Abstract
Our purpose is to answer the question of Leader [1, p. 2341, by producing an example of a sigma algebra a of subsets of a set X and a universally a-integrable function f (i.e., ffdu exists for every countably additive set function u on a) which is not measurable with respect to a. We shall show that the sigma algebra a of Borel subsets of the interval [0, 1] admits nonmeasurable universally a-integrable functions (R. E. Zink suggested that this Borel algebra might yield an example). Lusin has shown [2] that, subject to the continuum hypothesis, there exists an uncountable subset V of X = [0, 1] such that each perfect nowhere dense subset G of X contains at most an enumerable set of points of V. Let H be a subset of V that is not a Borel set. We shall now show that the characteristic function x(H) of H is universally sigma integrable. To this end suppose u is a countably additive set function on a-. Then [5] u = c +j where c is continuous and j= Ijk where jk is a two valued jump function, the range of k is at most enumerable, and the variation of j is the sum of the variations of the jk'S; moreover, since the variation of u is the variation of j plus the variation c of c, each of j, c, and c is countably additive on a. Corresponding to each jk there is [4, Theorem 27.1] a point Xk such that jk(E) =jk(En [Xk ]) for every Borel subset E of X. Hence fX(H)dj= Ejk(HCn[Xk]) and it suffices to show that there exists a Borel subset B of X such that HCB and c(B) = 0 (then fX(H)dc= 0). There exists [3 ] a set K = U i Fi, F1 closed for i > 1, of the first category in X and a set L such that c(L) = 0 and KUL = X. For each positive integer i there exists a perfect set Gi and an at most enumerable set Hi such that Fi = Gi U Hi. Thus H n K = HC(Ui Fi) = Hf (Ui (GiUkHi)) = Ui (Hrl (GiU-Hi)) CUi((HnGi) UHi) is at most enumerable since every HrlGi (Fi is a closed subset of X and K is of the first category in X; hence Fi is nowhere dense in X which implies that G, is a perfect nowhere dense subset of X) and every H, is at most enumerable. Since e is countably additive on o and c vanishes on one point sets, c vanishes on countable sets. Hence H=Hr)(KUL) C(HnK)UL and c(HnK)UL) <?(HnK) +Z(L) =0.
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