A CROSS SECTION THEOREM. I

A CROSS SECTION THEOREM. I

PETER GRAY

The theory surrounding a one-parameter stochastic process(Xt)t∈R+ has appli- cations in many elds. The predictable projection (^pXt)t∈R+and the predictable dual projection(X_t^p)t∈R+ for the process(Xt)t∈R+ both play roles in many such applications. Several results concerning these projections are derived using the Cross Section Theorem for predictable subsets of the setΩ×R⁺,which states that for any predictable setA and any >0there exists a predictable stopping timeS such that [S]⊂A and P(π(A))−P(π([S]))≤, where [S] denotes the graph ofS, π represents the projection onto the setΩ,and P is the probability measure onΩ.The goal of this paper is to extend the Cross Section Theorem to predictable subsets of the setΩ×R²+,thereby paving the way for the derivation of results pertaining to a two-parameter stochastic process (Xs,t)s,t∈R+.Such an extension of the Cross Section Theorem is dicult because, while the setR⁺ of positive real numbers is totally ordered, the set R²+ of ordered pairs of positive real numbers is not. We combat this diculty by rst collecting results using the double ltrationFs,t =Fs−∩ F.Once this is done, we will be able to consider the more general double ltrationFs,t=Fs∨t−.The main result of this paper is a cross section theorem for predictable subsets ofΩ×R²+ relative to the double ltration (Fs,t)s,t∈R+satisfyingFs,t=Fs−fors, t≥0.This theorem states that for any predictable subset A of Ω×R²+ and any > 0 there is a predictable stopping timeZ such that[Z]⊂AandP(π(A))−P(π([Z]))≤.

AMS 2000 Subject Classication: Primary 60G20; Secondary 60G40.

Key words: two-parameter stochastic process, cross section.

1. PRELIMINARIES

Notation and Terminology 1.1. The following will be used in the sequel.

1.1a (Ω,F, P) is a probability space.

1.1b R+ is the set of non-negative real numbers. N is the set of natural numbers.

1.1cB(R+)is the Borelσ-algebra generated by the intervals (s, t] ofR+. B R²₊

is the Borelσ-algebra generated by the rectangles(s, t]×(u, v]ofR²₊. 1.1d A functionX: Ω×R²₊→Ris called a two-parameter process, and is denoted (Xs,t). For literature pertaining to two-parameter processes, the reader is referred to [Dz], [C-W], [L], and [W].

REV. ROUMAINE MATH. PURES APPL., 53 (2008), 4, 297316

1.1e F ⊗ B(R₊) is theσ-algebra generated by the semiring F × B(R₊).

F ⊗ B(R²+) is theσ-algebra generated by the semiringF × B(R²+).

1.1f (F_t)t∈R+ is a ltration; that is,

• for eacht≥0,F_t is aσ-algebra contained inF, and

• F_s⊂ F_t ifs≤t.

We will simply write(F_t), and will assume that the ltration satises the usual conditions:

• F₀ contains all the negligible sets (that is,(F_t)is complete), and

• F_t= T

s>t

F_s for everyt≥0 (that is,(F_t)is right continuous).

See 1.1j below for another assumption about(F_t).

1.1g A functionS : Ω→[0,∞]is a stopping time for the ltration(F_t) if {S≤t} ∈ F_t for every t≥0.

LetS, T be two stopping times. The stochastic interval (S, T]is the set {($, t)∈Ω×R+|S($)< t≤T($)},

while [S, T)is the set

{($, t)∈Ω×R+| S($)≤t < T($)}.

The stochastic intervals (S, T) and[S, T]are dened in a similar fashion.

1.1h The predictableσ-algebraP₁ of subsets ofΩ×R+ is theσ-algebra generated by the sets A×(s, t] and B × {0}, where A ∈ F_s− and B ∈ F₀. A stopping time S is predictable if the stochastic interval [S,∞) is a pre- dictable set.

1.1i LetS be a stopping time. F_S denotes the σ-algebra {A∈ F |A∩ {S ≤t} ∈ F_t for everyt≥0}

while F_S− denotes theσ-algebra generated by the sets inF₀ as well as sets of the form A∩ {S > t}, wheret≥0 andA∈ F_t.

1.1j(G_s)_s∈

R+ is the ltration dened by

• G₀=F₀ and

• G_s=F_s− for everys≥0.

We will simply write(G_s), and will assume that the ltration(F_t)is such that G_S = F_S− for every predictable stopping time S for the ltration (G_s).

For example, this is the case when F_s =F_bsc for every s∈ R+, where bsc is the largest integer less than or equal to s.

1.1k(F_s,t)_s,t∈

R+ is a double ltration; that is,

• for eachs, t≥0,F_s,t is aσ-algebra contained inF, and

• F_s,t⊂ F_u,v ifs≤u andt≤v.

We will simply write (F_s,t), and will assume that F_s,t = G_s for every s, t≥0.

1.1l LetB ⊂Ω×R²₊. B($) is the set{(s, t)∈R²₊ |($, s, t)∈B}. π[B], called the projection ofB, is the set{$∈Ω|($, s, t)∈Bfor somes, t∈R+}. 1.1m The point(∞,∞)will be denoted by∞. Therefore, the inequality (s, t)<∞ means thats <∞ and t <∞.

1.1n Let g : Ω → [0,∞]×[0,∞] be a function. [g] denotes the set {($, s, t)∈Ω×R²+|g($) = (s, t)}, called the graph of g.

We now commence our study of the real-valued two-parameter process (Xs,t).

Denition 1.2. LetX: Ω×R²₊→Rbe a two-parameter process.

1.2a(Xs,t) is left continuous if for everys0, t0∈R+and$∈Ω, we have

(s,t)→(slim0,t0) 0≤s≤s₀ 0≤t≤t0

Xs,t($) =Xs0,t0($).

