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Comparison with the related literature

Stability of Backward Stochastic Differential Equations with Jumps

IV.6. Comparison with the related literature

IV.6. Comparison with the related literature

In this section we are going to compare our result with the existing literature. More precisely, the comparison will be made with Madan et al. [50] which is the most general result for discrete-time approximations of BSDEs. However, we need to underline that the work of Briand et al. [15] has greatly inspired the author. In the following we will translate the notation to the one we have used.

We start with the Condition(B1), i.e. the limit-filtration has to be quasi-left-continuous, which is common between the two works. Actually, we may recall that M´emin [53, Counter-example 3] which justifies the necessity of (B1). However, in our case we are not restricted in (square-integrable) L´evy martingale, but it can be an arbitrary (square-integrable) one.

We proceed to the conditions imposed on the sequence of driving martingales (ĎXk,, Xk)k∈N. In [50] the integrators can be multidimensional, while we have presented the real-valued case. However, in view of the results in Chapter III, we can readily adapt the dimension of the integrators to a higher one. The same holds true for the dimension of the solution of the BSDE, in view of the main theorem of Chapter II. In Chapter IV we preferred to present the result in an as simple as possible manner. In both cases, they are square-integrable martingales whose terminal values converge in mean to the respective terminal values of the natural pair of X. Now we proceed to describe a few improvements that take place under our framework.

The time horizon in our case can be infinite, while in [50] it is assumed finite.

The terminal values in [50] need to converge in a stronger sense than inL2−mean, see [50, Conditions (2.2) and (2.5)], while in our case the convergence is in L2mean. Actually, we have explained at the beginning of Subsection III.6.1 the role of Lemma III.39 in order to obtain this sharp convergence.

In our framework we need MμXk,

ΔXk, !PGk = 0 for every k N, which is a weaker condition compared to independence assumption imposed in [50]. Let us argue about it. To this end, let us fix a finite time horizon T, a k N and a partition (tm)m=0,...,M of [0, T]. We will assume that Xk,, Xk, have independent increments and are constant on the interval [tm, tm+1) form = 0, . . . , M1. In this setting we have for every boundedGk−predictable functionU

MμXk,

U MμXk,

ΔXk, !PGk =MμXk,

UΔXk,

=EM−1

m=0

U(tm,ΔXtk,m)ΔXtk,m

= 0.

Finally, regarding the properties the generator should possess, we will restrict ourselves on mentioning that Madan et al. [50, Assumption 1] demands the generators to be uniformly Lipschitz. In Subsection IV.5.6 we have commented that the uniform Lipschitz condition translates the pointwise convergence into seem-ingly stronger kind of convergence. For this reason, we could say that we have used finally the usual convergence assumptions as they translate in a weaker framework.

APPENDIX A

Auxiliary results

A.1. Auxiliary results of Chapter II

Proof of Lemma II.13. Let (γ, δ)∈ Cβ.We shall begin by obtaining the critical points of the map ΠΦ.We have

∂γΠΦ(γ, δ) = (2 + 9δ) e(δ−γ)ΦΦγ2+ (2−δΦ)γ−δ γ2(δ−γ)2 ,

∂δΠΦ(γ, δ) =−9

δ2 + e(δ−γ)Φ

( 9 + (2 + 9δ)Φ (δ−γ)

γ(δ−γ)2 − 2 + 9δ γ(δ−γ)2

) .

The only possible critical points for ΠΦare therefore such thatδ=−2/9 orγ= δΦ−2±

4+δ2Φ2

. However, the values δ=−2/9 andγ= δΦ−2−

4+δ2Φ2

are ruled out as negative. For 0< δ≤β we have δΦ−2 +√

4 +δ2Φ2

2Φ , δ

∈ Cβ. Let us define γΦ(δ) := δΦ−2+

4+δ2Φ2

, for 0< δ≤β.It is easy to verify thatγΦ(δ)∈(0, δ).Then, some tedious calculations yield that

∂ΠΦ

∂δ γ¯Φ(δ), δ

=−9

δ2 −exp[ δ−¯γΦ(δ) Φ]

¯

γΦ(δ) δ−¯γΦ(δ)2 ·2¯γΦ(δ)Φ + 9¯γΦ(δ) + 2 (¯γΦ(δ)Φ + 1)2 <0

therefore ΠΦdoes not admit any critical point onCβ, for which 0< γ < δ < β. Hence, the infimum on this set is necessarily attained on its boundary. The cases where at least one amongδandγgoes to 0, or where their difference goes to 0, lead to the value∞. The only remaining case is therefore 0< γ < δ=β, where β is fixed. Then we get

d

dγΠΦ γ, β

= (2 + 9β) e(β−γ)ΦΦγ2+ (2−βΦ)γ−β γ2(β−γ)2 ,

and ΠΦ(γ, β) viewed as a function ofγ attains its minimum at its critical point given by γΦ(β), since

Φ γ, β

<0 on (0, γΦ(β)) and Φ γ, β

>0 on (γΦ(β), β).

Now, we proceed to the second case, and start by determining the critical points of ΠΦ?. It holds

∂γΠΦ?(γ, δ) =−8

γ2 + 9δe(δ−γ)ΦΦγ2−(δΦ−2)γ−δ γ2(δ−γ)2 ,

∂δΠΦ?(γ, δ) =−9

δ2+ 9e(δ−γ)Φ(1 +δΦ)(δ−γ)−δ γ(δ−γ)2 .

Following analogous computations as above we can prove that, for (γ, δ)∈ Cβ, the equation

∂γΠΦ?(γ, δ) = 0⇔Pδ(γ) := 8(δ−γ)2−9δe(δ−γ)Φ Φγ2−(δΦ−2)γ−δ

= 0

has a unique root, sayγΦ?(δ),which moreover satisfiesγΦ?(δ)∈(γΦ(δ), δ).This can be proved because the functionPδ: (0, δ)→Ris decreasing, for each fixedδ∈(0, β), withPδ γΦ(δ)

>0 andPδ(δ)<0. Now observe that for γ > 1+δΦδ2Φ it holds ∂δ ΠΦ?(γ, δ)<0 and that Pδ 1+δΦδ2Φ

>0. Using the monotonicity of Pδ we have thatγΦ?(δ)> 1+δΦδ2Φ and therefore also ∂δ ΠΦ?Φ?(δ), δ)<0.Arguing as above we can conclude that the infimum is attained forδ=β at the point γΦ?(β), β

.

Finally, the limiting statements follow by straightforward but tedious computations.

133

A.1.1. The non-exponential submultiplicative case. In this subsection we provide thea priori estimates of the semi–martingale decomposition (II.20) in the case hζ :R→[1,∞), where hζ(x) = 1 for x < 0 and hζ(x) := (1 +x)ζ, for x >0 where ζ is a fixed number greater than 1, with the additional assumption that the process A defined in (F4) is P−a.s. bounded by Ψ. However, using completely analogous arguments thea priori estimates can be obtained for a submultiplicative function h. In [64, Proposition 25.4] (see the proof of (iii)) we can find an easy criterion for an increasing functionh:R→R+ to be submultiplicative. Namely, it suffices for the function h: R → (0,∞) to be flat on (−∞, b] and the function log(h(·)) to be concave on (b,∞), for some b ∈ R+. It is immediate that the functionhζ

as defined above is submultiplicative. Since ζ is fixed, we will simply denotehζ as hin the rest of the section.

