Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain. If $u\in C^2(\Omega)\cap C_0(\overline{\Omega})$ is a superharmonic function ($-\Delta u\ge 0)$ then, Hopf Boundary Lemma does implies that $$\frac{\partial u}{\partial \eta}(x)<0,$$
for every $x$ on the boundary where the normal derivative $\frac{\partial u}{\partial \eta}$ is defined.
Now assume that $u\in W_0^{1,2}(\Omega)$ is a superharmonic function ($-\Delta u \ge 0$ in the distributional sense) and $\|\Delta u\|_{\mathcal{M}(\Omega)}<\infty$ (so $\Delta u$ is a bounded Radon measure). There is $f\in L^1(\partial \Omega)$ (which depends on $u$) such that $$\int_\Omega \varphi \Delta u=-\int_\Omega \nabla \varphi\nabla u+\int_{\partial\Omega}\varphi f,\ \forall\ \varphi\in C^\infty(\overline{\Omega})\tag{1}$$
We will denote this $f$ by $\frac{\partial u}{\partial\eta}$ (note that when we have enough regularity, this is true in the classical sense). My question is: can we conclude that $$\frac{\partial u}{\partial \eta}<0,\ a.e.\ x\in \partial \Omega$$
Remark 1: I don't know if in this setting $f\in L^2(\partial\Omega)$. If anyone know it, please point me out.