1.2b (X_s,t) is adapted to the ltration (F_s,t) if for each s, t ∈ R+ the random variable Xs,t isF_s,t-measurable.

1.2c The predictable σ-algebra of subsets of Ω×R²+ is the σ-algebra generated by left continuous two parameter processes(X_s,t)which are adapted to (F_s,t). We denote thisσ-algebra byP.

Proposition 1.3. P is generated by the sets (S,∞)×[0, r₁] and A× {0} ×[0, r₂], where r₁, r₂ ∈R+, S : Ω → [0,∞] is a stopping time for (G_s), and A∈G0.

Proof. Let(X_s,t)be left continuous and adapted to(F_s,t). Set Y_s,tⁿ :=X_0,01{0}×{0}(s, t) +

∞

X

k=0

X_0,k

n1_{0}×(^k

n,^k+1_n ](s, t)+

+

∞

X

m=0

X^m

n,01₍^m

n,^m+1_n ]×{0}(s, t) +

∞

X

k,m=0

Xm n,_n^k1₍^m

n,^m+1_n ]×(_n^k,^k+1_n ](s, t), where 1 is the indicator function, and n ∈N. Since (X_s,t) is left continuous, Yⁿ→X pointwise asn→ ∞.

Fixm, n,and k,and consider the process R= (R_s,t), where Rs,t :=Xm

n,^k_n1₍^m

n,^m+1_n ]×(^k_n,^k+1_n ](s, t).

SinceXm

n,_n^k isG^m

n-measurable,Ris the pointwise limit of processes of the form

p

P

i=1

α_i1_Ai×(^m

n,^m+1_n ]×(^k

n,^k+1_n ],where α_i ∈Rand A_i ∈ G^m

n for every index i. But

for every index iwe have Ai×(^m_n,^m+1_n ] = (Si, Ti], where Si =

m

n on A_i

∞ on A^c_i, which is a stopping time for (G_s), and

T_i =

( _m+1

n on A_i

∞ on A^c_i,

which also is a stopping time for (G_s). Therefore, R is the pointwise limit of processes of the form P^p

i=1

α_i1_(Si,Ti]×(^k

n,^k+1_n ], where for every index i we have α_i ∈ R, and S_i, T_i are stopping times for (G_s). By considering in a similar manner the processes X_0,k

n1_{0}×(^k

n,^k+1_n ]andX^m

n,01₍^m

n,^m+1_n ]×{0},we obtainP ⊂ σ{(S₁, T1]×(r₁, s1], A× {0}×(r₂, s2],(S2, T2]×{0}},wherer1, r2, s1, s2 ∈R+, A ∈ G₀, and S₁, S₂, T₁, and T₂ are stopping times for (G_s) ; = σ{(S,∞)× [0, r1], A× {0} ×[0, r2]}, wherer1, r2 ∈R+,A∈ G₀, and S is a stopping time for(G_s). In fact we haveP =σ{(S,∞)×[0, r₁], A× {0} ×[0, r₂]}because the sets (S,∞)×[0, r1]and A× {0} ×[0, r2]are elements ofP.

Remark. For alternative generators ofP, see [D], p. 323.

2. STOPPING TIMES

Denition 1.4. LetZ : Ω→[0,∞]×[0,∞]be a function.

1.4a The functionZ is a stopping time for the ltration (F_s,t) if {Z ≤ z} ∈ F_z for every z∈R²₊.

1.4b The stopping timeZ = (S, T) is predictable if

• S is a predictable stopping time for(G_s), and

• {S <∞} ⊂ {T <∞}.

Proposition 1.5. Let Z_n= (S_n, T_n)be a sequence of stopping times for any double ltration (F_s,t⁰ ) that is right continuous in t. Assume that (Sn) is increasing. Then Z:= (sup_nS_n,lim sup_nT_n)is a stopping time for (F_s,t⁰ ).

Proof. Let(s, t)∈R²₊. We must show that {Z ≤(s, t)} ∈ F_s,t. For each index n and each(r, u)∈R²+ we have {Z_n≤(r, u)}={S_n≤r} ∩ {T_n≤u} ∈ F_r,u by hypothesis. SetT_n⁰ :=

∞

W

i=n

T_i.Then lim sup_nT_n= limnT_n⁰.We have {Z ≤(s, t)}=

n sup

n

Sn,lim

n T_n⁰

≤(s, t) o

=

= n

sup

n

Sn≤s o

∩n

limn T_n⁰ ≤t o

=

∞

\

n=1

{S_n≤s} ∩

∞

\

p=N

∞

[

k=1

∞

\

m=k

{T_m⁰ ≤t+p}, where N is any natural number, and (_p) is any sequence of positive numbers decreasing to 0;

=

∞

\

p=N

∞

[

k=1

" _∞

\

n=1

{S_n≤s} ∩

∞

\

m=k

{T_m∨Tm+1∨ · · · ≤t+p}

#

=

∞

\

p=N

∞

[

k=1

" _∞

\

n=1

{S_n≤s} ∩

∞

\

m=k

{T_m ≤t+p} ∩ {T_m+1 ≤t+p} ∩ · · ·

#

=

∞

\

p=N

∞

[

k=1

∞

\

m=k

[{S_m ≤s} ∩ {S_m+1 ≤s} ∩ · · · ∩ {T_m ≤t+_p} ∩ · · ·]

since (Sn) is increasing;

=

∞

\

p=N

∞

[

k=1

∞

\

m=k

[{Z_m ≤(s, t+p)} ∩ {Z_m+1 ≤(s, t+p)} ∩ · · ·]∈ F_s,t+⁰

N,

N ∈N being arbitrary. Since F_s,t⁰

is right continuous in t, we conclude that {Z ≤(s, t)} ∈ F_s,t⁰ .

