Boundary layers of finite metric outer measure exhaust the complement of a closed set in outer measure

If $F=\mathrm{\varnothing }$ or $A\setminus F=\mathrm{\varnothing }$, the asserted constant layers give the result without set-distance notation. Otherwise [F2] licenses $d(x,F)$, the layers are increasing, and closedness means every $x\mathrm{\notin }F$ has a ball disjoint from $F$, hence $d(x,F)>0$ and $x$ enters some $B_n$; thus $B_n\uparrow A\setminus F$. Put $C_n=B_{n+1}\setminus B_n$.

The function $x\mapsto d(x,F)$ is $1$-Lipschitz, since the triangle inequality gives $d(x,F)\le d(x,y)+d(y,F)$ and symmetrically. Therefore two nonempty annuli $C_i,C_j$ of the same parity with $i<j$ are positively separated: their distance-to-$F$ ranges are separated by the positive gap between $1\mathrm{/}(i+2)$ and $1\mathrm{/}(j+1)$. Repeated use of [F1] on finite parity unions gives ${\sum }_{k<m}{\mu }^{\ast }(C_{2k})\le {\mu }^{\ast }(A)$ and ${\sum }_{k<m}{\mu }^{\ast }(C_{2k+1})\le {\mu }^{\ast }(A)$, with empty annuli omitted.

By [F3], each parity series is the supremum of its increasing finite partial sums, and step 2.1 bounds that supremum by the finite number ${\mu }^{\ast }(A)$. Given $\epsilon >0$, choose a partial sum within $\epsilon \mathrm{/}2$ of each supremum; every later tail is then below $\epsilon \mathrm{/}2$, so both parity tails tend to zero. Now $(A\setminus F)\setminus B_n={\bigcup }_{k\ge n}{C}_k$, so subadditivity bounds its outer measure by those two tails. Thus all sufficiently large $n$ satisfy ${\mu }^{\ast }(B_n)\le {\mu }^{\ast }(A\setminus F)\le {\mu }^{\ast }(B_n)+\epsilon$, and taking the supremum over $n$ proves the stated equality.