= Solution
Write $J(A)$ for the partition supremum. Empty cells are permitted, so $J\geq\mu_i$ for each $i$, and $J(A)\leq\sum_i\mu_i(A)$ is finite. Also $J(\varnothing)=0$.
Let $A=\bigsqcup_{j\geq1}B_j$. Any measurable partition $A=\bigsqcup_iA_i$ yields
$$
\sum_i\mu_i(A_i)=\sum_j\sum_i\mu_i(A_i\cap B_j)\leq\sum_jJ(B_j).
$$
Taking its supremum gives one inequality. Conversely, for any $\varepsilon>0$ choose for each $j$ a partition $B_j=\bigsqcup_iB_{ji}$ whose sum exceeds $J(B_j)-\varepsilon2^{-j}$. The sets $A_i=\bigsqcup_jB_{ji}$ partition $A$, and <countable additivity> of each $\mu_i$ gives $J(A)\geq\sum_jJ(B_j)-\varepsilon$. Let $\varepsilon\downarrow0$. Thus \b[$J$ is a <finite measure>].
If another <measure> $\nu$ dominates every $\mu_i$, then for every such partition $\sum_i\mu_i(A_i)\leq\sum_i\nu(A_i)=\nu(A)$. Consequently $J\leq\nu$. Hence \b[$J=\bigvee_i\mu_i$ is the least dominating <measure>], the <finite supremum of positive measures>. It generally differs from the pointwise maximum of the numbers $\mu_i(A)$: the maximizing index may differ between pieces of $A$.
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