Proposition 1.6. Let n ∈ N, and let Zi = (Si, Ti), i = 1, . . . , n, be a nite set of stopping times for (F_s,t) such that each Si is a stopping time for (G_s). Set A_i :={inf_k≤nS_k=S_i} ∩ {S_i 6=S₁} ∩ {S_i6=S₂} ∩ · · · ∩ {S_i 6=Si−1}, i= 1, . . . , n,and set S :=

n

P

i=1

Si1Ai and T :=

n

P

i=1

Ti1Ai.

Denote (S, T) by minf_i(Z_i), and setZ := minf_i(Z_i).Then

• S is a stopping time for (G_s),

• Z is a stopping time for(F_s,t), and

• if T_i($) = ∞ ⇒ S_i($) = ∞ for each index i then T($) = ∞ ⇒ S($) =∞.

Proof. We rst prove thatS is a stopping time for(G_s).

Note that the sets Ai, i = 1, . . . , n, are pairwise disjoint. Further, we have Ai ∈ G_Si for each i(see [M], p. 20), and S^∞

i=1

Ai = Ω.It follows that S is a stopping time for (G_s).

Next, we prove that Z is a stopping time for(F_s,t). Let (s, t)∈R²+.We must show that {Z ≤(s, t)} ∈ G_s.By hypothesis, {Z_i ≤(s, t)} ∈ G_s for eachi.

So, we have {Z ≤(s, t)}=

n

[

i=1

{Z_i ≤(s, t)} ∩A_i =

n

[

i=1

({S_i≤s} ∩A_i)∩ {Z_i ≤(s, t)} ∈ G_s. Lastly, we prove the third assertion of the proposition. We haveT($) =

∞ ⇒ T_i($) = ∞ for some index i such that $ ∈ A_i; ⇒ S_i($) = ∞ by hypothesis ⇒S($) =∞ since $∈Ai.

3. PROJECTIONS π[A]

In this section we establish a key result concerning the projectionπ[A]of an F ⊗ B(R²₊)-measurable set A. It will emerge that π[A]is F-measurable.

Proposition 1.7. Let A∈ F ⊗ B(R²+)and let g be a function such that π[[g]]∈ F, and [g]⊂A on π[A]∩π[[g]]. Then π[[g]∩A]∈ F.

Proof. Let A be the collection of sets B ∈ F ⊗ B(R²₊) such that for any function h : Ω → [0,∞]×[0,∞] satisfying π[[h]] ∈ F and [h] ⊂ B on π[B]∩π[[g]] we have π[[h]∩B]∈ F. Let Rbe the ring generated by the sets (C×(s, t]×(u, v])∩Ω×R²+ andΩ×R²+, whereC∈ F and s, t, u, v∈R.

First, we show that A contains R. Let B ∈ R and let a function h be such that π[[h]]∈ F and [h]⊂B on π[B]∩π[[h]]. Without loss in generality we consider B =C×(s, t]×(u, v], where C ∈ F and s < t and u < v. We have π[[h]∩B] = (π[[h]])∩C ∈ F. Hence, B ∈ A.

Next, we show that A is a monotone class. Let (B_n) be a monotone sequence from A. Assume rst that (Bn) is increasing. Set B := S

n

Bn. Let a function h1 satisfy π[[h1]] ∈ F and [h1] ⊂ B on π[B]∩π[[h1]]. Let h⁰ : Ω → [0,∞]×[0,∞] be a function that satises the conditions set out below. We require that π[[h⁰]] = π[[h1]], h⁰ = h1 on (π[B])^c, [h⁰] ⊂ B1 on (π[B₁])∩π[h₁], and[h⁰]⊂B_i on(π[B_i]\π[B_i−1])∩π[h₁]fori= 2,3,4, . . .. Such a function h⁰ exists. Note, we haveπ[[h⁰]]∈ F, and[h⁰]⊂Bi on[π[Bi]∩π[[h⁰]]

for i ∈ N. Therefore, for each index i we have π[[h⁰]∩Bi] ∈ F (since each B_i is in A). Hence we have π[[h₁]∩B] = π[[h⁰]∩B] from the denition of h⁰; = π[[h⁰]∩S

i

Bi] = S

i

π[[h⁰]∩Bi] ∈ F. Second, assume that (Bn) is decreasing. Set B⁰ :T

n

B_n. Let a functionh₂ satisfy π[[h₂]]∈ F and[h₂]⊂B⁰ on π[B⁰]∩π[[h2]]. Let h⁰⁰ : Ω → [0,∞]×[0,∞] be a function that satises the conditions set out below. We require that π[[h⁰⁰]] = π[[h₂]], h⁰⁰ = h₂ on (π[B1])^c∪π[B⁰], and[h⁰⁰]⊂Bi on (π[Bi]\π[Bi+1])∩π[h2] for i∈N. Such a function h⁰⁰ exists. Note, we haveπ[[h⁰⁰]]∈ F, and [h⁰⁰]⊂B_i onπ[B_i]∩π[[h⁰⁰]]

for i∈N. Therefore, for each indexi we haveπ[[h⁰⁰]∩Bi]∈ F (since eachBi

is inA). Hence we haveπ[[h2]∩B⁰] =π[[h⁰⁰]∩B⁰](from the denition ofh⁰⁰);

=π[[h⁰⁰]∩T

i

B_i] =T

i

π[[h⁰⁰]∩B_i]∈ F (since[h⁰⁰]is a graph).

We have shown that A is a monotone class which contains the ring of generators ofF ⊗ B(R²₊). Because this ring contains the whole spaceΩ×R²₊, we conclude from the Monotone Class theorem (see [D-M], p. 13-I) that A is equal to F ⊗ B(R²₊). The statement of the proposition is now seen to be true.

