Analytic function outside a bounded domain

Suppose $$D\subseteq\mathbb C$$ is a bounded domain, with $$\partial D$$ smooth. Then suppose $$g:\mathbb C\backslash\overline D\to\mathbb C$$ is analytic and extends continuously to $$\partial D$$. If $$\lim_{z\to\infty}g(z)$$ is finite, show that $$g(u)=\lim_{z\to\infty}(g(z))-\frac1{2\pi i}\int_{\partial D}\frac{g(z)dz}{z-u}$$ for all $$u\in\mathbb C\backslash\overline D$$, assuming that $$\partial D$$ is oriented as the boundary of $$D$$.

This problem is confusing to me, because the integral term looks like the result of the Cauchy Integral Formula. But we are considering points outside $$\overline D$$. Would anyone be able to give some intuition for how to solve this problem?

Let $$\Omega$$ be a bounded domain with smooth boundary. Then for any $$h \in \mathcal{O}(\Omega)$$ which extends continuously to $$\partial{\Omega}$$, we have $$\forall u \in \Omega,\ h(u) = \frac{1}{2\pi i} \int_{\partial{\Omega}} \frac{h(z)}{z-u} dz.$$
We can assume that $$D$$ is contained in the unit disk $$\Delta$$. Set $$\Omega$$ the image of $$\mathbb{C}\backslash \overline{D}$$ by $$z \mapsto \frac{1}{z}$$ and for all $$z \in \Omega$$, set $$h(z) = g(\frac{1}{z})$$. Notice that $$\Omega$$ is a bounded domain since $$D \subset \Delta$$. Alas, $$h$$ is not holomorphic on $$\Omega$$ but it is holomorphic on $$\Omega\backslash\{0\}$$. Since the boundary of this latter domain is not smooth, we replace it by $$\Omega\backslash \Delta_{\varepsilon}$$ where $$\Delta_{\varepsilon}$$ stands for the disc centered at the origin with radius $$\varepsilon$$, which is chosen small enough so that $$\Delta_{\varepsilon} \subset \Omega$$. Now you can apply the Cauchy integral formula above and let $$\varepsilon$$ go to zero to conclude.