30.3. FILTRATIONS 819

because each set in the union is in B ([0, t])×Ft . If E is progressively measurable, is EC?

EC ∩ ([0, t]×Ω)∪

∈B([0,t])×Ft︷ ︸︸ ︷(E ∩ ([0, t]×Ω)) =

∈B([0,t])×Ft︷ ︸︸ ︷[0, t]×Ω

and so EC ∩ ([0, t]×Ω) ∈B ([0, t])×Ft . Thus the progressively measurable sets are a σ

algebra.Another observation of interest is in the following lemma.

Lemma 30.3.4 Suppose Q is in B ([0,a])×Fr. Then if b≥ a and t ≥ r, then Q is alsoin B ([0,b])×Ft .

Proof: Consider a measurable rectangle A×B where A ∈B ([0,a]) and B ∈Fr. Is ittrue that A×B∈B ([0,b])×Ft? This reduces to the question whether A∈B ([0,b]). If A isan interval, it is clear that A∈B ([0,b]). Consider the π system of intervals and let G denotethose Borel sets A ∈B ([0,a]) such that A ∈B ([0,b]). If A ∈ G , then [0,b]\A ∈B ([0,b])by assumption (the difference of Borel sets is surely Borel). However, this set equals

([0,a]\A)∪ (a,b]

and so[0,b] = ([0,a]\A)∪ (a,b]∪A

The set on the left is in B ([0,b]) and the sets on the right are disjoint and two of themare also in B ([0,b]). Therefore, the third, ([0,a]\A) is in B ([0,b]). It is obvious that Gis closed with respect to countable disjoint unions. Therefore, by Lemma 9.3.2, Dynkin’slemma, G ⊇ σ (Intervals) = B ([0,a]).

Therefore, such a measurable rectangle A×B where A ∈B ([0,a]) and B ∈Fr is inB ([0,b])×Ft and in fact it is a measurable rectangle in B ([0,b])×Ft . Now let Kdenote all these measurable rectangles A×B where A ∈B ([0,a]) and B ∈Fr. Let G (newG ) denote those sets Q of B ([0,a])×Fr which are in B ([0,b])×Ft . Then if Q ∈ G ,

Q∪ ([0,a]×Ω\Q)∪ (a,b]×Ω = [a,b]×Ω

Then the sets are disjoint and all but [0,a]×Ω\Q are in B ([0,b])×Ft . Therefore, this oneis also in B ([0,b])×Ft . If Qi ∈ G and the Qi are disjoint, then ∪iQi is also in B ([0,b])×Ft and so G is closed with respect to countable disjoint unions and complements. HenceG ⊇ σ (K ) = B ([0,a])×Fr which shows

B ([0,a])×Fr ⊆B ([0,b])×Ft ■

A significant observation is the following which states that the integral of a progres-sively measurable function is progressively measurable.

Proposition 30.3.5 Suppose X : [0,T ]×Ω→ E where E is a separable Banach space.Also suppose that X (·,ω) ∈ L1 ([0,T ] ,E) for each ω . Here Ft is a filtration and withrespect to this filtration, X is progressively measurable. Then

(t,ω)→∫ t

0X (s,ω)ds

is also progressively measurable.