Corollary 1.8. Let A∈ F ⊗ B(R²+). Then π[A]∈ F.

Proof. We are able to nd a function h: Ω → [0,∞]×[0,∞]such that π[[h]] = Ω and [h] ⊂ A on π[A]∩π[[h]]. From Proposition 1.7 we obtain π[[h]∩A] ∈ F. For such a function h we have π[[h]∩A] = π[A]. Hence we have π[A]∈ F.

4. SETS K_δ

We continue to prepare for the Cross Section theorem by introducing a special collection K_δ of predictable sets with compact cross sections.

Denition 1.9. LetK⊂Ω×R²+.

1.9a We say thatK has compact cross sections if for each$∈π[K]the cross section K($)⊂R²+ is compact.

1.9b The cross sectional closure ofK, denotedK^∗, is given byK^∗($) = (K($))^∗ for every $ ∈ Ω, where (K($))^∗ denotes the closure of the cross section K($)⊂R²+.

Proposition 1.10. Let N ∈N, and let (K_n)be a sequence of subsets of Ω×R²+with compact cross sections. Assume that each set Knhas the following properties: π[B^∗∩K_n]∈ F for every F ⊗ B(R²+)-measurable setB; and there is a stopping time Zn= (Sn, Tn) such that

•S_n is a predictable stopping time for (G_s),

•[Zn]⊂Kn,

• {Z_n<∞}=π[K_n],

•T_n($) =∞ ⇒S_n($) =∞ for every $∈Ω, and

•(s, t)∈Kn($)⇒s≥Sn($) for every$∈Ω. Set K :=

N

S

n=1

Kn andK⁰ :=

∞

T

n=1

Kn. Then K andK⁰ both have compact cross sections, and K has all the above properties of the sets Kn.Further, if the sequence (K_n) is decreasing, then the set K⁰ also has all the above properties.

Proof. The nite union and the countable intersection of compact subsets of R²+ are compact. Therefore, K and K⁰ have compact cross sections.

Next, let B ∈ F ⊗ B(R²+). We have π[B^∗ ∩K] = π[B^∗ ∩ S^N

n=1

K_n] = SN

n=1

π[B^∗∩K_n] ∈ F, and π[B^∗ ∩K⁰] = π[B^∗ ∩

∞

T

n=1

K_n] =

∞

T

n=1

π[B^∗ ∩K_n] if (K_n) is decreasing, since each setB^∗∩K_n has compact cross sections;∈ F.

We now show thatK has the second property set out in the proposition.

Set Z := minfn=1,...,N(Zn). (See Proposition 1.6 for the denition of minf.) From Proposition 1.6 we know that Z is a stopping time, and that if we write Z = (S, T), then S is a stopping time for (G_s). Note, the stopping time S is a predictable stopping time because S is the minimum of nitely many predictable stopping times Sn,n= 1, . . . , N.Next, let$∈[Z]. ThenZ($)<

∞. There is an index n such that Z($) = Zn($). From this equality we deduce that Z_n($) < ∞. By hypothesis, we have ($, Z_n($)) ∈ K_n. Hence ($, Z($)) = ($, Zn($))∈Kn⊂K.

We now show that {Z < ∞} = π[K]. We have π[K] = π[

N

S

n=1

Kn] =

N

S

n=1

π[Kn] =

N

S

n=1

{Z_n < ∞} by a property of the set Kn =

N

S

n=1

{S_n < ∞} ∩ {T_n < ∞}

=

N

S

n=1

{S_n <∞} by a property of the function Zn; =

N

S

n=1

[{S_n <

∞} ∩ {inf_n⁰S_n⁰ =Sn} ∩ T

n⁰⁰<n

{S_n6=S_n⁰⁰}] ={Z <∞}.

Penultimately, note that by Proposition 1.6 we have T($) = ∞ ⇒ S($) =∞ for every $∈Ω.

Lastly, we show that (s, t) ∈ K($) implies that s ≥ S($) for every

$ ∈ Ω. We have (s, t) ∈ K($) ⇒ ($, s, t) ∈ K ⇒ ($, s, t) ∈ K_n for some index n⇒s≥Sn($) by hypothesis;≥S($)since S = infn=1,...,N Sn.

To complete the proof, assume that the sequence (K_n) is decreasing.

We verify that the set K⁰ has all ve properties that were listed. Let Z :=

(sup_nS_n,lim sup_nT_n). Note that Z is a stopping time. To see this, it is enough to show (by Proposition 1.5) that the sequence (S_n) is increasing.

Let $ ∈ Ω and n0 ∈ N, and assume rst that Zn0+1($) < ∞. Then ($, S_n0+1($), T_n0+1($)) ∈ K_n0+1 by a property of the set K_n0+1 ⊂ K_n0. So,($, Sn0+1($), Tn0+1($))∈Kn0($).By a property of the setKn0, we have S_n0+1($)≥S_n0($). Next, assume thatZ_n0+1($) =∞. ThenS_n0+1($) =∞ or Tn0+1($) = ∞. We have Sn0+1($) = ∞ by a property of the set Kn0+1. HenceS_n0+1($)≥S_n0($),and we have shown thatZ is a stopping time. Now note, the stopping time sup_nS_n is a predictable stopping time for (G_s). This follows since each Sn is a predictable stopping time for(G_s), and since(Sn)is increasing.

