)+
ψ (t)−ψ (s)=ψ(t)−ψ(s)+ ψ (t)−ψ (s) ≥ ψ(t)−ψ(s)
2 2 1 1
since both ψ (t)−ψ (s) and ψ (t)−ψ (s) are non-negative. Hence, for any
2 2 1 1
a≤s<t≤b and any finite, non-overlapping, exact cover C of [s,t],
(cid:88) (cid:88)(cid:0) (cid:1)+
ψ (t)−ψ (s)= ∆ ψ ≥ ∆ ψ ;
2 2 I 2 I
I∈C I∈C
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andsoψ (t)−ψ (s)≥ψ (t)−ψ (s). Since∆ ψ −∆ ψ =∆ ψ −∆ ψ for
2 2 + + I 1 I − I 2 I +
anyintervalI ⊆J, wenowseethatψ −ψ andψ −ψ arenon-decreasing.
2 + 1 −
Turning to the statements about the relationship between the jumps of ψ,
ψ , and ψ , begin by observing that
+ −
(cid:12) (cid:12) (cid:0) (cid:1)±
(cid:12)ψ+(x±)−ψ+(x)(cid:12)= ψ(x±)−ψ(x)
(cid:12) (cid:12) (cid:0) (cid:1)∓
=⇒ (cid:12)ψ−(x±)−ψ−(x)(cid:12)= ψ(x±)−ψ(x) .
Further,ifψˇ(x)≡ψ(a+b−x), x∈[a,b],thenψˇagainhasboundedvariation,
ψ (x)=ψˇ (a+b−x), ψ(x−)=ψˇ(cid:0)(a+b−x)+(cid:1),
± ∓
and ψ (x−)=ψˇ (cid:0)(a+b−x)+(cid:1).
± ∓
(cid:0) (cid:1)+
Thus, we need only show that ψ (x+)−ψ (x) = ψ(x+)−ψ(x) , and,
+ +
because we can always restrict everything to the interval [x,b], it suffices to
(cid:0) (cid:1)+
handle x=a. To this end, note that ψ (a+)−ψ (a)= ψ(a+)−ψ(a) is
+ +
equivalent to β ≡ψ (a+)∧ψ (a+)=0. But, if
+ −
(cid:26) 0 if x=a
ψ (x)=
1
ψ (x)−β if x∈(a,b]
−
and
(cid:26) ψ(a) if x=a
ψ (x)=
2
ψ(a)+ψ (x)−β if x∈(a,b],
+
thenψ andψ arebothnon-decreasingandψ =ψ −ψ . Hence, bythefirst
1 2 2 1
part of this exercise, ψ ≤ψ , and so ψ (b)≤ψ (b)−β. That is, β =0.
− 1 − −
(1.2.23):
(i) For (cid:15) > 0, set D = {x ∈ (c,d] : |ψ(x)−ψ(x−)| ≥ (cid:15)}. Then D is
(cid:15) (cid:15)
finite. Choose c = x < ··· < x ≤ d so that D = {x ,...,x }, and set δ =
0 (cid:96) (cid:15) 1 (cid:96)
1min (x −x ). DefineI =(cid:2)x − δ,x (cid:3)andI(cid:48) =(cid:2)x ,x − δ(cid:3)
2 1≤k≤(cid:96) k k−1 k,n k n k k,n k−1 k n
for 1≤k ≤(cid:96). Then C ≡{I : 1≤k ≤(cid:96)}∪{I(cid:48) : 1≤k ≤(cid:96)}∪{[x ,d]} is
n k,n k,n (cid:96)
a finite, non-overlapping cover of J, and so
(cid:96)
Var (ψ;J)≥(cid:88)(∆ ψ)± −→ (cid:88)(cid:0)ψ(x)−ψ(x−)(cid:1)±
± Ik,n
k=1 x∈D(cid:15)
as n→∞. Hence, after letting (cid:15)(cid:38)0, we have proved (i).
Solutions Manual 9
(ii) Assume that ψ is a pure jump function. By (i) we know that
(cid:88) (cid:0) (cid:1)±
Var (ψ;J)≥ ψ(x)−ψ(x−) .
±
x∈D(ψ)
To prove the opposite inequality, let C be given, and for each I ∈ C let
I(cid:48) =I\{I−}. Then
 ±
(cid:88)(∆Iψ)± =(cid:88) (cid:88) (cid:0)ψ(x)−ψ(x−)(cid:1)
I∈C I∈C x∈D(ψ)∩I(cid:48)
(cid:88) (cid:0) (cid:1)±
≤ ψ(x)−ψ(x−) .
x∈D(ψ)
Hence, after taking the supremum over C’s, the result follows.
