> A set B ⊆ R is Bernstein, if it intersects all perfect sets in R, but does not contain any of them. A Bernstein set is a classical example of a nonmeasurable set; a Bernstein set cannot be Lebegsue measurable and cannot have the Baire property. Moreover, if A is an algebra of subsets of R, having the property that every set which is in A but not hereditary in A, contains a perfect set, then no Bernstein set can be a member of A. Most natural algebras of subsets of R have this property (see [5]), therefore Bernstein sets are in a sense universal examples of nonmeasurable sets.