We now prove that [Z] ⊂ K. Let Z($) < ∞. We must show that ($, Z($)) ∈ K⁰. Since Z($) < ∞, we have sup_nS_n($) < ∞. Therefore, Sn($)<∞for every indexn. It follows from a property ofKn thatZn($)<

∞ for every n. So, ($, Zn($)) ∈ Kn for every n, since [Zn] ⊂ Kn for all n. There is a subsequence (T_nk($))_k of (T_n($))_n (that depends on $) such that Tnk($) → lim sup_nTn($) ask → ∞. Since (Sn($)) is increasing, we have S_nk($) → sup_nS_n($) as k → ∞. Thus, we have ($, Z_nk($)) = ($, S_nk($), T_nk($))→($, Z($))ask→ ∞. For eachkwe have($, Z_nk($))

∈ K_nk. Since by assumption the sequence (K_n) is decreasing, ($, Z_nk($))_k is a sequence that eventually belongs to each of the sets Kn_k, k ∈ N. By the compactness of the cross sections involved, ($, Z($)) = lim_k($, Z_nk($)) belongs to each of the sets Knk,k∈N. So, we have($, Z($))∈T

k

Knk =K⁰. We now show that {Z < ∞} = π[K⁰]. First, we establish that {Z <

∞} = T

n

{Z_n < ∞}. In fact, let $ ∈ {Z < ∞}. Suppose there is a number n₀ ∈ N such that Z_n0($) ≮ ∞. Then either S_n0($) = ∞ or T_n0($) =

∞. Since Tn0($) = ∞ ⇒ Sn0($) = ∞, we obtain Sn0($) = ∞. Since (S_n) is increasing, we have S_m($) = ∞ for each m ≥ n₀. It follows that sup_nSn($) =∞. Therefore,Z($) = (sup_nSn($),lim sup_nTn($))≮∞, and we have reached a contradiction. Because of this contradiction, we conclude that Z_n($)<∞ for everyn∈N. Hence, we have{Z <∞} ⊂T

n

Z_n<∞}. Next, let$∈T

n

{Z_n<∞}. Then for eachnwe haveZn($)<∞. Note, for each n we have [Z_n] ⊂ K_n. Therefore, for each n we have ($, Z_n($)) ∈ K_n ⊂ K₁. Since K₁ has compact cross sections, there is M ∈ R²₊ (that depends on$)such that(s, t)≤M ∀(s, t)∈K1($). Thus, for eachnwe have Z_n($) ≤M. Consequently, we have Z($) = (sup_nS_n($),lim sup_nT_n($))≤ M. Accordingly, we have $ ∈ {Z < ∞}. Hence, T

n

{Z_n < ∞} ⊂ {Z < ∞}. Therefore, we have {Z < ∞} = T

n

{Z_n < ∞} (by the preceding lines); = T

n

π[K_n]since {Z_n <∞}=π[K_n]for all n;=π[T

n

K_n]since all sets K_n have compact cross sections; =π[K⁰].

We now show that lim sup_nT_n($) = ∞ ⇒ sup_nS_n($) = ∞ for every

$∈Ω. In fact, for each$∈Ωwe havelim sup_nT_n($) =∞ ⇒Z($)≮∞ ⇒ Zn0($) ≮ ∞ for some index n0, since {Z < ∞} = T

n

{Z_n < ∞}; ⇒ either Sn0($) =∞ or Tn0($) =∞ ⇒ Sn0($) =∞ (as Tn0($) = ∞ ⇒ Sn0($) =

∞);⇒sup_nS_n($) =∞.

Lastly, we prove that(s, t)∈K⁰($)⇒s≥sup_nSn($) for every $∈Ω. In fact, for each $∈Ω we have (s, t)∈K⁰($)⇒($, s, t)∈K⁰ =

∞

T

n=1

Kn⇒

($, s, t) ∈ Kn for every n; ⇒ (s, t) ∈ Kn($) for every n; ⇒ s ≥ Sn($) for everyn by a property ofK_n;⇒s≥sup_nS_n($).

Denition 1.11. We dene the setK to be the collection of nite unions

N

S

i=1

[S_i+_i, T_i∧n_i]×[r_i, r⁰_i], where n_i, N ∈N, _i > 0, r_i, r⁰_i ∈ R+, and S_i, T_i are stopping times for (Gs) for every index i. We dene the set Kδ to be the collection of countable intersections of sets from K.

Proposition 1.12. The set K_δis closed under nite union and countable intersection. Further, the elements of K_δ have all the properties that were presented in Proposition 1.10.

Proof. It is evident thatK_δ is closed under countable intersection.

Next, we show that K_δ is closed under nite union. Let M ∈ N and let (Ki)i=1,...,M be a nite family from K_δ. For each index i we may write K_i =

∞

T

j=1

K_i,j, where each set K_i,j is an element of K. Then we have ^MS

i=1

K_i=

M

S

i=1

∞

T

j=1

Ki,j =

∞

T

j1,j2,...,jM=1

K1,j1 ∪K2,j2 ∪ · · · ∪ KM,jM, which is a countable intersection of elements of K, since each set K_1,j1∪K_2,j2 ∪ · · · ∪K_M,jM is an element of K. Accordingly, the set ^MS

i=1

Ki is an element ofK_δ.

We now prove that the elements ofK_δpossess all the properties that were presented in Proposition 1.10. Let S, T be stopping times for (G_s), let >0, let n ∈ N, and let r, r⁰ ∈ R+. In view of Proposition 1.10, it is enough to show that the set K₀:= [S+,T∧n]×[r, r⁰] possesses all the aforementioned properties. Without loss in generality, we assume that S + ≤ T ∧n and r ≤ r⁰. Let B ∈ F ⊗ B(R²+). We will show that π[B^∗∩K₀] ∈ F. Let (δ_m) be a sequence of positive numbers decreasing to 0. For each indexm, denote by K_m the set [S+−δ_m, T ∧n+δ_m]×[r−δ_m, r⁰+δ_m], and by K_m^◦ the set (S +−δ_m, T ∧n+δ_m) ×(r −δ_m, r⁰ +δ_m). We have π[B^∗ ∩K₀] =

