 Solution

Write $J(A)$ for the partition supremum. Empty cells are permitted, so $J\ge {\mu }_i$ for each $i$, and $J(A)\le {\sum }_i{\mu }_i(A)$ is finite. Also $J(\varnothing )=0$.

Let $A={⨆}_{j\ge 1}{B}_j$. Any measurable partition $A={⨆}_i{A}_i$ yieldsTaking its supremum gives one inequality. Conversely, for any $\epsilon >0$ choose for each $j$ a partition $B_j={⨆}_i{B}_{ji}$ whose sum exceeds $J(B_j)-\epsilon 2^{-j}$. The sets $A_i={⨆}_j{B}_{ji}$ partition $A$, and countable additivity of each ${\mu }_i$ gives $J(A)\ge {\sum }_{j}J(B_j)-\epsilon$. Let $\epsilon \downarrow 0$. Thus $J$ is a finite measure.

If another measure $\nu$ dominates every ${\mu }_i$, then for every such partition ${\sum }_i{\mu }_i(A_i)\le {\sum }_i\nu (A_i)=\nu (A)$. Consequently $J\le \nu$. Hence $J={\bigvee }_i{\mu }_i$ 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 $A$.