By Forman Sinnickson Acton

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**Example text**

Thus [S ≤ T ] ∈ FR . By symmetry [T ≤ S] ∈ FR and so [S = T ] ∈ FR . (g) Set R = S ∧ T and let A ∈ FT and n ≥ 1. Then A ∩ [T ≤ S] ∩ [R = n] = A ∩ [T ≤ S] ∩ [T = n] = A ∩ [T = n] ∩ [n ≤ S] ∈ Fn . Thus A ∩ [T ≤ S] ∈ FR . Intersecting this with the set [S ≤ T ] ∈ FR we obtain A ∩ [T = S] ∈ FR . (h) Set G = n FTn . We have to show that FT = G. According to (d), FTn ⊆ FT , for all n ≥ 1, and consequently G ⊆ FT . To see the reverse inclusion, let A ∈ FT . We wish to show that A ∈ G. According to (c) all the Tk are G-measurable and hence so is the limit T = limk Tk .

On D and X2 ≤ X1 , P -as. on D. Now use (e). (f) −|X| ≤ X ≤ |X and so, using (c) and (d), −EG (|X|) ≤ EG (X) ≤ EG (|X|), that is, EG (X) ≤ EG |X| , P -as. on Ω. (g) Let Z1 , Z2 be conditional expectations of X1 , X2 given G respectively and assume that E(X1 ) + E(X2 ) is deﬁned. Then X1 + X2 is deﬁned P -as. 0). Consequently the conditional expectation EG (X1 + X2 ) is deﬁned. Moreover E(X1 ) > −∞ or E(X2 ) < +∞. Consider the case E(X1 ) > −∞. Then Z1 , Z2 are deﬁned everywhere and G-measurable and E(Z1 ) = E(X1 ) > −∞ and so Z1 > −∞, P -as.

Taking the union over all sets Dm gives the desired inequality on the set D = m Dm = EG (h) > −∞ . (b) follows from (a). 9 Dominated Convergence Theorem. Assume that Xn , X, h ∈ E(P ), |Xn | ≤ h and Xn → X, P -as. Then EG |Xn − X|) → 0, P -as. on the set EG (h) < ∞ . Remark. If E(Xn ) − E(X) is deﬁned, then EG (Xn ) − EG (X) ≤ EG |Xn − X| , for all n ≥ 1, and it follows that EG (Xn ) − EG (X) → 0, that is EG (Xn ) → EG (X), P -as. Proof. (b), 0 ≤ limn EG |Xn − X| ) ≤ EG limn |Xn − X| = 0, P -as.