Haar covering functionals are asymptotically additive

If either $f_i=0$, the error is zero. Otherwise set $s=f_1+f_2$, choose $0\le F\le 1$ compactly supported with $F=1$ on $\mathrm{supp}s$, and put $B=(F:f_0)>0$, $A=(s:f_0)>0$. Choose $0<\delta <ϵ\mathrm{/}(4(B+1))$, then $0<\eta <ϵ\mathrm{/}(4(A+\delta B+1))$. Thus $\delta B+2\eta (A+\delta B)<ϵ$. AC implies the DC used for the cutoff.

Put $q=s+\delta F$ and $h_i=f_i\mathrm{/}q$ where $q>0$, and zero elsewhere. On $\mathrm{supp}{f}_i$, $q\ge \delta$; a point where $q=0$ is outside that closed support and has a neighbourhood where $f_i=0$. Hence $h_i$ is continuous, supported in $\mathrm{supp}{f}_i$, and $h_1+h_2\le 1$. Choose $U$ so $\mathrm{\mid }{h}_i(xz)-h_i(x)\mathrm{\mid }<\eta$ for all $x$, both $i$, and $z\in U$, using uniform right-translation continuity.

For a finite cover $q\le \sum c_j{L}_{x_j}\varphi$, a nonzero term at $y$ has $x_j^{-1}y\in U$. Consequently $h_i(y)\le h_i(x_j)+\eta$ and $f_i(y)\le \sum c_j(h_i(x_j)+\eta )L_{x_j}\varphi (y)$. These coefficients are positive. The sum of the two coefficient sums is at most $(1+2\eta )\sum c_j$. Taking the infimum over covers of $q$, then dividing by $(f_0:\varphi )$, gives $I_{\varphi }(f_1)+I_{\varphi }(f_2)\le (1+2\eta )(I_{\varphi }(s)+\delta I_{\varphi }(F))$.

Subtracting $I_{\varphi }(s)$ and using the coordinate bounds yields an upper error at most $2\eta A+\delta (1+2\eta )B<ϵ$. Subadditivity gives its nonnegativity. The choices of $\delta ,\eta ,U$ preceded $\varphi$, so the bound holds uniformly for every allowed test function.