The obstacle reaction is supported on the contact set under measure regularity

($O$ is open and the two-sided test vanishes there) Because $u-\psi$ is continuous, $O$ is open in $\mathrm{\Omega }$, hence in ${\mathbb{R}}^n$; and for every real $\chi \in C_c^{\mathrm{\infty }}(O;\mathbb{R})$ one has ${\mathrm{\Lambda }}_u(\chi )=0$. Indeed, if $\chi =0$ this is trivial, so let $\chi \mathrm{\ne }0$, put $S:=\mathrm{supp}\chi \subseteq O$, a nonempty compact set, and let $m:={\min }_S(u-\psi )>0$, which is positive because $u-\psi$ is continuous and positive on $S$. For every $0<\epsilon <m\mathrm{/}\mathrm{\parallel }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}$ define $v_{\pm }:=u\pm \epsilon \chi$; then $v_{\pm }\in H_0^1(\mathrm{\Omega })$ by [F1]. On the set $\{\chi \mathrm{\ne }0\}\subseteq S$ one has $v_{\pm }-\psi \ge (u-\psi )-\epsilon \mathrm{\mid }\chi \mathrm{\mid }\ge m-\epsilon \mathrm{\parallel }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}>0$, while off $\{\chi \mathrm{\ne }0\}$ one has $v_{\pm }=u\ge \psi$ almost everywhere by [F1]; hence $v_{\pm }\in K$. Testing the variational inequality [F1] at $v_{\pm }$ gives $a(u,\pm \epsilon \chi )\ge F(\pm \epsilon \chi )$, that is $\pm \epsilon {\mathrm{\Lambda }}_u(\chi )\ge 0$, so ${\mathrm{\Lambda }}_u(\chi )=0$. For a complex test $\chi ={\chi }_1+i{\chi }_2$ on $O$, apply this argument to its real and imaginary parts; complex linearity [F2] gives ${\mathrm{\Lambda }}_u(\chi )=0$ as well.

(Compact subsets of the noncontact set are $\mu$-null) Let $K_0\subseteq O$ be compact. By [F3] there is $\chi \in C_c^{\mathrm{\infty }}(O)$ with $0\le \chi \le 1$ and $\chi =1$ on a neighborhood $U$ of $K_0$; then $K_0\subseteq U\subseteq \{\chi =1\}\subseteq \{\chi \mathrm{\ne }0\}$. Step 1.1 and the measure representation give ${\int }_{\mathrm{\Omega }}\chi d\mu ={\mathrm{\Lambda }}_u(\chi )=0$, and $\chi \ge 0$ is continuous, hence $\mu$-measurable; by [F6] $\chi =0$ $\mu$-almost everywhere, that is $\mu (\{\chi \mathrm{\ne }0\})=0$. Monotonicity [F7] applied to $K_0\subseteq \{\chi \mathrm{\ne }0\}$ gives $\mu (K_0)=0$.

($\mu (O)=0$ by inner regularity) By [F4] inner regularity on the open set $O$ gives $\mu (O)=\sup \{\mu (K_0):K_0\subseteq O\text{ compact}\}$, and every term of this supremum is $0$ by step 2.1, so $\mu (O)=0$; this includes the case $O=\mathrm{\varnothing }$, where the only compact subset is the empty set, whose measure is $0$.

(The complementary open set is empty) The set $\{u<\psi \}$ is open in $\mathrm{\Omega }$ and has Lebesgue measure zero, because $u\ge \psi$ almost everywhere by [F1]; were it nonempty it would contain a Euclidean ball of positive measure by [F5]. Hence $\{u<\psi \}=\mathrm{\varnothing }$ and therefore $\mathrm{\Omega }\setminus \{u=\psi \}=\{u>\psi \}\cup \{u<\psi \}=O$ with $\mu (\mathrm{\Omega }\setminus \{u=\psi \})=\mu (O)=0$; equivalently, $\mu$ is concentrated on the contact set $\{u=\psi \}$.

(The integral vanishes) The functions $(u-\psi {)}^{+}$ and $(u-\psi {)}^{-}$ are nonnegative and continuous, hence $\mu$-measurable; $(u-\psi {)}^{+}$ vanishes outside $O$, which is $\mu$-null by step 3.1, and $(u-\psi {)}^{-}$ vanishes outside $\{u<\psi \}=\mathrm{\varnothing }$ by step 4.1, so [F6] gives ${\int }_{\mathrm{\Omega }}(u-\psi {)}^{+}d\mu =0={\int }_{\mathrm{\Omega }}(u-\psi {)}^{-}d\mu$; both integrals being finite, ${\int }_{\mathrm{\Omega }}(u-\psi )d\mu =0$. This proves all the asserted conclusions: $\mu (\{u>\psi \})=0$, concentration on $\{u=\psi \}$ and vanishing of the integral. The positive-measure ball supplier [F5] uses Countable Choice; the cutoff [F3] is constructed in ZF and inner regularity is part of [F4], so the local argument makes no further selections. The assumed Axiom of Choice and Countable Choice also cover the inherited obstacle setting and existence of $u$ [F1].