By Wehrfritz B.A.F.

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1. 15. Proposition. Let X and Y be Hausdorffspaces, let /l be a Radon measure on X and suppose that f: X --. Y is continuous. ~f /l(f - 1 (K» < 00 for each compact set K £; Y then III is a Radon measure. e. f - I(K) is compact for each compact set K £; Y. PROOF. 1 for /lI, since condition (i) is part of the assumptions. Let B E 8l(Y) be given. For any a < /lI(B) = /l(f - l(B» there exists a compact set K £; f -1(B) such that a < /l(K). Now C := f(K) is a compact subset of Band /lI(C) = /l(j' -1(C» ~ /l(K) > a D which shows condition (ii).

4): If Q is a nonempty set and ~ is a family of subsets of Q closed under finite intersections, then the smallest Dynkin class containing ~ equals the a-algebra generated by ~. f: Z --. [0, 00]. The extension to the case where Il and v are a-finite is completely straightforward and therefore omitted. 13. It is, of course, a natural question to ask if equality in (12) holds for more general functions than just nonnegative lower semicontinuous ones. The following example shows that one cannot, in general, hope for too much.

PROOF. We define A: $'(X) --. [0, w[ by A(K) := inf {T(f) 11 K ~ f E re} and we shall show that A is a Radon content. e. A(K u L) ~ A(K) + A(L) for all K, L E $'(X). f 1\ g = 0. e. A is additive on disjoint compact sets, and this implies for compact sets C l £; C 2 that sup{A(K)IK £; C 2 \C l , K E $'(X)} ~ A(C 2 ) - A(C l ). f E re such that leI ~ f and T(f) ~ A(C l ) + e. We also fix some number a E JO, 1[ and define K ex := C 2 n {f ~ a}. Certainly K ex is a compact subset of C 2 \C l , and if 1K(J( ~ g E rc then implying A(C 2 ) ~ A(K ex ) I T(f) a +- ~ A(K ex ) 1 [A(C l ) a +- + eJ ° Taking now on the right-hand side the limit for a --.