Ahlfors-type theorem for Hausdorff measures

Suppose that $\Delta\subset\mathbb{C}$ is a domain, a function $f$ is analytic in $\Delta$, $D=f(\Delta)$ is viewed as a Riemann surface. We put $l_{R}=\{z\in\Delta: |f(z)|=R\}$. Let $E\subset\Delta$ be a closed set. Put $h_{\alpha,\beta}(r)=r^{\alpha}|\ln{r}|^{\beta},$ $0<\alpha<1,$ $0<\beta<1$. Let $\Lambda_{\alpha,\beta}(\cdot)$, $\Lambda_{\alpha+1,\beta}(\cdot)$ be the Hausdorff measures with respect to the functions $h_{\alpha,\beta}$, $h_{\alpha+1,\beta}$. Assume that $\Lambda_{\alpha+1,\beta}(E)<\infty$. We introduce the sets $l_{R,\varepsilon}=\{z\in l_{R}: \mathrm{dist} (z,\partial\Delta)\geq\varepsilon, |z|\leq\frac{1}{\varepsilon}\}$ and $T_{R,\varepsilon}=f(l_{R,\varepsilon}\cap E)$, $T_{R,\varepsilon}\subset D$. Put
$$ G_{\varepsilon}(R)=\begin{cases} 0& \text{ if } \Lambda_{\alpha,\beta}(T_{R,\varepsilon})=0 \text{ or } \Lambda_{\alpha,\beta}(T_{R,\varepsilon})=\infty, \\ \frac{\Lambda_{\alpha,\beta}^{\frac{1+\alpha}{\alpha}}(E\cap l_{R,\varepsilon})}{\Lambda_{\alpha,\beta}^{\frac{1}{\alpha}}(T_{R,\varepsilon})}& \text{ if } 0<\Lambda_{\alpha,\beta}(T_{R,\varepsilon})<\infty.\end{cases} $$
We define the upper Lebesgue integral $\underset{0 }{\overset{\infty}{\int^{\ast}}}g \text{d}m$ for a function $g$, ${g(x) \geq 0}$, $x>0$ in the following way: let $U(y)\overset{\text{def}}{=}\{x>0: g(x)>y\},$ $H(y)=m^{*}U(y)$. Then we put $\underset{0 }{\overset{\infty}{\int^{\ast}}}g \text{d}m \overset{\text{def}}{=}\int\limits_{0}^{\infty}H(y)\text{d}y.$
We prove the following result.
Theorem. The condition $\Lambda_{\alpha,\beta}(T_{R,\varepsilon})<\infty$ is fulfilled for almost all $R$ with respect to the $1$-Lebesgue measure and
$$ \underset{0 }{\overset{\infty}{\int^{\ast}}}\underset{\varepsilon\to+0}{\underline\lim}G_{\varepsilon}(R)\text{d}R\leq2\Lambda_{1+\alpha,\beta}(E). $$