Before we proceed to the proof we provide the properties of the submultiplicative function we will need, presented however for the function hζ, as well as some notation. For We start with properties.

(i) h:R+→[1,∞) andhis non-decreasing.

(ii) his submultiplicative,i.e. there existsch>0 such that

h(x+y)≤chh(x)h(y), for allx, y∈R+. (iii) It holdsx≤h(x) for everyx∈R+.

(iv) The indefinite integral R

h(s)−γdscan be explicitly calculated. We define forγ > 1ζ Hγ(x) :=

Z

(0,x]

h−γ(s) ds= 1 1−γζ

(1 +x)1−γζ−1

= 1

−γζ+ 1

(1 +x)−γζ+1−1

= 1

−γζ+ 1

h(x)−γζ+1ζ

−1

= 1

−γζ+ 1

h(x)−γ+1ζ

−1

= 1

−γζ+ 1

hϑ(γ)(x)−1

, (A.1)

where ϑ(γ) :=−γ+1ζ <0.Moreover, Z

(t,u]

h−γ(s)ds=Hγ(u)−Hγ(t) =|Hγ(t)| − |Hγ(u)|

= 1

| −γζ+ 1|[hϑ(γ)(t)−hϑ(γ)(u)] = 1

γζ−1[hϑ(γ)(t)−hϑ(γ)(u)]. (A.2) (v) The set ΓH :=

γ∈R+, Hγ :R+→(−∞,0] is increasing is non–empty. In view of the restriction in (iv), the set ΓH is indeed non-empty,i.e.

ΓH = (1

ζ,∞). (A.3)

(vi) There existsβ ∈R+such that E

Z

(0,T]

hβ(As)kfsk22 α2s dCs

<∞.

By [64, Proposition 25.4] we can calculate the submultiplicative constant ch= exp

ln h(2·0)

−ln h(0)

= 1.

Lemma A.1. Assume the framework of Lemma II.16 except for the definition of the β−norms, where we substitute the exponential function with the functionh, forζ≥1;e.g see(vi). Moreover, assume that A is bounded by the positive constant Ψ. Then, it holds for 1ζ < γ < δ <βˆ

kαyk2H2 δ

+kηk2H2 δ

"

2h(Ψ)hδ(Φ) + 9 +9[h(Ψ)]1−1ζ[h(Φ)]δ−1ζ δ−1ζ + 1

# kξk2L2

δ

+

"

2[h(Ψ)]1+ζ1[h(Φ)]δ−γ+1ζ

δ−γ+ζ1+ 1 +9h(Ψ)[h(Φ)]δ−γ δ−γ+ 1 + 9

δζ−1

# f α H2δ

, and

kyk2S2 δ +kηk2

H2δ

8 + 2h(Ψ)hδ(Φ) kξk2

H2δ

A.1. AUXILIARY RESULTS OF CHAPTER II 135

+

"

[h(Ψ)]1ζ

γζ−1 +2[h(Ψ)]1+1ζ[h(Φ)]δ−γ+1ζ δ−γ+1ζ + 1

# f α

2

H2δ

. Proof. Recall the identity

yt=ξ+ Z

(t,T]

fsdCs− Z

(t,T]

s=E

"

ξ+ Z

(t,T]

fsdCs

Gt

#

, (A.4)

and introduce the anticipating function

Ft= Z

(t,T]

fsdCs. Forγ∈ΓH, we have by the Cauchy-Schwarz inequality,

kFtk22 ≤ Z

(t,T]

h−γ(As) dAs Z

(t,T]

hγ(As)kfsk22 α2s dCs

= Z

(At,AT]

h−γ(ALs) ds Z

(t,T]

hγ(As)kfsk22 α2s dCs Lemma I.34

(i)

Z

(At,AT]

h−γ(s) ds Z

(t,T]

hγ(As)kfsk22 α2s dCs

(A.2)

= 1

γζ−1[hϑ(γ)(At)−hϑ(γ)(AT)]

Z

(t,T]

hγ(As)kfsk22 α2s dCs

≤ 1

γζ−1hϑ(γ)(At) Z

(t,T]

hγ(As)kfsk22

α2s dCs. (A.5)

Forδ∈R+ and by integrating (A.5) w.r.t. hδ(At) dAtit follows Z

(0,T]

hδ(At)kFtk22dAt (A.5)

≤ 1

γζ−1 Z

(0,T]

hδ+ϑ(γ)(At) Z

(t,T]

hγ(As)kfsk22

α2s dCsdAt

= 1

γζ−1 Z

(0,T]

hγ(As)kfsk22 α2s

Z

(0,s)

hδ+ϑ(γ)(At) dAtdCs

≤ 1

γζ−1 Z

(0,T]

hγ(As)kfsk22 α2s

Z

(0,s]

hδ+ϑ(γ)(At) dAtdCs

≤ 1

γζ−1 Z

(0,T]

hγ(As)kfsk22 α2s

Z

(A0,As]

hδ+ϑ(γ)(ALt) dtdCs, (A.6) where in the equality we used Tonelli’s Theorem. By [64, Proposition 25.4] we have that hδ+ϑ(γ) re-mains submultiplicative if δ+ϑ(γ) > 0, which is equivalent to δ > |ϑ(γ)|. Therefore, we can apply Lemma I.34.(vii) for the functiong(x) =hδ+ϑ(γ)(x). Then, Inequality (A.6) becomes

Z

(0,T]

hδ(At)kFtk22dAtLemma I.34≤ cδ+ϑ(γ)h hδ+ϑ(γ)(Φ) Z

(0,T]

hγ(As)kfsk22 α2s

Z

(A0,As]

hδ+ϑ(γ)(t) dtdCs

ch=1

= hδ+ϑ(γ)(Φ) δ+ϑ(γ) + 1

Z

(0,T]

hγ(As)kfsk22

α2s hδ+ϑ(γ)+1(As)−hδ+ϑ(γ)+1(A0) dCs A0=0

h(A0)=1

hδ+ϑ(γ)(Φ) δ+ϑ(γ) + 1

Z

(0,T]

h(As)δ+1ζ+1kfsk22

α2s dCs, (A.7)

which is integrable ifδ+1ζ + 1≤β; see (vi). Define

%:=δ+1

ζ + 1 and Λγ,δ,Φ:= hδ+ϑ(γ)(Φ) δ+ϑ(γ) + 1. To sum up, for

γ∈ΓH, δ > γ−1

ζ and 0< %≤β, (A.8)

we have

E hZ

(0,T]

hδ(At)kFtk22dAti

≤Λγ,δ,Φ f α

2 H2%

. (A.9)

For the estimate of kαykH2

δ we first use the fact that kαyk2

H2δ =E hZ

(0,T]

hδ(At)kytk22dAti

≤2E hZ

(0,T]

E

hδ(At)kξk22+hδ(At)kFtk22 Gt

dAt

i

= 2E hZ

(0,∞)

E

hδ(At)kξk22+hδ(At)kFtk22 Gt

dATti

=2E hZ

(0,∞)

hδ(At)kξk22+hδ(At)kFtk22dATti

= 2E hZ

(0,T]

hδ(At)kξk22dAt

i + 2E

hZ

(0,T]

hδ(At)kFtk22dAt

i

= 2E hZ

(A0,AT]

hδ(ALs)kξk22dsi + 2E

hZ

(0,T]

hδ(At)kFtk22dAt

i

= 2cδhhδ(Φ)E hkξk22

Z

(A0,AT]

hδ(s) dsi + 2E

hZ

(0,T]

hδ(At)kFtk22dAt

i

=

ch=1

2hδ(Φ)E

hkξk22hδ(λ)(AT −A0)i + 2E

hR

(0,T]hδ(At)kFtk22dAt

i

, forλ∈[A0, AT] or

2

δζ+1hδ(Φ)E

hkξk22 hδ+1ζ(AT)−hδ+1ζ(A0)i + 2E

hR

(0,T]hδ(At)kFtk22dAt

i

(A.10)

(i),(iii)

A0=0,(A.9)





2hδ(Φ)kξk2

L2δ+1+ 2Λγ,δ,Φ,ch

f α

2 H2%

.

