Ahlfors-Type Theorem for Hausdorff Measures

Suppose that Δ ⊂ \({\mathbb{C}}\) $\mathbb{C}$ 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| ≤ \(\frac{1}{\varepsilon }\) $\frac{1}{\epsilon }$ } and T_R,ε = f(l_R,ε ∩ E), T_R,ε ⊂ D. Put \({G}_{\varepsilon }\left(R\right)=\left\{\begin{array}{l}0\ \ \ if\, {\Lambda }_{\alpha ,\beta }\left({T}_{R,\varepsilon }\right)=0\ or\ {\Lambda }_{\alpha ,\beta }\left({T}_{R,\varepsilon }\right)=\infty ,\\ \frac{{\Lambda }_{\alpha ,\beta }^{\frac{1+\alpha }{\alpha }}\left(E\cap {l}_{R,\varepsilon }\right)}{{\Lambda }_{\alpha ,\beta }^{\frac{1}{\alpha }}\left({T}_{R,\varepsilon }\right)}\ \ if\, 0<{\Lambda }_{\alpha ,\beta }\left({T}_{R,\varepsilon }\right)<\infty .\end{array}\right.\) $$G_{\epsilon }\left(R\right)=\left\{\begin{array}{c}0if{\mathrm{\Lambda }}_{\alpha ,\beta }\left(T_{R,\epsilon }\right)=0or{\mathrm{\Lambda }}_{\alpha ,\beta }\left(T_{R,\epsilon }\right)=\infty ,\hfill \\ \frac{{\mathrm{\Lambda }}_{\alpha ,\beta }^{\frac{1+\alpha }{\alpha }}\left(E,\cap ,l_{R,\epsilon }\right)}{{\mathrm{\Lambda }}_{\alpha ,\beta }^{\frac{1}{\alpha }}\left(T_{R,\epsilon }\right)}if0<{\mathrm{\Lambda }}_{\alpha ,\beta }\left(T_{R,\epsilon }\right)<\infty .\hfill \end{array}\right)$$

Define the upper Lebesgue integral \(\underset{0}{\overset{{\infty }_{*}}{\int }}g\) $\underset{0}{\stackrel{\infty _{\ast }^{}}{\int }}g$ dm for a function g, g(x)≥0, x > 0 in the following way: let U(y) \(\stackrel{\scriptscriptstyle\mathrm{def}}{=}\) $\stackrel{\mathrm{def}}{=}$ {x > 0 : g(x) > y}, H(y) = m^*U(y). Then put \(\underset{0}{\overset{{\infty }_{*}}{\int }}g\) $\underset{0}{\stackrel{\infty _{\ast }^{}}{\int }}g$ dm \(\stackrel{\scriptscriptstyle\mathrm{def}}{=}\underset{0}{\overset{\infty }{\int }}H\left(y\right)dy\) $\stackrel{\mathrm{def}}{=}\underset{0}{\stackrel{\infty }{\int }}H\left(y\right)dy$ .

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 \(\underset{0}{\overset{{\infty }_{*}}{\int }}\underset{\varepsilon \to +0}{\underline{\mathrm{lim}}}{G}_{\varepsilon }\left(R\right)dR\le {2\Lambda }_{1+\alpha ,\beta }\left(E\right).\) $$\underset{0}{\stackrel{\infty _{\ast }^{}}{\int }}\underset{\epsilon \to +0}{\underline{\lim }}{G}_{\epsilon }\left(R\right)dR\le {2\mathrm{\Lambda }}_{1+\alpha ,\beta }\left(E\right).$$