The total variation of a signed or complex measure is a positive measure

The set function $\mathrm{\mid }\nu \mathrm{\mid }$ is nonnegative by [L1]. Also $\mathrm{\mid }\nu \mathrm{\mid }(\mathrm{\varnothing })=0$, [L1, L2] because the only countable measurable partition of $\mathrm{\varnothing }$ has every part equal to $\mathrm{\varnothing }$ and hence partition sum $0$.

Let $(E_m)$ be pairwise disjoint measurable sets and put $E={\bigcup }_m{E}_m$. [L1] For each $m$, choose a countable measurable partition $(A_{m,k}{)}_k$ of $E_m$. Then the doubly indexed family $(A_{m,k}{)}_{m,k}$ is a countable measurable partition of $E$, so [L1] gives ${\sum }_m\mathrm{\mid }\nu \mathrm{\mid }(E_m)\le \mathrm{\mid }\nu \mathrm{\mid }(E)$ after taking suprema over all admissible partitions of the pieces.

Conversely, let $(B_j)$ be a countable measurable partition of $E$. Then [L1, step 1.2] each $(B_j\cap E_m{)}_j$ is a countable measurable partition of $E_m$, and $\mathrm{\mid }\nu (B_j)\mathrm{\mid }\le {\sum }_m\mathrm{\mid }\nu (B_j\cap E_m)\mathrm{\mid }$ by the triangle inequality applied to the disjoint decomposition $B_j={\bigcup }_m(B_j\cap E_m)$. Summing over $j$ and using [L1] on each piece gives ${\sum }_j\mathrm{\mid }\nu (B_j)\mathrm{\mid }\le {\sum }_m\mathrm{\mid }\nu \mathrm{\mid }(E_m)\mathrm{.}$ Taking the supremum over all partitions $(B_j)$ of $E$ yields $\mathrm{\mid }\nu \mathrm{\mid }(E)\le {\sum }_m\mathrm{\mid }\nu \mathrm{\mid }(E_m)$.

Steps 1.2 and 2.1 prove countable additivity. Together with step 1.1 and [L2, step 1.1, step 1.2, step 2.1] ∎ [L2], this shows that $\mathrm{\mid }\nu \mathrm{\mid }$ is a measure.