2

δζ+1hδ(Φ)kξk2

L2δ+ 1

ζ

+ 2Λγ,δ,Φ,ch

f α

2 H2%

.

(A.11)

In the second equality we have used that the processes kξk221Ω×[0,∞](·) and |F(·)|2 are uniformly in-tegrable, hence their optional projections are well defined. Indeed, using (A.5) and recalling that E[kF0k22]<∞, we can conclude the uniform integrability of |F(·)|2. Then, it holds that

ΠGo hδ(A·)kξk22+hδ(A·)|F(·)|2

t=hδ(AtGo kξk221Ω×[0,∞](·)

t+hδ(AtGo |F(·)|2

t

=hδ(At)E kξk22

Gt

+hδ(At)E kFtk22

Gt

=E

hδ(At)kξk22+hδ(At)kFtk22 Gt

, which justifies the use of Corollary I.23.

For the estimate of kykS2

δ we have kykS2

δ =E h

sup

0≤t≤T

hδ2(At)kytk22i

≤E h

sup

0≤t≤T

hδ2(At)E

kξk2+kFtk2

Gt2i

=E h

sup

0≤t≤TE

hδ2(At)kξk2+hδ2(At)kFtk2

Gt2i

≤2E

"

sup

0≤t≤TE q

hδ(At)kξk22+hδ(At)kFtk22

Gt

2#

(A.5)

≤ 2E

 sup

0≤t≤TE

" s

hδ(AT)kξk22+ 1

γζ−1hδ+ϑ(γ)(At) Z

(t,T]

hγ(As)kfsk22 α2s dCs

Gt

#2

≤ 2E

 sup

0≤t≤TE

" s

hδ(AT)kξk22+ 1 γζ−1

Z

(t,T]

[h(As)]γ+δ+ϑ(γ)kfsk22 α2s dCs

Gt

#2

≤ 2E

 sup

0≤t≤TE

" s

hδ(AT)kξk22+ 1 γζ−1

Z

(0,T]

[h(As)]δ+1ζkfsk22 α2s dCs

Gt

#2

A.1. AUXILIARY RESULTS OF CHAPTER II 137

≤ 8E

"

hδ(AT)kξk22+ 1 γζ−1

Z

(0,T]

[h(As)]δ+1ζkfsk22 α2s dCs

#

= 8kξk2H2 δ

+ 1

γζ−1 f α

2

H2δ+ 1

ζ

. (A.12)

where in the second and third inequalities we used the inequalitya+b≤p

2(a2+b2) and (A.5) respec-tively.

What remains is to controlkηkH2

δ. We remind once more the reader that Z T

t

s=ξ−yt+Ft, hence

E

kξ−yt−Ftk22 Gt

=E

"

Z

(t,T]

dTr[hηis]

Gt

#

. (A.13)

In addition, we have

hδ(As) = (1 +As)δζ =δζ Z

(A0,As]

(1 +x)δζ−1dx+ 1 =δζ Z

(A0,As]

hδ−1ζ(x)dx+ 1 (A.14)

Z

(0,T]

hδ(As) dTr[hηis]

(A.14)

≤ δζ Z

(0,T]

Z

(A0,As]

hδ−1ζ(t) dtdTr[hηis] + Tr[hηiT]

Lemma I.34

(i)

δζ Z

(0,T]

Z

(A0,As]

hδ−1ζ(ALt) dt dTr[hηis] + Tr[hηiT]

= δζ

Z

(0,T]

Z

(0,s]

hδ−1ζ(At) dAtdTr[hηis] + Tr[hηiT]

≤ δζ Z

(0,T]

hδ−1ζ(At) Z

(t,T]

dTr[hηis]dAt+ Tr[hηiT], so that

kηkH2 δ ≤δζE

"

Z

(0,T]

hδ−1ζ(At) Z

(t,T]

dTr[hηis] dAt

#

+E[Tr[hηiT]]. (A.15) We now need an estimate for E

hR

(0,T]dTr[hηis]i

, which is given by E[Tr[hηiT]] = E

|ξ−y0+F(0)|2

≤3E

kξk22+ky0k22+kF0k22

(A.4)

≤ 9E kξk22

+ 9E

kF0k22(A.5)

(i)

9kξkL2

δ+ 9 1 δζ−1

f α

H2δ

forδ∈ΓH (A.5)

≤ 9kξkL2

δ+ 9

δζ−1 f α

H2δ

, (A.16)

where we used the fact thatE ky0k22

≤2E

kξk22+kF0k22 .

For the first summand on the right-hand-side of (A.15), we have E

"

Z

(0,T]

hδ−1ζ(At) Z

(t,T]

dTr[hηis] dAt

#

Lemma I.23

= E

"

Z

(0,T]

hδ−ζ1(At)E

"

Z

(t,T]

dTr[hηis]

Gt

# dAt

#

(A.13)

= E

"

Z

(0,T]

hδ−1ζ(At)E

|ξ−yt+Ft|2 Gt

dAt

#

≤ 3E

"

Z

(0,T]

hδ−1ζ(At)E

kξk22+kytk22+kFtk22 Gt

dAt

#

(A.4)

≤ 3E

"

Z

(0,T]

hδ−1ζ(At)kξk22dAt

# + 3E

"

Z

(0,T]

hδ−1ζ(At)kFtk22dAt

#

+ 6E

"

Z

(0,T]

hδ−ζ1(At)E

kξk22+kFtk22 Gt

dAt

#

Lem. I.23

= 9E

"

Z

(0,T]

hδ−1ζ(At)kξk22dAt

# + 9E

"

Z

(0,T]

hδ−1ζ(At)kFtk22dAt

#

(γ∈ΓH)

= 9E

"

Z

(A0,AT]

hδ−1ζ(ALt)kξk22dt

# + 9E

"

Z

(0,T]

hδ−1ζ(At)kFtk22dAt

#

Cor. I.36

(A.9)

9hδ−1ζ(Φ)E

"

kξk22 Z

(A0,AT]

hδ−1ζ(t) dt

#

+ 9Λγ,δ−1ζ f α H2δ+1

forδ+ 1≤β

≤ 9hδ−1ζ(Φ) δ−1ζ + 1E

hkξk22

hδ−1ζ+1(AT)−1 dti

+ 9Λγ,δ−1ζ f α

H2δ+1

≤ 9hδ−1ζ(Φ) δ−1ζ + 1 kξkδ−1

ζ+1+ 9Λγ,δ−1ζ f α

H2δ+1

. (A.17)

In total, Inequality (A.15) can be written kηkH2

δ ≤ δζE

"

Z T 0

hδ−1ζ(At) Z

(t,T]

dTr[hηis] dAt

#

+E[Tr[hηiT]]

(A.16)

(A.17)

9hδ−1ζ(Φ) δ−1ζ + 1 kξkδ−1

ζ+1+ 9Λγ,δ−ζ1 f α

H2δ+1

+ 9kξkL2

δ+ 9

δζ−1 f α

H2δ

(A.18) In order to be able to construct a contraction, we need to have on the right hand side of (A.11), resp.

