S
n=1
An = Ω. Fortunately, this follows immediately from the second part of statement .4 and
|Q(Ω)−Q(An)| ≤
βρ(Λ) +ρ(0) + inf
1An≥Z∈Xρ(Z)−ρ(λ)
λ for allλ >0;
Note thatβρ(Λ)≥EQ[−λ1An]− inf
1An≥Z∈Xρ(λZ) + (λQ(Ω) +ρ(λ)−ρ(0)).
9 Proofs of results from section 2
Proof of Lemma 2.1:
It remains to show the if part. For this purpose let (Xi)i∈I denote a uniformly bounded net inXbwith pointwise limitX∈Xb.Settingc:= lim inf
i ρ(Xi) and fixingε >0,we may find a subnet (Xi(k))k∈K withρ(Xi(k))< c+ε for everyk∈K.Hence (Xi(k)+c+ε)k∈K is a net in someArwhich converges pointwise toX+c+ε.So in view of condition (2.1)X+c+ε is also the pointwise limit of some sequence (Yn)n fromAr. Ifρsatisfies the Fatou property we may concludeρ(X+c+ε)≤lim inf
n→∞ ρ(Yn)≤0,and henceρ(X)≤c+ε.This completes the proof.
Proof of Theorem 2.2:
Let us retake assumptions and notations from Theorem 2.2
The implication.2⇒.3 is always valid as indicated in Proposition 1.4.
Let us now introduce the space ˆF consisting of all real linear forms onXbwhich are representable by someµ∈F.
The operator normk · kFˆ on ˆF w.r.t. the sup normk · k∞satisfieskR
·dµkFˆ =kµkF for everyµ∈F.Since F is supposed to be complete w.r.tk · kF,( ˆF ,k · kFˆ) is a Banach space. The topological dual ˆF0 of ˆF will be endowed with the respective operator normk · k,andBFˆ0 denotes the unit ball in ˆF0.
Since F contains all Dirac measures, Xb may be embedded isometrically into ˆF0 w.r.t. k · k∞ and k · k by the evaluation mapping ˆe:Xb→Fˆ0.Next let us fix an arbitraryJ ∈Fˆ0 outside the closure cl(ˆe(Xb)∩BFˆ0) of ˆ
e(Xb)∩BFˆ0w.r.t. the weak * topologyσ( ˆF0,F).ˆ By Hahn-Banach theorem we may find someσ( ˆF0,Fˆ)−continuous real linear form Λ on ˆF0with
sup{Λ( ˜J)|J˜∈cl(ˆe(Xb)∩BFˆ0)}<Λ(J).
In addition there is someµ∈F with Λ( ˜J) = ˜J(R
·dµ) for any ˜J∈Fˆ0.Without loss of generality we may assume kµkF = 1.Since ˆeis isometric, we obtain then
kJk>sup{
Z
X dµ|X∈Xb,sup
ω∈Ω
|X(ω)| ≤1}=kµkF = 1.
Hence ˆe(Xb)∩BFˆ0 isσ( ˆF0,Fˆ)−dense inBFˆ0.
Now let condition (2.2) be valid, and let us assume thatρsatisfies the nonsequential Fatou property. Using Dirac measuresδω (ω ∈Ω),we may define for anyJ ∈Fˆ0 a mappingXJ ∈ RΩ viaXJ(ω) := J(R
·δω).EachXJ is bounded because |XJ(ω)| ≤ kJk holds for every ω ∈ Ω. Furthermore for any J ∈ Fˆ0 there exists a uniformly bounded net (Xi)i∈I inXbsuch that (ˆe(Xi))i∈I converges toJ w.r.t. σ( ˆF0,Fˆ).In particularXJ is the pointwise limit of (Xi)i∈I,which means that it belongs toXbdue to (2.2). Hence the mapping ˆρ: ˆF0→R, J 7→ρ(XJ) is well defined with ˆρ(ˆe(X)) =ρ(X) forX∈Xb.
For everyr >0 and any net (Ji)i∈I in ˆρ−1(]− ∞,0])∩rBFˆ0 we may select by Banach-Alaoglu theorem a subnet (Ji(k))k∈Kand someJ∈rBF0such that (Ji(k))k∈Kconverges toJw.r.t. σ( ˆF0,Fˆ).Then (XJi(k))k∈Kis a uniformly bounded net inXbwhich converges pointwise toXJ.Sinceρfulfills the nonsequential Fatou property, we obtain
ˆ
ρ(J) =ρ(XJ)≤lim inf
k ρ(XJi(k)) = lim inf
k ρ(Jˆ i(k))≤0.
Thus the sets ˆρ−1(]− ∞,0])∩rBFˆ0(r >0) areσ( ˆF0,Fˆ)−compact, which means that ˆρ−1(]− ∞,0]) is closed w.r.t.
