Denote the indicator function on a meausurable set $A$ as $\chi_A$.
On page 11 of the book Regularity of Free Boundaries in Obstacle-type Problems, the author states that:
Consider the problem (1) $$\begin{cases} \Delta u = f\chi_\Omega, & \text{in }B_r(x_0) \\ u = |∇u| = 0, & \text{on }B_r(x_0) \setminus Ω \\ u\ge 0, & \text{in }B_r(x_0) \\ \end{cases}$$ where $\Omega$ is a given open set and $x_0\in\partial\Omega$ is a given point. Then, any solution $u$ to problem (1) is also a solution to problem (2), aka the obstacle problem: $$\begin{cases} \Delta u = f\chi_{\{u>0\}}, & \text{in }B_r(x_0) \\ u\ge 0, & \text{in }B_r(x_0) \\ \end{cases}$$
The author seems to imply that $B_r(x_0)\,\cap\,\Omega =\{u>0\}$. But how to prove it?
Since $u\ge 0$ in problem (1), it seems to be some kind of (strong) minimum principle - $u(x)>\inf_{\partial(B_r(x_0)\,\cap\,\Omega)} u =0$. However, since $u$ is not harmonic ($f$ is not necessarily the zero function), I have no idea how to proceed.
I have tried considering the function $\tilde u(x)=u(x)-\frac{||f||_\infty+\epsilon}{2n}||x||^2$, where $\Delta \tilde u = f\chi_{\Omega}-(||f||_\infty+\epsilon)<0$ and thus $\tilde u$ does not attain any minimum in the interior.
However, this does not lead to any conclusion about $u$ being positive in the interior.