(A.12), and (A.18) only kfαkH2

δ. One way to overcome this problem is to assume that A is P−a.s.

bounded, say by Ψ.Then, the Inequality (A.11) can be written as

kαyk2

H2δ





2hδ(Φ)kξk2

L2δ+1+ 2Λγ,δ,Φ,ch

f α

2 H2%

2

δζ+1hδ(Φ)kξk2

L2δ+ 1

ζ

+ 2Λγ,δ,Φ,ch

f α

2 H2%





2h(Ψ)hδ(Φ)kξk2

L2δ+2[h(Ψ)]

1+ 1ζ[h(Φ)]δ+ϑ(γ) δ+ϑ(γ)+1

f α

H2δ

,

2

δζ+1hζ1(Ψ)hδ(Φ)kξk2

L2δ

+2[h(Ψ)]

1+ 1ζ[h(Φ)]δ+ϑ(γ) δ+ϑ(γ)+1

f α

H2δ

.

(A.19)

the Inequality (A.12) can be written as kykS2

δ ≤8kξk2

H2δ+ 1 γζ−1

f α

2

H2δ+ 1

ζ

≤8kξk2

H2δ+[h(Ψ)]1ζ γζ−1

f α

2

H2δ

(A.20) and the Inequality (A.18) can be written as

kηkH2

δ ≤ 9 + 9[h(Ψ)]1−ζ1[h(Φ)]δ−1ζ δ−1ζ + 1

! kξkδ+

9h(Ψ)Λγ,δ−1ζ+ 9 δζ−1

f α

2 H2δ

= 9 + 9[h(Ψ)]1−ζ1[h(Φ)]δ−1ζ δ−1ζ + 1

! kξkδ+

9h(Ψ)[h(Φ)]δ−γ δ−γ+ 1 + 9

δζ−1

f α

2 H2δ

. (A.21)

A.2. Auxiliary results of Chapter III

Lemma A.2. Let(Lk)k∈Nbe a sequence ofRp−valued processes and(Nk)k∈Nbe a sequence ofRq−valued processes such that

(i) Tr

[Lk]

k∈Nis uniformly integrable and (ii) Tr

[Nk]

k∈N is bounded inL1(G;R),i.e. sup

k∈N

E Tr

[Nk]

<∞.

A.2. AUXILIARY RESULTS OF CHAPTER III 139

Then

Var [Lk, Nk]

1

k∈N is uniformly integrable.

Proof. By Corollary I.29.(iv), it is sufficient to prove that the sequence Var [Lk,i, Nk,j]

k∈N

is uniformly integrable, for every i = 1, . . . , p and j = 1, . . . , q. We will use Theorem I.25 in order to prove the required property. Let i, j be arbitrary but fixed. The L1–boundedness of the sequence (Var([Lk,i, Nk,j]))k∈N is obtained by means of the Kunita–Watanabe inequality in the form I.26 and the L1−boundedness of the sequences [Lk,i]

k∈N and [Nk,j]

k∈N; the former is L1−bounded as uniformly integrable. By the Kunita–Watanabe inequality again, but now in the form (I.24), and the Cauchy–Schwarz inequality, we obtain for anyA∈ G

Z

A

Var [Lk,i, Nk,j]

dP

K-W ineq.

≤ Z

A

[Lk,i]

1

2[Nk,j]

1

2 dP

C-S ineq.

≤ Z

A

[Lk,i] dP 12Z

A

[Nk,j]dP 12

≤ Z

A

[Lk,i] dP 12

sup

k∈N

Z

[Nk,j] dP 12

≤ CZ

A

[Lk,i] dP 12

, (A.22)

where C2 := sup

k∈N E hTr

[Nk]i

. Note that for the third inequality we used that [Nk,j] ≥ 0 for all k∈N.

Now we use [35, Theorem 1.9] for the uniformly integrable sequence [Lk,i]

k∈N. For everyε >0, there existsδ >0 such that, wheneverP(A)< δ, it holds

sup

k∈N

Z

A

[Lk,i] dP< ε2

C2 ⇐⇒sup

k∈N

Z

A

[Lk,i] dP 1/2

< ε C. This further implies sup

k∈N

E

1AVar [Lk,i, Nk,j](A.22)

< ε, which is the required condition.

The proof for the predictable quadratic variation is completely analogous. We state the result, however, for our convenience.

Lemma A.3. Let(Lk)k∈Nbe a sequence ofRp−valued processes and(Nk)k∈Nbe a sequence ofRq−valued processes such that

(i) hLkiandhNkiare well-defined for every k∈N, (ii) Tr

hLki

k∈Nis uniformly integrable and (iii) Tr

hNki

k∈N is bounded inL1(G;R),i.e. sup

k∈N

E Tr

[Nk]

<∞.

Then

Var hLk, Nki

1

k∈N is uniformly integrable.

In the following we provide some results which can be seen as simple exercises in measure theory.

Lemma A.4. LetD be a countable and dense subset ofR.Then every openU ⊂Ris the countable union of intervals with endpoints in D, i.e. there exists a sequence of intervals (ak, bk)

k∈N with ak, bk ∈D, for every k∈N,such that U =∪k∈N(ak, bk).

Proof. LetUbe an open subset ofR.For everyx∈Uthere exists anrx>0 such thatB(x, rx)⊂U, whereB(y, r) ={z∈R,|y−z|< r}.Due to the density ofDwe can chooseax, bx∈D such thatax< bx

and (ax, bx) ⊂ B(x, rx) ⊂U. Then clearly ∪x∈U(ax, bx) ⊂ U. On the other hand, D is countable and thereforeD×Dis also countable. Moreover, we can naturally associate to each ordered pair{{a},{a, b}}

the interval (a, b) ifa < bor the empty set ifa≥b, which proves that there are at most countably many

intervals with endpoints inD.

Corollary A.5. Let U ⊂R\ {0} be open. Then for everyi= 1, . . . , `there exists a countable subfamily of I(X∞,i), which has been introduced in Section III.4, name(Ui,k)k∈N,such that U =∪k∈NUi,k.

Proof. LetU be an open subset ofR\ {0} andi∈ {1, . . . , `}. Then, there exist open setsU+, U such thatU+⊂(0,∞),U⊂(−∞,0) andU =U+∪U.On the other hand, by [41, Lemma VI.3.12] we can conclude that the setI(X∞,i) is at most countable. Therefore, the complement ofI(X∞,i), denote it byI(X∞,i)c, is dense inR.Indeed, if I(X∞,i)c was not dense, there would exist an open subsetV of Rsuch thatI(X∞,i)c∩V =∅ or equivalentlyV ⊂ I(X∞,i). However this is a contradiction, sinceV is uncountable and I(X∞,i) is countable.

