Uniqueness of a Radon measure representing a distribution

If $\mathrm{\Omega }=\mathrm{\varnothing }$, equality is immediate. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, its singleton indicator is a smooth compactly supported test and determines every measure. Assume henceforth $n\ge 1$. Set $\lambda :=\mu -\nu$. If $K\subset \mathrm{\Omega }$ is compact and nonempty, then for every sufficiently large $m$, rational inner balls with doubled closures in $\mathrm{\Omega }\cap \{x:\mathrm{dist}(x,K)<1\mathrm{/}m\}$ cover $K$; [F2] gives a finite subcover, and ordering finite rational lists by natural-number codes gives a least such cover. Put the fixed bump [F3] on each selected ball and combine them as ${\eta }_m=1-{\prod }_j(1-{\psi }_{m,j})$. Then $0\le {\eta }_m\le 1$, it equals one on $K$, is supported in the $1\mathrm{/}m$-neighborhood of $K$, and ${\eta }_m(x)\to {\mathbf{1}}_K(x)$. Since each ${\eta }_m$ is a smooth compactly supported test function, $\int {\eta }_{m}d\lambda =0$; dominated convergence [F4] with respect to the finite regular total-variation measure [F1] gives $\lambda (K)=0$. The empty compact set has measure zero directly.

For any Borel $A\subseteq \mathrm{\Omega }$ and $\epsilon >0$, inner regularity of $\mathrm{\mid }\lambda \mathrm{\mid }$ [F1] gives a compact $K\subseteq A$ with $\mathrm{\mid }\lambda \mathrm{\mid }(A\setminus K)<\epsilon$. Since step 1.1 gives $\lambda (K)=0$, we have $\mathrm{\mid }\lambda (A)\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }(A\setminus K)<\epsilon$. Thus $\lambda (A)=0$ for every Borel $A$, so $\mu =\nu$.