62.3. FILTRATIONS 2063

Proof: Suppose Q ∈ [0,T ]×Ω is progressively measurable. This means for each t,

Q∩ [0, t]×Ω ∈B ([0, t])×Ft

What about(s,ω) ∈ [0, t]×Ω, (s,ω)→

∫ s

0XQdr?

Is that function on the right B ([0, t])×Ft measurable? We know that Q∩ [0,s]×Ω isB ([0,s])×Fs measurable and hence B ([0, t])×Ft measurable. When you integrate aproduct measurable function, you do get one which is product measurable. Therefore, thisfunction must be B ([0, t])×Ft measurable. This shows that (t,ω)→

∫ t0 XQ (s,ω)ds is

progressively measurable. Here is a claim which was just used.Claim: If Q is B ([0, t])×Ft measurable, then (s,ω)→

∫ s0 XQdr is also B ([0, t])×Ft

measurable.Proof of claim: First consider A×B where A ∈B ([0, t]) and B ∈Ft . Then∫ s

0XA×Bdr =

∫ s

0XAXBdr = XB (ω)

∫ s

0XA (s)dr

This is clearly B ([0, t])×Ft measurable. It is the product of a continuous function of swith the indicator function of a set in Ft . Now let

G ≡{

Q ∈B ([0, t])×Ft : (s,ω)→∫ s

0XQ (r,ω)dr is B ([0, t])×Ft measurable

}Then it was just shown that G contains the measurable rectangles. It is also clear thatG is closed with respect to countable disjoint unions and complements. Therefore, G ⊇σ (Kt) = B ([0, t])×Ft where Kt denotes the measurable rectangles A×B where B ∈Ftand A ∈B ([0, t]) = B ([0,T ])∩ [0, t]. This proves the claim.

Thus if Q is progressively measurable, it follows that (s,ω)→∫ s

0 XQ (r,ω)dr≡ f (s,ω)is progressively measurable because for (s,ω) ∈ [0, t]×Ω,(s,ω)→ f (s,ω) is B ([0, t])×Ft measurable. This is what was to be proved in this special case.

Now consider the conclusion of the proposition. By considering the positive and neg-ative parts of φ (X) for φ ∈ E ′, and using Pettis theorem, it suffices to consider the casewhere X ≥ 0. Then there exists an increasing sequence of progressively measurable simplefunctions {Xn} converging pointwise to X . From what was just shown,

(t,ω)→∫ t

0Xnds

is progressively measurable. Hence, by the monotone convergence theorem, (t,ω) →∫ t0 Xds is also progressively measurable.

What else can you do to something which is progressively measurable and obtain some-thing which is progressively measurable? It turns out that shifts in time can preserve pro-gressive measurability. Let Ft be a filtration on [0,T ] and extend the filtration to be equalto F0 and FT for t < 0 and t > T , respectively. Recall the following definition of progres-sively measurable sets.

62.3. FILTRATIONS 2063Proof: Suppose Q € [0,7] x Q is progressively measurable. This means for each f,QN(0,t1] x QE B([0,t]) x FWhat about 5(s,@) € [0,t] x Q, (s,@) + | Xodr?0Is that function on the right 4([0,t]) x ¥, measurable? We know that QM [0,5] x Q is&([0,5]) x Fs measurable and hence #([0,t]) x A; measurable. When you integrate aproduct measurable function, you do get one which is product measurable. Therefore, thisfunction must be #([0,t]) x F, measurable. This shows that (t,@) > {} 2o(s,@)ds isprogressively measurable. Here is a claim which was just used.Claim: If Q is F ((0,r]) x Y; measurable, then (s,@) > fj 2odr is also B (([0,t]) x F,measurable.Proof of claim: First consider A x B where A € #((0,t]) and B € F;. Then[ Gara [2% 2dr = 2s (0) [° Ba (shar0 0 0This is clearly #([0,t]) x A, measurable. It is the product of a continuous function of swith the indicator function of a set in .¥;. Now letG= {0 € B((0,t]) x F; : (s,@) > [ Xo (r,@)dr is B((0,t]) x F; measurable}Then it was just shown that Y contains the measurable rectangles. It is also clear that@G is closed with respect to countable disjoint unions and complements. Therefore, Y D0 (.4;) = B((0,t]) x FA, where .% denotes the measurable rectangles A x B where B € .F;and A € &((0,¢]) = A([0,T]) 1 [0,t]. This proves the claim.Thus if Q is progressively measurable, it follows that (s,@) > Jy Zo (r,@) dr= f (s,@)is progressively measurable because for (s,@) € [0,t] x Q,(s,@) > f (s,@) is B([0,t]) xF#, measurable. This is what was to be proved in this special case.Now consider the conclusion of the proposition. By considering the positive and neg-ative parts of @ (X) for @ € E’, and using Pettis theorem, it suffices to consider the casewhere X > 0. Then there exists an increasing sequence of progressively measurable simplefunctions {X,,} converging pointwise to X. From what was just shown,t(.0) > | X,dsis progressively measurable. Hence, by the monotone convergence theorem, (t,@) >fo Xds is also progressively measurable. §JWhat else can you do to something which is progressively measurable and obtain some-thing which is progressively measurable? It turns out that shifts in time can preserve pro-gressive measurability. Let .¥; be a filtration on [0,7] and extend the filtration to be equalto Fo and “7 fort <0 andt > T, respectively. Recall the following definition of progres-sively measurable sets.