The above allow us to apply Lemma A.4 for D =I(X∞,i) in order to find two countable families (U+,i,k)k∈N and (U−,i,k)k∈Nsuch that

U+= [

k∈N

U+,i,kandU= [

k∈N

U−,i,k.

The required countable family is (U+,i,k)k∈N∪(U−,i,k)k∈N. Lemma A.6. It holdsσ J(X)

=B(R`), whereJ(X)has been introduced in Subsection III.4 and σ J(X)

denotes the σ−algebra inR` generated by the familyJ(X).

Proof. The spaceR`is a finite product of the second countable metric spaceR. Therefore, it holds B(R`) =B(R)⊗ · · · ⊗ B(R)

| {z }

`−times

and to this end, it is sufficient to prove thatσ(I(X∞,i)) =B(R) for everyi= 1, . . . , `.

By Corollary A.5 we have

σ(I(X∞,i)) =σ({U ⊂R\ {0}, U is open}) =σ({U ⊂R\ {0}, U is open} ∪ {R} ∪ {0})

=σ({U ⊂R, U is open}) =B(R),

which allows us to conclude.

Lemma A.7. Let (Σ,S)be a measurable space and %1, %2 be two finite signed measures on this space.

Let Abe a family of sets with the following properties:

(i) Ais aπ−system,i.e. if A, B∈ A, thenA∩B∈ A.

(ii) It holdsσ(A) =S

(iii) %1(A) =%2(A) for everyA∈ A.

Then it holds%1=%2.

Proof. Let %+i , %i be the (positive) measures obtained by the Jordan decomposition of %i, for i= 1,2.By (iii) we obtain

%+1(A)−%1(A) =%+2(A)−%2(A), for everyA∈ A.

However, the above equality can be translated into equality of measures, i.e.

%+1(A) +%2(A) =%+2(A) +%1(A), for everyA∈ A.

The class

C={A∈ S, %+1(A) +%2(A) =%+2(A) +%1(A)}

can be proven to be aλ−system.1 Now we can conclude that the (positive) measures%+1+%2 and%+2+%1 coincide on D(A)⊂ C, where we have denoted by D(A) the λ−system generated by A.Using now the property (i) and theπ−λLemma,2see [4, Lemma 4.11], we can conclude that the (positive) measures

%+1 +%2 and%+2 +%1 coincide onσ(A) =S.

Let, now, P1, N1, resp. P2, N2, theHahn decomposition of the signed measure %1, resp. %2, where with Pi we denote the set of positive mass of %i and with Ni we denote the set of negative mass of %i, fori= 1,2. In view of the %+i ⊥%i , fori= 1,2 and of the following equalities

Ω = (P1∩P2)∪(P1∩N2)∪(N1∩P2)∪(N1∩N2)

%+1(P1∩P2) =%+1(P1∩P2) +%2(P1∩P2) =%+2(P1∩P2) +%1(P1∩P2) =%+2(P1∩P2)

%+1(P1∩N2) +%2(P1∩N2) =%+2(P1∩N2) +%1(P1∩N2) = 0 0 =%+1(N1∩P2) +%2(N1∩P2) =%+2(N1∩P2) +%1(N1∩P2)

%2(N1∩N2) =%+1(N1∩N2) +%2(N1∩N2) =%+2(N1∩N2) +%1(N1∩N2) =%1(N1∩N2) we can conclude thatP1∩P2,N1∩N2is a common Hahn decomposition of the signed measures%1 and

%2. Therefore

1Aλ−system is also referred to as Dynkin system or d−system or Sierpi´nski class. In [4, Foonote 4, p. 135] the interested reader can find references for this ambiguity in terminology. In [10] is used the termσ−additive class.

2Also known asDynkin’s Lemma.

A.3. AUXILIARY RESULTS OF CHAPTER IV 141

For every A ∈ {B ⊂(P1∩P2), B ∈ S} we obtain %+1(A) = %+2(A), hence we can conclude that

%+1 =%+2.

For every A∈ {B ⊂(N1∩N2), B ∈ S} we obtain %1(A) =%2(A), hence we can conclude that

%1 =%2.

A.3. Auxiliary results of Chapter IV

A.3.1. Moore–Osgood Theorem. The Moore–Osgood Theorem, whose well-known form is The-orem A.8, provides sufficient conditions for the existence of the limit of a doubly-indexed sequence and it can be seen as a special case of Rudin [63, Theorem 7.11]. Here we provide a second form, since the second time3we need to apply the aforementioned theorem we need to relax the existence of the pointwise limits; see Condition (ii) of Theorem A.8. For more comments on the existence of the iterated limits and of the joint limit of a doubly-indexed sequence the interested reader should consult Hobson [37, Chapter VI, Sections 336-338]. Specifically for the validity of the Theorem A.9 see [37, Chapter VI, Section 337, p. 466] and the reference therein.

Theorem A.8. Let (Γ, dΓ) be a metric space and(γk,p)k,p∈

Nbe a sequence. If (i) lim

p→∞sup

k∈N

dΓk,p, γk,∞) = 0 and (ii) lim

k→∞dΓk,p, γ∞,p) = 0 for allp∈N, then the joint limit lim

k,p→∞γn,k exists. In particular holds lim

k,p→∞γk,p= lim

p→∞γ∞,p= lim

k→∞γk,∞. Theorem A.9. Let (R,| · |)and(ϑk,p)k,p∈

N be a sequence. If (i) lim

p→∞sup

k∈N

k,p−ϑk,∞|= 0and (ii) lim

p→∞ lim sup

k→∞

ϑk,p−lim inf

k→∞ ϑk,p

= 0, then the joint limit lim

k,p→∞ϑk,p exists.

A.3.2. Weak convergence of measures on the positive real line.

Definition A.10. Let (µk)k∈

Nbe a countable family of measures on R+,B(R+)

. We will say that the sequence (µk)k∈Nconverges weakly to the measureµif for everyf : R+,k · k

→ R,| · |

continuous and bounded holds

k→∞lim Z

[0,∞)

f(x)µk(dx)− Z

[0,∞)

f(x)µ(dx)

= 0.

We denote the weak convergence of (µk)k∈Ntoµ byµk

−−w→µ. In this section we will use the set

W0,∞+ :=

C∈D(R)

C is increasing, withC0= 0 and lim

t→∞Ct∈R .

Remark A.11. We can extend every element of W0,∞+ , name C its arbitrary element, on [0,∞] such that it is left-continuous at the symbol ∞by definingC:= limt→∞Ct.

We provide in the following proposition some convenient equivalence for the weak convergence of finite measures. The statement is tailor-made to our later needs, but the interested reader may consult Bogachev [10, Section 8.1 - Section 8.3]. Then we provide in Theorem A.14 a new, to the best knowledge of the author, characterisation of weak convergence of finite measures to an atomless measure defined on the positive real line, which uses relatively compact sets of the Skorokhod space instead of relatively compact sets of the space of continuous functions defined onR+ endowed with the k · k−topology.

