Vector measure

Remarks

Example: let $\mathcal{F}$ be the Lebesgue measurable subsets of $[0, 1]$, let $1 \le p < \infty$, and let $F(E) = \mathbf{1}_E \in L^p([0, 1])$. Then $F$ is a countably additive vector measure. Finite additivity holds because $\mathbf{1}_{E_1 \cup E_2} = \mathbf{1}_{E_1} + \mathbf{1}_{E_2}$ for disjoint sets. For a disjoint sequence, $\norm{F(E) - \sum_{n \le N} F(E_n)}_p = \lambda\big( \bigcup_{n > N} E_n \big)^{1/p} \to 0$.

The same map with values in $L^\infty([0, 1])$ is a vector measure that is not countably additive. For $E_n = [2^{-n}, 2^{-n+1})$, the remainder $\mathbf{1}_{(0, 2^{-N})}$ has $L^\infty$ norm $1$ for every $N$.

Non-example: $G(E) = \lambda(E)^2 \in \R$ is not a vector measure, since $G([0, 1]) = 1$ but $G([0, \tfrac12)) + G([\tfrac12, 1]) = \tfrac12$.

A real or complex measure that is finite is a countably additive vector measure with $B = \R$ or $B = \C$.