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By Michael Rockner

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19 below). 10. Proposition. 6) is linear. 8). e. STJ(€) is represented by the harmonic function (iii) Let cp € P(U C ) £ € 8(U) . Then, if x -> y (£) on U . e. (iv) iyO Let bourhood C € fi(U) and V of P(V) . Then 3U with n € fl(U) and = C n € P1 V (v) with UC . be such that there exists an open neigh- . i on = (p € P(U U V) . e. = J 1^(5^) cp dX d for every such that (p 6 P(U U V) . £ € Pf V of 3U ,. (U)) . ' be such that £ € fl(U) and (vii) Let Then U c V £t is representd by a harmonic function.

I. a.. u + Z bL. u + c u ifj iJ iJ i with sufficiently smooth coefficients (cf. [7]). Then the harmonic measures u y , x € U, of (P,H) above may always be replaced by *m p , x € U . 4. THE PWB-SOLUTION FOR DISTRIBUTIONS GIVEN AN ARBITRARY OPEN SET In this section let exhaustion of U U be a fixed open subset of by compact sets. 1. Remark. 3]). Hence, if M (M £i, C (M_,|| l|_) E E Let for ,|| II ) to fc. 2 (L (P),ll II- 0 ) V open, with p (£) , x € V, £ € V* , reasons we set for r, Z 3V Define in C (V) (cf.

Let P under P on T . % 1 : P -* V by n T <£) := £ + n , C € Pf n 6 P' F € F) . 8 . and P a probability measure on be a bijective map such that T and T are both 8/8-measurable and F/F-measurable. Then T(F P ) - F T(P > . Proof. By symmetry it suffices to prove: T(F P ) c F T ( P ) . Clearly for every probability measure P1 on (P',8) we have that p» F = (B € 8 : there exists B. € F such that B ^ B U BNBj € W p f } . Let B € FP and B} € F be such that B ^ B U B^Bj € M . Then TCBj^TCB) U T(B)^T(Bj) = T(B,^B U B^Bj) € W T ( p ) .