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 $\mathrm{\mid }\nu \mathrm{\mid }(\mathrm{\varnothing })=0$, because every member of any countable measurable partition of $\mathrm{\varnothing }$ is empty and its partition sum is $0$.

Let $(E_m)$ be pairwise disjoint measurable sets and put $E={\bigcup }_m{E}_m$. Fix $N$ and suppose $\mathrm{\mid }\nu \mathrm{\mid }(E_m)<+\mathrm{\infty }$ for $m\le N$. For any $\epsilon >0$, the definition of each supremum gives a countable partition of $E_m$ whose sum exceeds $\mathrm{\mid }\nu \mathrm{\mid }(E_m)-\epsilon \mathrm{/}(N+1)$. Choose these finitely many partitions using [L3], enumerate all their cells in one countable partition, and append $E\setminus {\bigcup }_{m\le N}{E}_m$ as one more cell. This is a countable measurable partition of $E$, so [L1] gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge {\sum }_{m\le N}\mathrm{\mid }\nu \mathrm{\mid }(E_m)-\epsilon$. Let $\epsilon \downarrow 0$. If some $\mathrm{\mid }\nu \mathrm{\mid }(E_m)=+\mathrm{\infty }$, for each finite threshold choose one partition of that $E_m$ with sum above the threshold and append $E\setminus E_m$; then $\mathrm{\mid }\nu \mathrm{\mid }(E)=+\mathrm{\infty }$. Hence in all cases $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge {\sum }_{m\le N}\mathrm{\mid }\nu \mathrm{\mid }(E_m)$ for every $N$, and taking the supremum of finite partial sums gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge {\sum }_m\mathrm{\mid }\nu \mathrm{\mid }(E_m)$. No partitions were chosen across infinitely many $m$.

Conversely, let $(B_j)$ be a countable measurable partition of $E$. For each $j$, countable additivity of $\nu$ on $B_j={⨆}_m(B_j\cap E_m)$ and the triangle inequality for finite partial sums and their limits give $\mathrm{\mid }\nu (B_j)\mathrm{\mid }\le {\sum }_m\mathrm{\mid }\nu (B_j\cap E_m)\mathrm{\mid }$. For each $m$, the cells $(B_j\cap E_m{)}_j$ partition $E_m$, so ${\sum }_j\mathrm{\mid }\nu (B_j\cap E_m)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(E_m)$ by [L1]. Summing the first inequality over $j$ and interchanging two nonnegative series yields ${\sum }_j\mathrm{\mid }\nu (B_j)\mathrm{\mid }\le {\sum }_m\mathrm{\mid }\nu \mathrm{\mid }(E_m)$. Taking the supremum over all partitions $(B_j)$ gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\le {\sum }_m\mathrm{\mid }\nu \mathrm{\mid }(E_m)$.

Steps 1.2 and 1.3 prove countable additivity. Together with step 1.1 and [L2], this shows that $\mathrm{\mid }\nu \mathrm{\mid }$ is a measure.