I know that harmonic measure $\omega$ in complex plane $\mathbb{C}$ is absolutely continuous with Hausdorff measure $\mathcal{H_{h_k}}$ $(\omega << \mathcal{H_{h_k}})$, where $$ h_k(t) = t e^{k\sqrt{\log\frac{1}{t}\log\log\log \frac{1}{t}}} $$ with some constant $k$. Why it follows from this that the Hausdorff dimension of the support $supp\,(\omega)$ is at least $1$?
This is a result of Makarov from 1985 and I have tried to find an explanation to this claim about Hausdorff dimension but haven't come up with anything. Mayby this is a simple thing but I just don't understand it...
How do different $\mathcal H_h$ measures compare? The larger $h$ is near zero, the larger is the measure. So, $\mathcal H_h\ll \mathcal H_g$ whenever $h(t)=O(g(t))$ as $t\to 0$.
For every $d\in (0,1)$, $h_k(t)=O(t^d)$ as $t \to 0$. Just take the logarithms of both sides to see this. Therefore, $\mathcal H_{h_k} \ll \mathcal H^d$ for every $d\in (0,1)$.
If $E$ is a set of Hausdorff dimension $\dim E<1$, then pick $d$ such that $\dim E<d<1$ and note that $\mathcal H^d(E)=0$. By the above, the harmonic measure of $E$ is zero.
Conclusion: any set of Hausdorff dimension less than $1$ has zero harmonic measure.