∞

T

m=1

π[B ∩K_m]. In fact, we have ^∞T

m=1

π[B ∩K_m] ⊂

∞

T

m=1

π[B^∗ ∩K_m] since B ⊂ B^∗; = π[B^∗ ∩

∞

T

m=1

K_m] since (K_m) is decreasing, and since each set B^∗∩K_m has compact cross sections; =π[B^∗∩K₀]. On the other hand, for each index mwe haveπ[B∗ ∩K₀]⊂π[B^∗∩K_m^◦]since K0⊂K_m^◦;=π[B∩K_m^◦] since every cross section of the set K_m^◦ is an open ball when S + < ∞;

⊂π[B∩K_m]sinceK_m^◦ ⊂K_m. Therefore,π[B^∗∩K₀]⊂

∞

T

m=1

π[B∩K_m]. So, we

haveπ[B^∗∩K₀] =

∞

T

m=1

π[B∩K_m]∈ F by Corollary 1.8, sinceK_m ∈ F ⊗B(R²+) for every index m.

We now reveal the remaining properties of the setK₀.SetZ := (S+, r). Note, Z is a stopping time. In fact, let (s, t) ∈ R²+. Then {Z ≤ (s, t)} = {S ≤s if r≤t

∅ otherwise . In either case we have{Z ≤(s, t)} ∈ G_s. We complete the proof by making the following ve observations. The stopping time S+ is a predictable stopping time since > 0. The graph [Z] is a subset of K₀, since by assumption S+≤T ∧nandr ≤r⁰. We have{Z <∞}={S+ <

∞} = π[[S +, T ∧n]], since S + ≤ T ∧n = π[K0]. For each $ ∈ Ω we have r($) = ∞ ⇒ S($) + = ∞, since r < ∞. For each $ ∈ Ω we have (s, t)∈K0($)⇒s∈[S($)+, T($)∧n]andt∈[r, r⁰]⇒s≥(S+)($).

5. THE CROSS SECTION THEOREM

We begin with some important precursors to the Cross Section theorem for predictable subsets of Ω×R²+.

Denition 1.13. LetRbe the ring of subsets ofR²+ generated by the sets (z, z⁰] such that z, z⁰ ∈ R²+, with z < z⁰. The measure µ is the set function fromR intoR+that is given byµ((z, z⁰]) =the area of the rectangle(z, z⁰]for everyz < z⁰, and which is additively extended toR.

Remark. The measure µ is the Lebesgue measure on R, and can be extended to a sigma-additive measure on B(R²+), with values in [0,∞]. We still denote by µ this sigma-additive extension.

Lemma 1.14. Let B ∈ F ⊗ B(R²₊) and $∈Ω. Then B($)∈ B(R²₊). Proof. LetF ⊂Ω, z, z⁰ ∈R²+, and(B_n)⊂Ω×R²+. Then(F×(z, z⁰])($)∈ B(R²₊), while(B1−B2)($) =B1($)−B2($) and(S

n

Bn)($) =S

n

Bn($).

Notation 1.15. Let A, B∈ F ⊗ B(R²+),λ∈R+, and z∈R²+. We denote by ABλ,z the set {$∈Ω|µ((A∩Ω×[0, z])($))> λµ((B∩Ω×[0, z])($))}. WhenA⊂B and 0≤λ <1, the reader might think of the setAB_λ,z as being the projection of the portion of A∩Ω×[0, z]that lls some Borel cross section of B∩Ω×[0, z]by a factor of100λpercent or more.

Proposition 1.16. Let A, B ∈ F ⊗ B(R²+), and let z ∈ R²+. Then for every λ∈R+ we have ABλ,z ∈ F.

Proof. Let R be the ring generated by the sets Sⁿ

i=1

F_i ×(z_i, z_i⁰], where n ∈ N, Fi ∈ F, and zi, z_i⁰ ∈ R²₊ for every index i. Fix A0 ∈ R. Write A0∩Ω×[0, z] =

n0

S

i=1

F0,i×B0,i, where the setsF0,i∈ F are mutually disjoint, and where each set B_0,i is an element of B(R²+). Let A be the collection of sets C in F ⊗ B(R²+) such that for every λ ∈ R+ the set A0C_λ,z is an element of F. We rst show that A contains the ring R. Let C₁ ∈ R and let λ ∈ R+. Write C₁∩Ω×[0, z] =

n1

S

i=1

F_1,i×B_1,i, where the sets F_1,i ∈ F are mutually disjoint, and where each set B1,i is an element ofB(R²+). Then A₀C_1λ,z = S

(i,j)

F_0,i∩F_1,j, where the union is taken over all pairs (i, j) such that µ(B0,i) > λµ(B1,j). Therefore, we have A0C1λ,z ∈ F. Accordingly, we have C1 ∈ A, since λ∈R+ was arbitrary. Next, let(Cn) be an increasing sequence from A. SetC :=

∞

S

n=1

Cn. We will show that C ∈ A. Note, for each $ ∈Ω we have

µ((C∩Ω×[0, z])($)) = lim

n→∞µ((Cn∩Ω×[0, z])($)),

since µis sigma-additive. Let λ∈R+, and let (δm) be a sequence of positive numbers decreasing to0, withλ−δ₁>0. We haveA0Cλ,z =

∞

S

m=1

∞

T

n=1

A0Cnλ−δm,z

(since µis nite on [0, z])∈ F. Since λ∈R+ was arbitrary, we conclude that C ∈ A. Now, let(C_n⁰) be a decreasing sequence fromA. SetC⁰:=

∞

T

n=1

C_n⁰. We will show that C⁰∈ A. Note, for each$∈Ωwe have

µ((C⁰∩Ω×[0, z])($)) = lim

n→∞µ((C_n⁰ ∩Ω×[0, z])($)),

sinceµis sigma-additive and sinceµ((C₁⁰∩Ω×[0, z])($))<∞for every$∈Ω. Let λ∈R+. We haveA₀C_λ,z⁰ =

