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By Bieri R.

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Extra info for A 1.5-dimensional version of Hopfs Theorem on the number of ends of a group

Example text

For let J1 = J11 - J12 with (positive) Radon measures J11' J12 being concentrated on disjoint Borel sets; then PROOF. At first we remark that IJ fdJ1 1 ~ flfldJ11 + f'f'dJ12 ~ Ilfll·IIJ111 showing also II TJlII ~ 11J11l· Let now T E (C(X))' be given. f, g, h E C+(X) such that h ~ f T+(f) + g. h(x)f(x) h'(x):= { ~(x) + g(x) h(x)g(x) ~(x) h"(x):= + g(x) { then h', h" E C+(X), h' ~ 00, f E and for f, g + C +(X). E C +(X) we have T+(g). f(x) = g(x) = 0, f, h" ~ g and h' T(h) ~ T+(f) + h" = h, implying + T+(g) so that finally T+ is additive on C +(X).

The" only if" part is obvious, so suppose that (Ili IG(l)i E I converges to (Ill G(l) for each tYw E D. Letf E CC(X) have the compact support C, and choose tYw 1 , . . , tYw n ED such that C ~ G(ll U G(l2 U ... u G(ln. 17 there exist compact sets C k ~ GCXk k = 1, ... , n such that C = C 1 U ... U Cn. By Urysohn's lemma there exist functions qJk E CC(X) such that 1Ck ~ qJk ~ 1Gock 54 2. Radon Measures and Integral Representations and SUPP(qJk) ~ G(lk' k = 1, ... , n. k(x) nqJk(X) , XE C, X E X\C, L qJj(x) := j= 1 0, belong to CC(X), sUPP(h) ~ G(lk andf = Lk= 1 fk' so we have for and the assertion follows.

The reverse inequality will follow immediately if we can show that A is additive on disjoint compact sets. Therefore let K, L E % with K n L = 0 be given. One direction, namely ,1(K u L) ~ ,1(K) + ,1(L) is obvious, so it remains to be shown that for arbitrary ,1(K) + ,1(L) ~ ,1(K u L) B > 0 + B. By definition there is an open set WEd containing K u L such that A(W) - ,1(K u L) < B. e. we have K £G, L £H, Gn H = 0. Hence ,1(K) + ,1(L) ~ A(G n W) + A(H n W) = A«G u H) n W) ~ thus finishing the proof.

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