Finite supremum of positive measures (${\bigvee }_{i=1}^N{\mu }_i$)

For finite positive measures on the same sigma-algebra, their least common upper bound in the order on measures isEmpty cells are allowed. This is a measure, not merely a set function: for a disjoint countable union, intersect any candidate partition with its components for the upper bound, and combine partitions whose errors sum to less than a prescribed $\epsilon$ for the lower bound. Countable additivity of each component gives both inequalities. Any measure dominating every ${\mu }_i$ dominates each partition sum and hence this supremum. The result is bounded above by ${\sum }_i{\mu }_i$ and so remains finite.

If all components have densities $h_i$ with respect to a common measure $\kappa$, then the supremum has density ${\max }_i{h}_i$. For the upper bound integrate the pointwise maximum. For equality, partition by the smallest index attaining that maximum; these are measurable sets. For example two measures supported on disjoint sets have their sum as supremum, rather than the numerical maximum of their masses on the whole space.

Take Lebesgue decompositions relative to a fixed finite positive measure $\kappa$. The finite union $S$ of null supports of their singular components is still $\kappa$-null. On $S$ each original measure equals its singular component; on $S^c$ it equals its absolutely continuous component. Restriction commutes with the partition formula defining the finite supremum of positive measures. Thus the supremum restricts to ${\bigvee }_i{\mu }_i^s$ on $S$ and to ${\bigvee }_i{\mu }_i^a$ off $S$. The latter is absolutely continuous with respect to $\kappa$ because it is bounded by the sum of the absolutely continuous components. Uniqueness of the Lebesgue decomposition proves the displayed identities, and the Radon-Nikodym derivative of the absolutely continuous component is the pointwise maximum of the individual derivatives.