Let $\Omega \subset \mathbb{R}^{3}$ be a $C^{1}$-domain, not necessarily bounded. Consider solutions $\phi : \overline{\Omega} \to \mathbb{R}$, $\phi \in C^{\infty}(\Omega) \cap C^{1}(\overline{\Omega})$, $\nabla\phi \in L^{2}(\Omega)$, to the Neumann problem \begin{align} \Delta \phi (x) &= 0 &&\text{ in } \Omega \\ \nabla_{\! n} \phi(x) &= g(x) &&\text{ in } \partial \Omega \end{align} I want to prove that $\nabla \phi$ is unique in this case.

The standard approach to proving this in bounded domains is to take two solutions to the above problem, say $\phi$ and $\psi$, and investigate the convergent integral \begin{align} I = \int_{\Omega} \lvert \nabla \phi - \nabla \psi \rvert^{2} \, d^{3}x. \end{align} Using the divergence theorem, we find that \begin{align} I = \int_{\partial \Omega} (\phi - \psi) (\nabla_{\! n} \phi - \nabla_{\! n} \psi) \, d^{2}x = 0, \end{align} such that we must have $\nabla \phi = \nabla \psi$ in $\Omega$, and by continuity, in $\overline{\Omega}$.

Is this proof valid when $\Omega$ is unbounded?