∞

S

n=1

A₀C_nλ,z⁰ . Sinceλ∈R+ was arbitrary, we conclude that C⁰ ∈ A. By applying the Monotone Class theorem we deduce that for everyA0 ∈R,C∈ F ⊗B(R²₊), andλ∈R+we haveA0Cλ,z ∈ F. Now, letA⁰ be the collection of setsA⁰ inF ⊗ B(R²+) such that for everyλ∈R+ the setA⁰Bλ,zis an element ofF. By the above we haveR⊂ A⁰. Next, let(A⁰_n)be an increasing sequence from A⁰. SetA⁰ :=

∞

S

n=1

A⁰_n. We will show thatA⁰ ∈ A⁰. Let λ ∈ R+. Since µ is sigma-additive, we have A⁰B_λ,z =

∞

S

n=1

A⁰_nB_λ,z ∈ F. Since λ∈ R+ was arbitrary, we conclude thatA⁰ ∈ A. Lastly, let (A⁰⁰_n) be a

decreasing sequence from A⁰. SetA⁰⁰ :=

∞

T

n=1

A⁰⁰_n. We will show that A⁰⁰ ∈ A⁰. Let λ∈ R+, and let (δm) be a sequence of positive numbers decreasing to 0. Since µ is sigma-additive, we have A⁰⁰B_λ,z =

∞

S

m=1

∞

T

n=1

A⁰⁰_nB_λ+δm,z ∈ F. Since λ ∈R+ was arbitrary, we conclude that A⁰⁰ ∈ A⁰. We have shown that A⁰ is a monotone class containing R.By applying the Monotone Class theorem we deduce that for every λ∈R+ the set AB_λ,z is an element ofF.

Proposition 1.17. Letz∈R²+, let(A_n) be an increasing sequence from F ⊗ B(R²+), and letA=

∞

S

n=1

A_n. LetB be a measurable subset ofA, let >0, and let 0≤λ <1. There is an indexN ∈N such thatP(π[B∩Ω×[0, z]])− P((B∩AN)Bλ,z)≤.

Proof. Sinceµis sigma-additive, for each$∈Ω we have

µ((B∩A_n∩Ω×[0, z])($))%µ((B∩A∩Ω×[0, z])($)) =µ((B∩Ω×[0, z])($)).

Therefore, since λ <1 and sincez ∈R²+ is nite, for each $∈Ω there is an indexN($)such thatµ((B∩A_N($)∩Ω×[0, z])($))> λµ((B∩Ω×[0, z])($)).

Hence, we have (B∩An)Bλ,z %(B∩A)Bλ,z. Since each set(B∩An)Bλ,z is an element of F (see Proposition 1.16), we have P((B∩A_n)B_λ,z) % P((B∩ A)Bλ,z) =P(BBλ,z) =P(π[B∩Ω×[0, z]]).

The statement of the proposition now follows.

Proposition 1.18. Let z ∈ R²₊, A₀ ∈ P, and ₀ > 0. Let B₀ be a measurable subset of A0. Then there is an element K0 of K_δ such that K0 ⊂ A₀∩Ω×[0, z] and

P(π[B₀∩Ω×[0, z]])−P(π[B₀∩K₀∩Ω×[0,z]])≤₀.

Proof. Let A be the collection of sets A ∈ P such that for every >0, 0≤λ <1, andB ⊂Athere is an elementK ofK_δsuch thatK ⊂A∩Ω×[0, z]

and

P(π[B∩Ω×[0, z]])−P((B∩K)B_λ,z)≤.

LetS denote the ring generated by the sets(S, T]×(s, t]andD× {0} × (u, v] such that S, T are stopping times for (G_s), D ∈ G₀, and s, t, u, v ∈R+. Note, we have P = σ(S). It is our goal to show that A is a monotone class containing S. It will follow from the Monotone Class theorem that A = P. Let A∈ S. We may write

A=

n⁰

[

i=1

(Si, Ti]×(ri, si]∪

m⁰

[

j=1

[0_Aj, Uj]×(vj, wj],

whereSi,Ti, andUj are stopping times for(G_s),Aj ∈ G₀, and ri, si, vj, wj are real numbers for every index iand every index j. Without loss in generality, we will consider A =

n

S

i=1

(S_i, T_i]×(r_i, s_i], and we will assume that the sets Aⁱ:= (Si,i]×(ri, si]are pairwise disjoint.

Let > 0, 0 ≤λ <1, and let B be a measurable subset of A. Let (_n) be a sequence of positive numbers decreasing to 0. For each L ∈N and each index i denote by K_L,i the set [S_i +_L, T_i∧L]×[r_i+_L, s_i]∈ K_δ. For each index iwe have K_L,i % Aⁱ asL → ∞. By Proposition 1.17, for each index i there is a number Li ∈N such that

P(π[B∩Aⁱ∩Ω×[0, z]])−P((B∩Aⁱ∩KLi,i)(B∩Aⁱ)_λ,z)≤ 2ⁱ. SetK:=

n

S

i=1

K_Li,i. We haveK∩Ω×[0, z]∈ K_δ,K∩Ω×[0, z]⊂A∩Ω×[0, z], and

P(π[B∩[Ω×[0, z] ])−P((B∩K∩Ω×[0, z])B_λ,z)

=P(π[B∩Ω×[0, z]]\(B∩K)Bλ,z) since

(B∩K∩Ω×[0, z])B_λ,z = (B∩K)B_λ,z⊂B∩Ω×[0, z];

=P π

h B∩

n

[

i=1

Aⁱ∩Ω×[0, z]

i

\ B∩

n

[

i=1

KLi,i

B∩

n

[

i=1

Aⁱ

λ,z

!