Proposition A.12. Let (µk)k∈

N be sequence of finite measures on R+,B(R+)

, and denote the as-sociated distribution functions by Ck, for k ∈ N, for which the finiteness of the measure translates to Ck := limt→∞Ctk∈R+ for every k∈N. We assume the following

(i) The sequence(µk)k∈Nis bounded, i.e. supk∈Nµk(R+)<∞. Equivalently,supk∈NCk <∞.

3The first one is in the proof of Theorem IV.10, while the second is in the proof of Proposition A.16

(ii) The sequence (µk)k∈N is tight. In other words, for every ε >0 there exists a compact setI such that supk∈Nµk(R+\I)< ε.4 Equivalently,supk∈N|Ck −CNk|< εfor someN ∈R+.

(iii) µk({0}) = 0 for everyk∈N. Equivalently,Ck ∈ W0,∞+ for everyk∈N.

(iv) µ is atomless,i.e. µ({t}) = 0for every t∈R+. Equivalently, C is continuous.

The following are equivalent:

(I) µk−−→w µ.

(II) Ctk−−−−→

k→∞ Ct, for everyt∈R+. (III) Ck−−−−→J1(R) C.

(IV) Ck−−→lu C. (V) sup

I∈IN

k(I)−µ(I)| −−−−→

k→∞ 0 for everyN ∈N, whereIN :=

I⊂[0, N] | I is interval . Proof. The equivalence between (I) and (II) is a classical result,e.g. see Bogachev [10, Proposition 8.1.8]. The equivalences between (II),(III) and (IV) are provided by Jacod and Shiryaev [41, Theorem VI.2.15.c.(i), Proposition VI.1.17 b)]. We obtain the equivalence between (IV) and (V) in view of the validity of the following inequalities for every N∈N:

sup

t∈[0,N]

|Ctk−Ct| ≤ sup

I∈IN

k(I)−µ(I)| (becauseC0k=C0= 0)

≤ sup

s,t∈[0,N]

|Ct−k −Csk−Ct+Cs|+ sup

s,t∈[0,N]

|Ctk−Csk−Ct+Cs| + sup

s,t∈[0,N]

|Ct−k −Cs−k −Ct−+Cs−|

≤3 sup

t∈[0,N]

|Ctk−Ct|+ 2 sup

t∈[0,N]

|Ct−k −Ct−|+ sup

t∈[0,N]

|Ct−k −Ct|

≤5 sup

t∈[0,N]

|Ctk−Ct|+ sup

t∈[0,N]

|Ct−k −Ctk|+ sup

t∈[0,N]

|Ctk−Ct| (Ckis c`adl`ag for everyk∈N)

≤6 sup

t∈[0,N]

|Ctk−Ct|+ sup

t∈[0,N]

|Ct−k −Ctk|.

Then Jacod and Shiryaev [41, Lemma VI.2.5] implies that lim sup

k→∞

sup

I∈IN

k(I)−µ(I)| ≤6 lim

k→∞ sup

t∈[0,N]

|Ctk−Ct|+ lim sup

k→∞

sup

t∈[0,N]

|Ct−k −Ctk|

≤ sup

t∈[0,N]

|Ct−−Ct|= 0.

The following lemma provides a useful criterion to conclude the equivalence betweenδk·k-convergence andδJ1(R+)-convergence inD [0,∞]

. Lemma A.13. Let (Ck)k∈

N⊂ W0,∞+ which satisfies the following properties:

(i) Ck−−−−→J1(R)

k→∞ C. (ii) Ck −−−−|·|

k→∞ C.

(iii) The limit functionC is continuous.

Then it holdsCk −−−−−→k·k

k→∞ C.

Proof. Let us fix an arbitraryε >0.We start by exploiting the fact thatC∈ W0,∞+ . Then, there exists K >0 such that

sup

t≥K

(C−Ct)< ε

4. (A.23)

By (ii), there existsk1=k1(ε)∈Nsuch that sup

k≥k1

|Ck −C|< ε

4. (A.24)

4Without loss of generality we can assume thatIis of the form [0, N], for someN >0.

A.3. AUXILIARY RESULTS OF CHAPTER IV 143

By (iii) there exists ¯s > K such thatC¯s= 12(CK+C). By (i), (iii) and Proposition I.113, there exists k2=k2(ε,¯s) such that

sup

k≥k2

sup

0≤t≤¯s

|Ctk−Ct|<1

4(C−CK)< ε

16, (A.25)

where the last inequality is valid in view of (A.23), since ¯s > K.Using the fact that the functions are increasing we derive

sup

k≥k1∨k2

sup

¯s≤t<∞

Ck −Ctk

≤ sup

k≥k1∨k2

Ck −C¯sk

≤ sup

k≥k1∨k2

Ck −C

+ sup

k≥k1∨k2

C−C¯s

+ sup

k≥k1∨k2

C¯s−C¯sk

(A.24)

<

(A.25)

ε 4 +ε

4+ ε 16 < ε

2. (A.26)

Therefore, by combining the above, we obtain for everyk≥k1∨k2

sup

t∈R+

|Ctk−Ct| ≤ sup

0≤t≤¯s

|Ctk−Ct|+ sup

¯s≤t≤∞

|Ctk−Ct|(A.25)<

(A.26)

< ε.

Recall that the Kolmogorov metric δKolm(C1, C2) := kC1−C2k, for C1, C2 ∈ W0,∞+ , dominates the L´evy metric, which metrises the weak topology on the space of finite measures. Hence the previous lemma can be seen as a criterion for the weak convergence µCk

−−w→µC, where µCk, for k∈ Nis the finite measures associated to the increasing functionCk∈ W0,∞+ .

The next step is to provide a characterisation of the aforementioned weak convergence of measures using relatively compact sets of D. Since, however, a relatively compact set of D is, in general, only locally uniformly bounded, see Theorem I.111, we have to restrict ourselves to those relatively compact subsets which are uniformly bounded, so that the integrals make sense. The following theorem can be regarded as an extension of Parthasarathy [56, Theorem 6.8] in the special case that the limiting measure is an atomless measure onR+. Forα∈D(Rp) the modulusw0N has been defined in Definition I.108.

Theorem A.14. Let the measurable space R+,B(R+)

, (µk)k∈

N be a sequence of finite measures such that µ is an atomless finite measure andA ⊂D(R). Assume that the following conditions are true

(i) µk−−→w µ.

(ii) Ais uniformly bounded, i.e. supα∈Akαk<∞and (iii) Ais relatively compact. In other words, lim

ζ↓0sup

α∈A

wN0 (α, ζ) = 0 for everyN ∈N. Then,

k→∞lim sup

α∈A

Z

[0,∞)

α(x)µk(dx)− Z

[0,∞)

α(x)µ(dx)

= 0. (A.27)

Proof. We will decompose initially the quantity whose convergence we intent to prove as follows sup

α∈A

Z

[0,∞)

α(x)µk(dx)− Z

[0,∞)

α(x)µ(dx)

≤sup

α∈A

Z

[0,N]

α(x)µk(dx)− Z

[0,N]

α(x)µ(dx) + sup

α∈A

Z

[N,∞)

α(x)µk(dx)

+ sup

α∈A

Z

[N,∞)

α(x)µ(dx)

(A.29)

for someN >0 to be determined and then the first summand of the right-hand side will be decomposed as follows

sup

α∈A

Z

[0,N]

α(x)µk(dx)− Z

[0,N]

α(x)µ(dx)

≤sup

α∈A

Z

[0,N]

α(x)µk(dx)− Z

[0,N]

α(x)µeαk(dx) + sup

α∈A

Z

[0,N]

α(x)eµαk(dx)− Z

[0,N]

α(x)eµα(dx) + sup

α∈A

Z

[0,N]

α(x)µeα(dx)− Z

[0,N]

α(x)µ(dx)

(A.31)

where the measureseµαk, fork∈Nandα∈ A, will be constructed given the measureµk and the function α. For the second and third summand of (A.29) we will prove that they become arbitrarily small for largeN >0. Then, we will conclude once we obtain the convergence to 0 ask→ ∞of the summands of (A.31). To this end, let us fix anε >0.

