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A counterexample is given to a conjecture of Val'skii that if K is a compact plane set with interior U and the continuous function f on K satisfies fj UER(U) and fj bKER(bK) then f ER(K). The conjecture is shown to be true when U is a disc. Let K be a compact subset of the complex plane C. We denote by C(K) the algebra of all continuous complex-valued functions on K, by R(K) the subalgebra of all uniform limits on K of rational functions with poles outside K, and by A (K) the algebra of all functions in C(K) which are analytic on the interior of K. In [2] Val'skii conjectured the following: Suppose U is the interior of K, and f EC(K) satisfies fj _U R(U) and fj bKER(bK). (U or cl(U) denote the closure of U, bK the boundary of K.) Then f ER(K). He also mentioned the special case when U is the open unit disc A. In the present paper a counterexample is given to the general conjecture but it is shown to be true in the case of the disc. THEOREM 1. There exists a compact set K with interior U and feC(K) suchthatf| (ER(U),f |bKfER(bK), butfEER(K). PROOF. Let P be an arc with endpoints -1 and 1 lying in A, with positive continuous analytic capacity [1, p. 203]. A\P has two components U and V say. Let {Ai } be a sequence of disjoint open discs in V such that V\U> 1 Aj is nowhere dense, and the sum of the radii of the discs is finite. Let K =-\U1 Ai. Then the interior of K is U and bK=bAkUJU cl(U, 1 bAj). Since a (P) >0, there exists an f continuous on C and analytic outside P such that f'(oc) s0. Let A be the measure on bK which coincides with dz on bA, with -dz on U =1 bAj and is zero on the rest of bK. Then f gdO = 0 for every rational function with poles outside K and hence f gdp = 0 for all geR(K). Since fbKfdc4t= fbAfdz=27ri.f'(oo) L0, f |KEER(K). Since U has connected complement, R(U) =A(U) and so f J UeR(U). Likewise C\(TVUb U) has just two components, so R(VkJb U) Received by the editors October 23, 1970. AMS 1969 subject classifications. Primary 3070.

Rational approximation on the union of sets