Let $G$ be a non-elementary, finitely generated Kleinian group, $\Lambda(G)$ its limit set and $\Omega(G) = \overline {\mathbb C} \backslash \Lambda(G)$ its set of discontinuity. Let $\delta(G)$ be the critical exponent for the Poincarè series and let $\Lambda_c$ be the conical limit set of $G$. Suppose $\Omega_0$ is a simply connected component of $\Omega(G)$. We prove that

1. $\delta(G) = \dim(\Lambda_c)$
2. A simply connected component $\Omega$ is either a disk or $\dim(\partial \Omega)>1$
3. $ \Lambda(G)$ is either totally disconnected, a circle or has dimension $>1$
4. $G$ is geometrically infinite iff $\dim(\Lambda)=2$
5. If $G_n \to G$ algebraically then $\dim(\Lambda)\leq \liminf \dim(\Lambda_n)$
6. The Minkowski dimension of $\Lambda$ equals the Hausdorff dimension
7. If $\text{area}(\Lambda)=0$ then $\delta(G) =\dim(\Lambda(G))$

The proof also shows that $\dim(\Lambda(G)) >1$ iff the conical limit set has dimension $>1$ iff the Poincarè exponent of the group is $>1$. Furthermore, a simply connected component of $\Omega(G)$ either is a disk or has non-differentiable boundary in the the sense that the (inner) tangent points of $\partial \Omega$ have zero $1$-dimensional measure. almost every point (with respect to harmonic measure) is a twist point.