= Solution
Start with $R_0=A$. Having chosen disjoint $A_1,\ldots,A_i$, put $R_i=A\setminus\bigcup_{j\leq i}A_j$ and
$$
s_i=\sup\{\mu(B):B\in\mathfrak M,\ B\subseteq R_i\}.
$$
The empty set shows $s_i\geq0$, and finite total variation bounds it above. If $s_i>0$, choose $A_{i+1}\subseteq R_i$ with $\mu(A_{i+1})\geq s_i/2>0$; if $s_i=0$, choose $A_{i+1}=\varnothing$. This is the required recursion, including the initial step $i=0$.
Disjointness and nonnegativity give $\sum_i\mu(A_i)\leq|\mu|(A)<\infty$, hence $\mu(A_i)\to0$. Write $R=A\setminus\bigcup_i A_i$. For every measurable $B\subseteq R$,
$$
\mu(B)\leq s_i\leq2\mu(A_{i+1})\longrightarrow0.
$$
Thus every measurable subset of $R$ has nonpositive <measure>. Each finite partition of $R$ consequently has variation sum exactly $-\mu(R)$, and therefore $|\mu|(R)=-\mu(R)$. It follows that $\mu^+(R)=0$. Finally <countable additivity> gives
$$
\boxed{\mu(R)=\mu(A)-\sum_i\mu(A_i)\leq\mu(A)<0,\qquad\mu^+(R)=0.}
$$
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