Condition (i) implies that supk∈Nµk(R+) < ∞as well as the tightness of the sequence by the convergenceµk(R+)−−−−→

k→∞ µ(R+). Hence, forM := supα∈Akαk<∞, which is Condition (ii), there exists N=N(ε, M)>0 such that

sup

k∈N

µk [N,∞)

< ε

2M, which implies sup

k∈N,α∈A

Z

[N,∞)

α(x)µk(dx)

< ε 2.

In other words, we have proven that the second and third summand of the right-hand side of (A.29) become arbitrarily small for large N.In the following N is assumed fixed, but large enough so that the above hold.

We proceed now to construct for everyk∈Nandα∈ Aa suitable measureeµαk on R+,B(R+) such that we can obtain the convergence of the summands in (A.31). DefineKN := supk∈

Nµk([0, N])<∞.

By (iii) there exists ζ=ζ(ε, KN)>0 such that for everyζ0 < ζ it holds supα∈Aw0N(α, ζ0)< 4Kε

N. By the definition of w0 (see Definition I.108) for everyα∈ Aand every ζ0 < ζ, there exists aζ0-sparse set PαN,ζ0, i.e.

PαN,ζ0:=n

0 =t(0;PαN,ζ0)< t(1;PαN,ζ0)< . . . < t(κα;PαN,ζ0) =N : min

i=1,...,κα−1 t(i;PαN,ζ0)−t(i−1;PαN,ζ0)

> ζ0o , such that

i=1,...,κmaxαsupn

|α(s)−α(u)|:s, u∈

t(i−1;PαN,ζ0), t(i;PαN,ζ0)o

< w0N(α, ζ0) + ε 4KN

< sup

α∈A

wN0 (α, ζ0) + ε 4KN

< ε 2KN

. (A.32) To sum up, for every (α, ζ0)∈ A ×(0, ζ) we can find a partition of [0, N], PαN,ζ0, satisfying (A.32). For the following, given a finite measure on R+,B(R+)

, nameλ, and a sparse setPαN,ζ0

. we will denoteA(i;PαN,ζ0) := [t(i−1;PαN,ζ0), t(i;PαN,ζ0)) for i= 1, . . . , κα and

. we define the finite measureeλα: R+,B(R+)

→[0,∞) by eλα(·) :=

κα

X

i=1

λ A(i;PαN,ζ0)

δt(i−1;PαN,ζ0)(·),

where δx is the Dirac measure sitting at the pointx. With the above notation we have

Z

[0,N]

α(x)λ(dx)− Z

[0,N]

α(x)λeα(dx)

=

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)λ(dx)−

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)λeα(dx)

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)λ(dx)− Z

A(i;PαN,ζ0)

α(x)eλα(dx)

=

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)λ(dx)−α t(i−1;PαN,ζ0 λ

[t(i−1;PαN,ζ0), t(i;PαN,ζ0))

=

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)−α t(i−1;PαN,ζ0 λ(dx)

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)−α t(i−1;PαN,ζ0 λ(dx)

A.3. AUXILIARY RESULTS OF CHAPTER IV 145

(A.32)

≤ ε

2KN κα

X

i=1

Z

A(i;PαN,ζ0)

λ(dx) = ε 2KN

λ([0, N]). (A.33)

For every k ∈ Nand α ∈ A we define µeαk := (µgk)α. Now, using the approximation (A.33) for µk, for k∈N, we obtain

sup

k∈N,α∈A

Z

[0,N]

α(x)µk(dx)− Z

[0,N]

α(x)eµαk(dx)

< ε

2. (A.34)

To sum up, we have constructed the measuresµeαk such that the first and third summand of (A.31) become arbitrarily small uniformly onk∈Nand onα∈ A.

We can conclude (A.27) once we obtain the validity of

k→∞lim sup

α∈A

Z

[0,N]

α(x)eµαk(dx)− Z

[0,N]

α(x)eµα(dx)

= 0.

Indeed, for fixedζ0∈(0, ζ),

Z

[0,N]

α(x)µeαk(dx)− Z

[0,N]

α(x)µeα(dx)

=

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)µeαk(dx)−

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)eµα(dx)

=

κα

X

i=1

Z

A(i;PαN,ζ0)

α(x)eµαk(dx)− Z

A(i;PαN,ζ0)

α(x)eµα(dx)

=

κα

X

i=1

α t(i;PαN,ζ0)h

µk(A(i;PαN,ζ0))−µ A(i;PαN,ζ0)i

(ii)

≤ kαk

κα

X

i=1

µk A(i;PαN,ζ0)

−µ A(i;PαN,ζ0)

≤ sup

α∈A

kαk·κα· sup

i=1,...,κα

µk A(i;PαN,ζ0)

−µ A(i;PαN,ζ0)

≤ sup

α∈A

kαk·κα· sup

I∈IN

µk(I)−µ(I)

. (A.35)

Let us now denotem(PαN,ζ0) := min

t(i;PαN,ζ0)−t(i−1;PαN,ζ0), i= 1, . . . , κα > ζ0 by the definition of PαN,ζ0. Therefore,

sup

α∈A

κα≤sup

α∈A

l N m(PαN,ζ0)

m5≤lN ζ0 m

<∞.

Hence, for every fixed sparse set PαN,ζ0 which satisfies (A.32) we obtain sup

α∈A

Z

[0,N]

α(x)µeαk(dx)− Z

[0,N]

α(x)µeα(dx)

≤sup

α∈A

kαk·sup

α∈A

κα· sup

I∈IN

µk(I)−µ(I)

−−−−(i)

k→∞ 0, where we have used Proposition A.12.(V) which is equivalent to (i).

Remark A.15. We have presented the previous theorem for A ⊂ D(R). However, the result can be readily adapted forA ⊂D(Rp).

Proposition A.16. Let the measurable space R+,B(R+)

and(µk)k∈

Nbe a sequence of finite measures such that µk({0}) = 0 for every k ∈ N and µ is atomless. Additionally, let A := (αk)k∈

N ⊂D(Rp).

Assume the following to be true:

(i) µk

−−→w µ.

(ii) Ais uniformly bounded, i.e. supα∈Akαk<∞.

(iii) AisδJ1(Rp)−convergent. In other words,αk J1(R

p)

−−−−−→

k→∞ α. Then,

k→∞lim Z

[0,∞)

αk(x)µk(dx)− Z

[0,∞)

α(x)µ(dx) 2

= 0.