σ( ˆF0,Fˆ) by Krein-Smulian theorem. Now it is easy to check that ρ−1(]− ∞,0])∩Xb is closed w.r.t. σ(Xb, F), which implies that all level setsρ−1(]− ∞, c])∩Xb(c∈R) areσ(Xb, F)−closed due to the translation invariance ofρ.This shows statement .2, drawing on Propositions 1.2, 6.6 and Lemma 6.5. As a further consequence we have equivalence of the statements .1 - .3 under (2.1), (2.2) in view of Lemma 2.1.
If we strengthen condition (2.2) by the assumption that the setsAr from (2.1) are compact w.r.t. σ(X, F), it remains to show the implication.2⇒.1.Indeed statement .2 implies thatρis lower semicontinuous w.r.t. the weak topologyσ(X, F) by Propositions 1.2. Furthermore for any uniformly bounded net (Xi)i∈I inXb with pointwise limit X ∈ Xb we may suppose without loss of generality that (Xi)i∈I is a net in some Ar due to translation invariance ofρ. Then, drawing on relativeσ(X, F)−compactness of Ar,the mapping X is theσ(X, F)−limit of (Xi)i∈I.This implies lim inf
i ρ(Xi)≥ρ(X),and completes the proof.
Proof of Remark 2.4:
Let ˆedenote the evaluation mapping fromXbinto the topological dualF0 ofF w.r.t. the norm of total variation.
It is isometric w.r.t. the sup normk · k∞ and the operator norm k · k.Then the if part is obvious in view of Banach-Alaoglu theorem. Conversely, translation invariance and relative σ(X, F)−compactness of the sets Ar
imply theσ(F0, F)−compactness of the sets ˆe(Xb)∩rBF0 (r > 0),where BF0 denotes the unit ball w.r.t. k · k.
This means that ˆe(Xb) is closed w.r.t. σ(F0, F) due to the Krein-Smulian theorem, and then ˆe(X) =F0 because Xb separates points inF,and thus ˆe(Xb) is dense w.r.t. σ(F0, F).The proof is finished.
A Appendix
Proposition A.1 Let(Ω,e F)e be a measurable space, and let {∅,Ω} ⊆ S ⊆e Fe be stable under finite union and countable intersection, generatingF.e Furthermore, for every A∈ S there exists a sequence (An)n in S such that Ωe\A=
∞
S
n=1
An. Then every probability contentQ onFeis a probability measure if and only if lim
n→∞Q(An) = 1 holds for any isotone sequence(An)n inS with
∞
S
n=1
An=Ω.e Proof:
Let Q be a probability content onF,e and let us denoteϕ:= Q|S.The only if part of the statement is obvious.
For the if part we want to show (*) lim we may apply a version of the general extension theorem by K¨onig (cf. [13], Theorem 7.12 with Proposition 4.5).
Hence condition (∗) together with the assumptions on S guarantee a probability measure P on theσ−algebra Fewith P|S =ϕ,and P(A) = sup
A⊇B∈S
P(B) for everyA ∈F.e In particular P≤Q which implies P = Q due to additivity of Q and P.Therefore it remains to prove the condition (∗).
proof of (*):
Let (An)nbe an isotone sequence in S with
∞
S
n=1
An=A∈ S.By assumption there exists some isotone sequence (Bn)ninSwith which shows (*), and completes the proof.
B Appendix
Proof of Proposition 1.1:
LetX∈X,and letρ0+(X,·) :X→Rdenote the respective rightsided derivative ofρatX defined byρ0+(X, Y) :=
h→0lim+
ρ(X+hY)−ρ(X)
h . It it well known from convex analysis thatρ0+(X,·) is well defined and sublinear satisfying ρ0+(X, Y −X)≤ρ(Y)−ρ(X) for all Y ∈X.Then we may choose by Hahn-Banach theorem some linear form ˜Λ onXwith ˜Λ≤ρ0+(X,·).Moreover, we obtain forZ≥0
Λ(Z˜ )≤ρ0+(X,(X+Z)−X)≤ρ(X+Z)−ρ(X)≤0.
Therefore Λ := −Λ is a positive linear form fulfilling Λ(X˜ −Y) ≤ ρ0+(X, Y −X) ≤ ρ(Y)−ρ(X) for Y ∈ X, and Λ(Y) ≤π(Y) forY ∈ C.This implies Λ|C=π due to linearity of Λ|C and π.Furthermore we have shown βρ(Λ) =−Λ(X)−ρ(X).
For the proof of the equivalence stated in Proposition 1.1 note that the only if part is obvious, the if part follows immediately from the Fenchel-Moreau theorem (cf. [2], Theorem 4.2.2), observing thatβρ(Λ) <∞only if Λ is positive linear and extendsπ(see also Proposition 3.9 in [9]). This completes the proof.
Acknowlegdements:
The author would like to thank Freddy Delbaen and Alexander Schied for helpful hints. He is also indepted to Heinz K¨onig for good advice and continuous exchange of ideas from the fields of measure theory and superconvex analysis.
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