The extension of measurable functions

Say a measurable space (Y, O) has the extensiotn property (resp. the extenisiont property in the restricted senise) if for every measurable space (X.S) and every subset A of X (resp. subset A of X with X\A singleton), each function f: A Y measurable for S(A) = (B r A: B E S) may be extended to a measurable function g: XY. A countably generated and separated ( Y,66) has the extension property if and only if it is a standard space, i.e. it is isomorphic to a Borel subset of the real line. The discrete space ( Y,2> ) has the extension property in the restricted sense if and only if the cardinality of Y is not two-valued measurable. We shall use the following notation and terminology: if (X,5s) is a measurable space, and A c X, let S(A) = (B n A: B E 5). A measurable space (X,6) is separable if G'J is countably generated and separated (contains singletons); (X,6'I) is standard if there is a complete separable (i.e. Polish) metric on X for which 'i becomes the associated Borel a-algebra; equivalently, (X,(31) is isomorphic with some Borel subset of the real numbers. The extension problem. Let there be given (1) measurable spaces ( X, 5) and (Y, 13), (2) a subset A of X, and (3) a function f: A -Y, measurable from S(A) to 0i. We ask whether there is an S-measurable g: X -Y extending f (i.e. g(a) = f (a) for a e A). Say that a measurable space (Y, q) has the extension property if such an extension exists for each (X,$-), A and f as above. If an extension exists whenever X \ A is a singleton set, then (Y, 'M ) has the extension property in the restricted sense. What seem to have been the first positive results for this problem were obtained by von Alexits [1] and Sierpiniski [8] in the case of real functions on subsets of Polish spaces; here, particular attention was paid to the Baire class of f and g. It is a small step from their results to the following characterization of separable spaces with the extension property: THEOREM 1. Among separable spaces, those (Y, 635) having the extension property are precisely the standard spaces. PROOF. To prove the extension property for standard spaces, obtain an extension for functions assuming finitely many values and pass to the limit (cf. exercise 13, p. 260 of Cohn [2]); alternatively, derive the result from 2.3 Lemma 1 of Lehmann [4]. Received by the editors November 24, 1981. 1980 Mathematics Subject Classification. Primary 28A20; Secondary 03E15, 03E55. Ke' words and phrases. Measurable space, extension property, separable space, standard space, measurable cardinal, two-valued measurable cardinal, separability character. T1983 American Mathematical Society 0002-9939/82/0000-0622/$02.00