(iii) To see that ψ is right-continuous, note that
d
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:88) (cid:0) (cid:1)(cid:12)
(cid:12)ψd(x+h)−ψd(x)(cid:12)=(cid:12) ψ(x)−ψ(x−) (cid:12)
(cid:12) (cid:12)
(cid:12)x∈D(ψ)∩(x,x+h] (cid:12)
(cid:88) (cid:12) (cid:12)
≤ (cid:12)ψ(x)−ψ(x−)(cid:12)≤Var(ψ;J)<∞.
x∈D(ψ)∩(x,x+h]
Hence, since D(ψ)∩(x,x+h] (cid:38) ∅ as h (cid:38) 0, ψ (x) = lim ψ (x+h).
d h(cid:38)0 d
Moreover, just as in (ii),
(cid:88) (cid:0) (cid:1)±
Var (ψ ;J)= ψ(x)−ψ(x−) ≤Var (ψ;J),
± d ±
x∈D(ψ)
and therefore ψ has bounded variation. Finally, because D(ψ) = D(ψ ) and
d
ψ (x)−ψ (x−)=ψ(x)−ψ(x−) for x∈D(ψ), we are done.
d d
(iv) To prove that ψ is continuous, suppose that ψ (x)(cid:54)=ψ (x−) for some
c c c
x∈(c,d]. Then we would have that ψ (x)−ψ (x−)(cid:54)=ψ(x)−ψ(x−), which,
d d
because D(ψ ) = D(ψ), would lead to the contradiction that x ∈ D(ψ) and
d
yet ψ (x)−ψ (x−)(cid:54)=ψ(x)−ψ(x−). To see that ψ is the only pure jump ϕ
d d d
for which ψ˜=ψ−ϕ is continuous, note that ψ˜−ψ =ψ −ϕ. Hence ψ −ϕ
c d d
would be a continuous, pure jump function, of which the function which is
identically 0 is the only one.
(v) First note that
ψ =ψ +ψ =⇒ Var (ψ;J)≤Var (ψ )+Var (ψ ).
c d ± ± c ± d
10 Solutions Manual
To prove the opposite inequality, use the preceding to see that it suffices to
show that, for each (cid:15)>0,
(cid:88)(cid:0) (cid:1)±
Var (ψ;J)≥Var (ψ ;J)+ ψ(x)−ψ(x−)
± ± c
(*) x∈D(cid:15)
(cid:88)
− |ψ(x)−ψ(x−)|,
x∈D(ψ)\D(cid:15)
where D = {x ∈ (c,d] : |ψ(x) − ψ(x−)| ≥ (cid:15)}. Next, given (cid:15) > 0 and
(cid:15)
n ≥ 1, choose C so that (cid:107)C (cid:107) ≤ 1 and, for each I ∈ C , either I ∩D = ∅
n n n n (cid:15)
or I contains precisely one element of D and this element is its right hand
(cid:15)
endpoint. Then
Var (ψ;J)≥ (cid:88) (cid:2)(∆ ψ )±−Var(ψ ;I)(cid:3)+ (cid:88) (∆ ψ)±
± I c d I
I∈Cn I∈Cn
I∩D(cid:15)=∅ I∩D(cid:15)(cid:54)=∅
(cid:88) (cid:88) (cid:88)
≥ (∆ ψ )±+ (∆ ψ)±−(cid:96)ω − |ψ(x)−ψ(x−)|,
I c I n
I∈Cn I∈Cn x∈D(ψ)\D(cid:15)
I∩D(cid:15)(cid:54)=∅
where (cid:96) = card(D ) and ω = sup{|ψ (y)−ψ (x)| : |y −x| ≤ 1}. Hence,
(cid:15) n c c n
since by Exercise 1.2.21, Var (ψ ;J)=lim (cid:80) (∆ ψ )±, (*) follows.
± c n→∞ I∈Cn I c
(1.2.24):
(cid:112)
(i) Because α2+β2 ≤α+β for all α, β ≥0,
(cid:88)(cid:112) (cid:88)(cid:0) (cid:1) (cid:0) (cid:1)
vol(I)2+(∆ ψ)2 ≤ vol(I)+|∆ ψ| ≤(d−c)+Var ψ;[c,d] ,
I I
I∈C I∈C
which proves the upper bound. The lower bound is a application of the
classical triangle inequality. Namely, for any n ≥ 1 and x ,...,x ∈ R2,
1 n
|(cid:80)n x | ≤ (cid:80)n |x |. Equivalently, for any {(a ,b ) : 1 ≤ m ≤ n} ⊆
m=1 m m=1 m m m
R2,
(cid:118)
n (cid:117)(cid:32) n (cid:33)2 (cid:32) n (cid:33)2
(cid:88) (cid:112) (cid:117) (cid:88) (cid:88)
a2 +b2 ≥(cid:116) |a | + |b | .
m m m m
m=1 m=1 m=1
Hence,
(cid:118)
(cid:117) (cid:32) (cid:33)2
(cid:0) (cid:1) (cid:88)(cid:112) (cid:117) (cid:88)
Arc ψ;[c,d] ≥ vol(I)2+(∆ ψ)2 ≥(cid:116)(d−c)2+ |∆ ψ| ,
I I
I∈C I∈C
from which the desired estimate follows when one takes the supremum over
C.