A function with nonnegative test pairings is nonnegative a.e.

(Local vanishing, uniformly in the open set and the class) Let $\omega \subseteq {\mathbb{R}}^n$ be open, let $z\in L^2(\omega )$ satisfy ${\int }_{\omega }z\phi \ge 0$ for every nonnegative $\phi \in C_c^{\mathrm{\infty }}(\omega )$, let $O^{\mathrm{\prime }}\subseteq \omega$ be open, and let $\eta \in C_c^{\mathrm{\infty }}(O^{\mathrm{\prime }})$ with $\eta \ge 0$; we claim ${\int }_{\omega }(z^{-}{)}^2\eta =0$. If $\eta =0$ this is trivial, so assume $\eta \mathrm{\ne }0$. Choose a real representative $f$ of the class $z$ and set $g:=z^{-}\eta$, the class of the measurable function $f^{-}\eta$; since $\eta$ is bounded with compact support and $f^{-}\le \mathrm{\mid }f\mathrm{\mid }$, this class lies in $L^2(\omega )$ and has a representative vanishing outside the compact set $S:=\mathrm{supp}\eta \subseteq O^{\mathrm{\prime }}$ [F10, F11]. If $O^{\mathrm{\prime }}={\mathbb{R}}^n$, choose any $\epsilon >0$; otherwise [F6] applies to $S$ and the nonempty closed set ${\mathbb{R}}^n\setminus O^{\mathrm{\prime }}$, giving $d:=\mathrm{dist}(S,{\mathbb{R}}^n\setminus O^{\mathrm{\prime }})>0$, and choose $0<\epsilon <d\mathrm{/}2$. Let ${\phi }_{\epsilon }$ be the smooth representative of ${\rho }_{\epsilon }\ast g$ from [F2]. It is compactly supported; when $O^{\mathrm{\prime }}\mathrm{\ne }{\mathbb{R}}^n$, its support lies in $\overline{S+\mathrm{supp}{\rho }_{\epsilon }}\subseteq O^{\mathrm{\prime }}$ because $\mathrm{supp}{\rho }_{\epsilon }\subseteq {\bar{B}}_{2\epsilon }(0)$ and $2\epsilon <d$ [F1]. Since ${\rho }_{\epsilon }\ge 0$ and $f^{-}\eta \ge 0$, monotonicity of the integral gives ${\phi }_{\epsilon }\ge 0$ throughout [F3], so ${\phi }_{\epsilon }\in C_c^{\mathrm{\infty }}(\omega )$ is a nonnegative test function and the hypothesis yields ${\int }_{\omega }z{\phi }_{\epsilon }\ge 0$; on the other hand $\mid {\int }_{\omega }z({\phi }_{\epsilon }-g)\mid \le \mathrm{\parallel }{\phi }_{\epsilon }-g{\mathrm{\parallel }}_2\mathrm{\parallel }z{\mathrm{\parallel }}_2\to 0$ as $\epsilon \downarrow 0$ by [F2] and [F4], so ${\int }_{\omega }zg\ge 0$. Finally $zg=z{z}^{-}\eta =-z^{+}{z}^{-}\eta -(z^{-}{)}^2\eta =-(z^{-}{)}^2\eta$ because $z^{+}{z}^{-}=0$ pointwise [F10], so ${\int }_{\omega }(z^{-}{)}^2\eta \le 0$; as the integrand is nonnegative this forces ${\int }_{\omega }(z^{-}{)}^2\eta =0$. The argument uses only the pairing hypothesis on the open set $\omega$.

