Ahlfors-Type Theorem for Hausdorff Measures

Suppose that Δ ⊂ C is a domain, f is an analytic function in Δ, D = f (Δ) is considered as a Riemann surface. Put l R = { z ∈ Δ : | f ( z )| = R }. Let E ⊂ Δ be a closed set. Put h α , β ( r ) = r α | ln r | β , 0 < α < 1, 0 < β < 1. Let Λ α , β (·), Λ α +1, β (·) be the Hausdorff measures with respect to the functions h α , β , h α +1, β . Assume that Λ α +1, β ( E ) < ∞. We introduce the sets l R , ε = { z ∈ l R : dist( z , ∂ Δ) ≥ ε , | z | ≤ 1 ε } and T R , ε = f ( l R , ε ∩ E ), T R , ε ⊂ D . Put G ε R = 0 i f Λ α , β T R , ε = 0 o r Λ α , β T R , ε = ∞ , Λ α , β 1 + α α E ∩ l R , ε Λ α , β 1 α T R , ε i f 0 < Λ α , β T R , ε < ∞ . Define the upper Lebesgue integral ∫ ∞ ∗ 0 g dm for a function g , g ( x )≥0, x > 0 in the following way: let U ( y ) = def { x > 0 : g ( x ) > y }, H ( y ) = m * U ( y ). Then put ∫ ∞ ∗ 0 g dm = def ∫ ∞ 0 H y d y . We prove the following result. Theorem . The condition Λ α , β ( T R , ε ) < ∞ is fulfilled for almost all R with respect to the 1-Lebesgue measure and ∫ ∞ ∗ 0 lim ̲ ε → + 0 G ε R d R ≤ 2 Λ 1 + α , β E .