=P

n

[

i=1

π[B∩Aⁱ∩Ω×[0, z]]\

n

[

i=1

(B∩K_Li,i)(B∩Aⁱ)_λ,z

!

since the sets Aⁱ are disjoint;

≤P

n

[

i=1

(π[B∩Aⁱ∩Ω×[0, z]]\(B∩K_Li,i)(B∩Aⁱ)_λ,z)

!

≤

n

X

i=1

P π[B∩Aⁱ∩Ω×[0, z]]\(B∩K_Li,i)(B∩Aⁱ)_λ,z

≤

n

X

i=1

2ⁱ ≤. Thus, A∈ A. SinceA∈ S was arbitrary, we haveS ⊂ A.

Next, let (A⁰_n) be an increasing sequence from A. Set A⁰ :=

∞

S

n=1

A⁰_n. We will show that A⁰ ∈ A. Let B⁰ be a measurable subset of A⁰, and let > 0 and 0 ≤ λ < 1. There are numbers λ1, λ2 such that 0 ≤ λ1, λ2 <

1 and λ₁λ₂ = λ. By Proposition 1.17 there is an index N ∈ N such that P(π[B⁰∩Ω×[0, z]])−P((B⁰∩A⁰_N)B_λ,z⁰ )≤ ₂. SinceA⁰_N ∈ A, for the measurable subset C := [((B⁰∩A⁰_N)B_λ⁰

1,z)×R²+]∩(B⁰∩A⁰_N)∩Ω×[0, z]of A⁰_N there is a set K ∈ K_δ such thatK⊂A⁰_N ∩Ω×[0, z]⊂A⁰∩Ω×[0, z]andP(π[C∩Ω×

[0, z]])−P((C∩K)Cλ2,z)≤ ₂. Butπ[C∩Ω×[0, z]] = (B⁰∩A⁰_N)B_λ⁰_1,z. Hence P((B⁰ ∩A⁰_N)B_λ⁰

1,z)−P((C∩K)C_λ2,z) ≤ ₂. Note, we have (B⁰ ∩K)B_λ,z⁰ ⊃ (C∩K)C_λ2,z. Hence

P(π[B⁰∩Ω×[0, z] ])−P((B⁰∩K)B⁰_λ,z)≤P(π[B⁰∩Ω×[0, z]])

−P((C∩K)Cλ2,z) =P(π[B⁰∩Ω×[0, z]])−P((B⁰∩A⁰_N)B_λ⁰_1,z)+

+P((B⁰∩A⁰_N)B⁰_λ1,z)−P((C∩K)C_λ2,z)≤ 2 +

2 =. We conclude that A⁰ ∈ A.

Lastly, let(A⁰⁰_n) be a decreasing sequence fromA. SetA⁰⁰:=

∞

T

n=1

A⁰⁰_n. We will show that A⁰⁰ ∈ A. Let B⁰⁰ be a measurable subset of A⁰⁰, and let > 0 and 0≤λ <1. Let λ⁰∈R+satisfy λ < λ⁰ <1. There is a sequence(λ_n) from [0,1) such that Q^∞

n=1

λ_n = λ⁰. Note, the setB⁰⁰ is a measurable subset of A⁰⁰₁. So, there is a set K1 ∈ K_δ such thatK1 ⊂A⁰⁰₁ ∩Ω×[0, z]and

P(π[B⁰⁰∩Ω×[0, z]])−P((B⁰⁰∩K1)B_λ⁰⁰_1,z)≤ 2.

The setC1:= [((B⁰⁰∩K₁)B_λ⁰⁰_1,z)×R²₊]∩B⁰⁰∩K₁∩Ω×[0, z]is a measurable subset ofA⁰⁰₂. So, there is a setK2∈ K_δsuch thatK2⊂A⁰⁰₂∩Ω×[0, z]andP(π[C1∩Ω×

[0, z]])−P((C₁∩K₂)C_1λ2,z)≤ ₂₂. Note thatπ[C₁∩Ω×[0, z]] = (B⁰⁰∩K₁)B⁰_λ,z. The set C2:= [((C1∩K2)C_1λ2,z)×R²+]∩C1∩K2∩Ω×[0, z]is a measurable subset of A⁰⁰₃. So, there is a set K₃ ∈ K_δ such that K₃ ⊂ A⁰⁰₃ ∩Ω×[0, z] and P(π[C2∩Ω×[0, z]])−P((C2∩K₃)C2λ3,z)≤ ₂3. Note, we haveπ[C2∩Ω×[0, z]] = (C₁∩K₂)C_1λ2,z. Continuing inductively, we obtain a sequence (K_n) from K_δ such that for each index nwe haveKn+1 ⊂A⁰⁰_n+1∩Ω×[0, z]and

P(π[C_n∩Ω×[0, z]])−P((C_n∩K_n+1)C_nλn+1,z)≤ 2ⁿ⁺¹, where

Cn= [((Cn−1∩Kn)Cn−1λn,z)×R²₊]∩Cn−1∩Kn∩Ω×[0, z]

and

π[C_n∩Ω×[0, z] ] = (Cn−1∩K_n)Cn−1λn,z, n= 2,3,4. . . . So, for each index nwe have

P(π[B⁰⁰∩Ω×0, z]])−P((C_n∩K_n+1)C_nλn+1,z) =

=P(π[B⁰⁰∩Ω×[0, z]])−P((B⁰⁰∩K1)B_λ⁰⁰_1,z)+

+P(π[C₁∩Ω×[0, z]])−P((C₁∩K₂)C_1λ2,z) +· · ·+ +P(π[C_n∩Ω×[0, z]])−P((C_n∩K_n+1)C_nλn+1,z)≤

2+

2² +· · ·+ 2ⁿ⁺¹ ≤.