5We denote bydxethe least integer greater than or equal tox.

Proof. Let us denote γk,m:=

Z

[0,∞)

αk(x)µm(dx)− Z

[0,∞)

αk(x)µ(dx) 2

fork∈Nandm∈N.

We are going to apply Theorem A.9 in order to obtain the required result. By Theorem A.14 we have that

m→∞lim sup

k∈N

γk,m = 0.

In other words, Condition (i) of Theorem A.9 is satisfied. Let us prove, now, that Condition (ii) of the aforementioned theorem is also satisfied, i.e. we need to prove that

m→∞lim lim sup

k→∞

γk,m−lim inf

k→∞ γk,m

= 0.

However, it is sufficient to prove that limm→∞lim supk→∞γk,m = 0, since the elements of the doubly-indexed sequence are positive. To this end, we have

lim sup

k→∞

γk,m

≤lim sup

k→∞

Z

[0,∞)

αk(x)µm(dx)− Z

[0,∞)

α(x)µm(dx) 2 + lim sup

k→∞

Z

[0,∞)

α(x)µm(dx)− Z

[0,∞)

α(x)µ(dx) 2 + lim sup

k→∞

Z

[0,∞)

α(x)µ(dx)− Z

[0,∞)

αk(x)µ(dx) 2

. We are going to prove now that the two last summands of the right-hand side of the above inequality are equal to zero. We start with the second and we realise that we have only to use thatµk −−w→µ and Theorem A.14 for the special case of a singleton, in order to verify our claim. The third summand is also equal to zero as an outcome of the bounded convergence theorem. Indeed, we have that the the sequence A is uniformly bounded and by Proposition I.121 we have the pointwise convergenceαk(x)−−→α(x) for k → ∞, at every point xwhich is point of continuity of α. Since the set{x∈R+,∆α(x) 6= 0}

is at most countable, we can conclude that it is a µ−null set. Therefore, since every element of A is Borel measurable, we can conclude.

Finally, we deal with the first summand. Let us provide initially some helpful results. Using again Proposition I.121 we have that

lim sup

k→∞

αk(x)−α(x) 2

∆α(x)

2 for everyx∈R+, (A.36) because the only possible accumulation points of the sequence αk,i(x)

k∈N areα∞,i(x) and α∞,i(x−), for everyi= 1, . . . , pand everyx∈R+; recall Remark I.119. Now observe that the functionR+3x7−→

∆α(x)

2 ∈ R+ is Borel measurable and µ−almost everywhere equal to the zero function. This observation allows us to apply Mazzone [51, Theorem 1] in order to conclude that

Z

[0,∞)

∆α(x)

2µm(dx)−−−−→

m→∞

Z

[0,∞)

∆α(x)

2µ(dx) = 0. (A.37) In view of the above and the boundedness of the sequenceA, we apply Fatou’s lemma and we obtain for every m∈N

lim sup

k→∞

Z

[0,∞)

αk(x)µm(dx)− Z

[0,∞)

α(x)µm(dx) 2

≤ Z

[0,∞)

lim sup

k→∞

αk(x)−α(x) 2

µm(dx)

(A.36)

≤ Z

[0,∞)

∆α(x)

2µm(dx).

Now, we can conclude that the Condition (ii) of Theorem A.9 is indeed satisfied by combining the above

bound with Convergence (A.37).

Remark A.17. Let us adopt the notation of Theorem A.14 and Proposition A.16 and assume that the measurable space is [0, T],B([0, T])

, for someT ∈R+. We claim that we can adapt the aforementioned results without loss of generality, which can be justified as follows. The limit measureµis atomless, so the generality is not harmed if in Theorem A.14 the integrandsαk,for somek∈N, have a jump at pointT.

A.3. AUXILIARY RESULTS OF CHAPTER IV 147

Indeed, by the weak convergenceµk ⇒µand Proposition A.12 we have that the sequence of distribution functions (Ck)k∈

N which is associated to the sequence (µk)k∈

N converges uniformly. Therefore, we can easily conclude that

lim sup

k→∞

∆αk(T)

2µk({T}) = lim sup

k→∞

∆αk(T)

2∆CTk

∆α(T)

2∆CT= 0.

In other words, we can simply assume that the distribution functions are constant after timeT in order to reduce the general case to the compact-interval case.

The following corollary is almost evident due to the fact that the limit measure is atomless. However we state it in the form we will need it in Subsection IV.5.2 and we provide its complete (rather trivial) proof.

Corollary A.18. Let the measurable space R+,B(R+)

,(µk)k∈Nbe a sequence of finite measures, where µk({0}) = 0for everyk∈Nandµis atomless. Let, moreover,A:= (αk)k∈N⊂D(Rp)and the following to be true:

(i) µk

−−→w µ.

(ii) Ais uniformly bounded, i.e. supα∈Akαk<∞and (iii) AisδJ1(Rp)−convergent.

Then,

Z

[0,·]

αk(x)µk(dx) J1(R

p)

−−−−−→

k→∞

Z

[0,·]

α(x)µ(dx). (A.38)

Proof. LetCk denote the distribution function associated to the measureµk for everyk∈Nand γk(t) :=

Z

[0,t]

αk(x)µk(dx) = Z

[0,t]

αk(s) dCsk ∈D(Rp) fork∈N.

We are going to apply Proposition I.118 in order to prove the required convergence. To this end, let us fix a t ∈ R+ and a sequence (tk)k∈N such that tk −−→ t as k → ∞. We need to prove that the requirements (i) - (iii) of the aforementioned theorem are satisfied.

(i) limk→∞

γk(tk)−γ(t) 2

γk(tk)−γ(t−) 2

= 0.However,γis continuous. Therefore, it is sufficient to prove

k→∞lim

γk(tk)−γ(t) 2= 0.

To this end, we have γk(tk) =

Z

[0,tk]

αk(s) dCsk= Z

[0,tk∧t]

αk(s) dCsk+ Z

[tk∧t,t]

αk(s) dCsk+ Z

[t,tk∧t]

αk(s) dCsk. The first summand converges toγ(t), because the sequence (Ck1[0,tk])k∈

Nisk · k−convergent and (αk)k∈

N is uniformly bounded δJ1(Rp)−convergent. In other words, the conditions of Proposition A.16 are satisfied. For the second summand we have

Z

[tk∧t,t]

αk(s) dCsk

≤Var Z

[0,·]

αk(s) dCsk

t

−Var Z

[0,·]

αk(s) dCsk

tk∧t

= Z

[tk∧t,t]

αk(s)∨0 dCsk+

Z

[tk∧t,t]

−αk(s)∨0 dCsk

≤sup

k∈N

αk

µk [tk∧t, t]

= sup

k∈N

αk

Ctk−Ctkk∧t

−−−−→

k→∞ 0,

where we used Proposition A.12 in order to conclude the convergence. Analogously we can prove that the third summand converges also to 0.

(ii) If limk→∞

γk(tk)−γ(t)

2= 0 and (sk)k∈Nsuch thattk≥skfor everyk∈Nandsk−−−−→

k→∞ t, then

k→∞lim

γk(sk)−γ(t) 2= 0.

We can repeat the exact same arguments as in (i) in order to prove the convergence for every (sk)k∈N such thatsk −−−−→

k→∞ t.