(From local test functions to a.e. vanishing) Let $\omega$ and $z$ be as in step 1.1. Let $O^{\mathrm{\prime }}\subseteq \omega$ be open and let $K\subseteq O^{\mathrm{\prime }}$ be compact; by [F7] there is $\chi \in C_c^{\mathrm{\infty }}(O^{\mathrm{\prime }})$ with $0\le \chi \le 1$ and $\chi =1$ on a neighbourhood $N$ of $K$, and step 1.1 with $\eta =\chi$ gives ${\int }_{\omega }(z^{-}{)}^2\chi =0$; as this integrand is nonnegative and measurable, [F5] gives $(z^{-}{)}^2\chi =0$ a.e. on $\omega$, hence $(z^{-}{)}^2=0$ a.e. on $K$. Now for arbitrary open $O^{\mathrm{\prime }}\subseteq \omega$ for $j\in \mathbb{N}$ take $K_j:=\{x:\mathrm{\mid }x\mathrm{\mid }\le j+1\}$ if $O^{\mathrm{\prime }}={\mathbb{R}}^n$ and $K_j:=\{x\in O^{\mathrm{\prime }}:\mathrm{\mid }x\mathrm{\mid }\le j+1\text{ and }\mathrm{dist}(x,{\mathbb{R}}^n\setminus O^{\mathrm{\prime }})\ge 1\mathrm{/}(j+1)\}$ otherwise; each $K_j$ is closed, bounded and contained in $O^{\mathrm{\prime }}$, hence compact by [F9], and the $K_j$ cover $O^{\mathrm{\prime }}$ (a point $x\in O^{\mathrm{\prime }}$ has a ball $B(x,r)\subseteq O^{\mathrm{\prime }}$, so $x\in K_j$ for every $j+1\ge \max \{\mathrm{\mid }x\mathrm{\mid },1\mathrm{/}r\}$). Since $(z^{-}{)}^2$ vanishes a.e. on each $K_j$, it vanishes a.e. on the union $O^{\mathrm{\prime }}$ by [F8]. Taking $O^{\mathrm{\prime }}=\omega$, $(z^{-}{)}^2=0$ a.e. on $\omega$, so $z^{-}=0$ a.e. on $\omega$ and $z\ge 0$ a.e. on $\omega$ by [F10].

(First assertion) Apply step 2.1 with $(\omega ,z)=(\mathrm{\Omega },\zeta )$: the nonnegative-pairing hypothesis holds by assumption, so $({\zeta }^{-}{)}^2=0$ a.e. on $\mathrm{\Omega }$, hence ${\zeta }^{-}=0$ a.e. on $\mathrm{\Omega }$ and $\zeta \ge 0$ a.e. on $\mathrm{\Omega }$ by [F10].

(Second assertion) Assume additionally that ${\int }_{\mathrm{\Omega }}\zeta \phi =0$ for every $\phi \in C_c^{\mathrm{\infty }}(O)$, and put $\omega :=O$ and $z:=\zeta {\mathrm{\mid }}_O$, the restriction of the class, which lies in $L^2(O)$ with ${\int }_O\mathrm{\mid }\zeta {\mathrm{\mid }}^2\le {\int }_{\mathrm{\Omega }}\mathrm{\mid }\zeta {\mathrm{\mid }}^2$ and ${\int }_O\zeta \phi ={\int }_{\mathrm{\Omega }}\zeta \phi$ for every $\phi \in C_c^{\mathrm{\infty }}(O)$ [F11]. For every nonnegative $\phi \in C_c^{\mathrm{\infty }}(O)$ one has ${\int }_O\zeta \phi =0\ge 0$ and also ${\int }_O(-\zeta )\phi =0\ge 0$, so step 2.1 applies with $(\omega ,z)=(O,\zeta {\mathrm{\mid }}_O)$ and with $(\omega ,z)=(O,-\zeta {\mathrm{\mid }}_O)$, whose negative parts are ${\zeta }^{-}$ and ${\zeta }^{+}$ respectively; hence $({\zeta }^{-}{)}^2=0$ and $({\zeta }^{+}{)}^2=0$ a.e. on $O$, so ${\zeta }^{+}={\zeta }^{-}=0$ a.e. on $O$ and $\zeta =0$ a.e. on $O$ by [F10].

Step 3.1 proves the first assertion and step 3.2 the second for arbitrary open $\mathrm{\Omega }$, class $\zeta$ and open $O\subseteq \mathrm{\Omega }$; the argument of steps 1.1-2.1 is uniform in the pair $(\omega ,z)$, so the second assertion needed no re-run of the estimates. Countable Choice was used in the approximate-identity interface of step 1.1 and in the countable-union step of step 2.1 [A1, F2, F8], while the localisation itself is choice-free.