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In this paper we define the consistent criteria of hypotheses such as the probability of any kind of errors is zero for given criteria. We prove necessary and sufficient conditions for the existence of such criteria.

Abstract. General statistical structures are considered in this paper. It is shown that a general statistical structure can be reduced to some orthogonal or absolutely continuous statistical structure.The problems of statistics of stochastic processes usually deal with a family of distributions of an observable trajectory of a stochastic process. It is natural to study the set of all possible trajectories and to associate an arbitrary distribution of the stochastic process with a probability measure in the space of all trajectories. It is important for such an association that all possible distributions have a common support. Note that a common support exists in all natural cases.If μ is a σ-finite measure, then one can construct a probability measure μ such that μ is equivalent to μ. Indeed, let (E, S) be a measurable space and let μ be a σ-finite measure in S. Then E = ∞ k=1 A k , where A k are S-measurable and disjoint sets and μ(A k ) < ∞ for all k. Now we choose a sequence of nonnegative numbers ρ k such thatThen μ ∼ μ and μ(E) = 1. This means that one can transfer the case of σ-finite measures to the case of equivalent probability measures in finite dimensional spaces. This is not always possible in infinite dimensional spaces.An exceptional role of absolutely continuous and orthogonal structures is well known in statistics (see, for example, [1], [2]). Below we consider some important cases where the general setting can be reduced to similar problems of estimation of parameters for other absolutely continuous or orthogonal statistical structures. First we discuss some examples. Example 1. Let Θ consist of two points θ 1 and θ 2 and let two probability measures μ θ 1 and μ θ 2 be defined in a measurable space (E, S). The set E can be represented as the union of three disjoint subsets A 1 , A 2 , and A 3 , that is, E = A 1 + A 2 + A 3 , and moreover, μ θ 1 ∼ μ θ 2 in the set A 2 , μ θ 2 (A 1 ) = 0, and μ θ 1 (A 3 ) = 0.2010 Mathematics Subject Classification. Primary 62H05, 62H12.