For a signed measure, total variation is nu-plus plus nu-minus, finite partitions suffice, and nu-plus and nu-minus are extremal

Let $(E_n)$ be a countable measurable partition of $E$. Using [L2] and the [L1, L2] triangle inequality, $\mathrm{\mid }\nu (E_n)\mathrm{\mid }=\mathrm{\mid }{\nu }^{+}(E_n)-{\nu }^{-}(E_n)\mathrm{\mid }\le {\nu }^{+}(E_n)+{\nu }^{-}(E_n)$ for each $n$. Summing and using countable additivity of the positive measures ${\nu }^{+}$ and ${\nu }^{-}$ gives ${\sum }_n\mathrm{\mid }\nu (E_n)\mathrm{\mid }\le {\nu }^{+}(E)+{\nu }^{-}(E)\mathrm{.}$ Hence [L1] yields $\mathrm{\mid }\nu \mathrm{\mid }(E)\le {\nu }^{+}(E)+{\nu }^{-}(E)$.

The two-piece partition $E=(E\cap P)\bigsqcup (E\cap N)$ from [L2] gives [L1, L2, step 1.1] $\mathrm{\mid }\nu (E\cap P)\mathrm{\mid }+\mathrm{\mid }\nu (E\cap N)\mathrm{\mid }={\nu }^{+}(E)+{\nu }^{-}(E)\mathrm{.}$ Therefore $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge {\nu }^{+}(E)+{\nu }^{-}(E)$, and together with step 1.1 this proves equality. Because this equality is already realized by a finite partition, finite partitions suffice for signed measures.

If $F\subseteq E$ is measurable, then [L2] gives [L2, step 2.1] $\nu (F)={\nu }^{+}(F)-{\nu }^{-}(F)\le {\nu }^{+}(F)\le {\nu }^{+}(E),$ so ${\sup }_{F\subseteq E}\nu (F)\le {\nu }^{+}(E)$. Taking $F=E\cap P$ gives equality: $\nu (F)={\nu }^{+}(E)$. The same argument with $E\cap N$ gives ${\inf }_{F\subseteq E}\nu (F)=-{\nu }^{-}(E)\mathrm{.}$

Steps 2.1 and 3.1 prove all four displayed formulas.