Haar candidate sets have the finite intersection property

For any $U$, a cutoff $v$ at $\{e\}\subseteq U$ has $v(e)=1$, $0\le v\le 1_U$ and compact support. Set $\varphi =(v-1\mathrm{/}2{)}^{+}$. Then $\varphi (e)=1\mathrm{/}2$ and its support lies in the compact set $\{v\ge 1\mathrm{/}2\}\subseteq U$. Thus $E_U$ contains an approximant. The coordinate intervals are nonempty because they contain every such vector, and they are finite closed real intervals.

For finitely many neighbourhoods $U_1,\dots ,U_n$, their intersection $W$ is an identity neighbourhood, and $E_W\subseteq {\bigcap }_j{E}_{U_j}$. Step 1.1 makes this intersection nonempty. For $n=0$ the intersection is $\mathcal{P}$, also nonempty by an approximant. The compact real intervals and AC give compactness of $\mathcal{P}$, so the closed FIP theorem supplies a common point $T$.

The closed equations in [F1] hold throughout each $E_U$, hence for $T$, and its strictly positive coordinate lower bounds persist. Set $T(0)=0$. For $f,g\ge 0$ and $ϵ>0$, [F2] gives $U$ such that $0\le I_{\varphi }(f)+I_{\varphi }(g)-I_{\varphi }(f+g)<ϵ$ on its approximants. This finite-coordinate condition with upper bound $\le ϵ$ is closed, so it holds for $T\in E_U$. An error in $[0,ϵ]$ for every positive $ϵ$ is zero. Thus $T$ is additive; when an argument is zero this is already the assigned zero value.