Every complex measure has finite total variation

Put $\alpha :=\mathrm{Re}\nu$ and $\beta :=\mathrm{Im}\nu$. By [L2], these are finite signed measures and $\mathrm{\mid }\nu (A)\mathrm{\mid }=\mathrm{\mid }\alpha (A)+i\beta (A)\mathrm{\mid }\le \mathrm{\mid }\alpha (A)\mathrm{\mid }+\mathrm{\mid }\beta (A)\mathrm{\mid }$ for every measurable $A$.

Let $(E_n)$ be any countable measurable partition of $E$. Step 1.1 and the one-piece lower bound in the definition of variation give ${\sum }_n\mathrm{\mid }\nu (E_n)\mathrm{\mid }\le {\sum }_n\mathrm{\mid }\alpha (E_n)\mathrm{\mid }+{\sum }_n\mathrm{\mid }\beta (E_n)\mathrm{\mid }\le {\sum }_n\mathrm{\mid }\alpha \mathrm{\mid }(E_n)+{\sum }_n\mathrm{\mid }\beta \mathrm{\mid }(E_n)\mathrm{.}$ By [L3], $\mathrm{\mid }\alpha \mathrm{\mid }={\alpha }^{+}+{\alpha }^{-}$ and $\mathrm{\mid }\beta \mathrm{\mid }={\beta }^{+}+{\beta }^{-}$, so countable additivity of the four positive Jordan parts turns the right side into $\mathrm{\mid }\alpha \mathrm{\mid }(E)+\mathrm{\mid }\beta \mathrm{\mid }(E)$.

The quantities $\mathrm{\mid }\alpha \mathrm{\mid }(E)$ and $\mathrm{\mid }\beta \mathrm{\mid }(E)$ are finite by [L2] and [L3]: both Jordan parts of a finite signed measure are finite on $E$. Thus step 2.1 gives the partition-independent bound ${\sum }_n\mathrm{\mid }\nu (E_n)\mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }(E)+\mathrm{\mid }\beta \mathrm{\mid }(E)<+\mathrm{\infty }\mathrm{.}$ Taking the supremum over all countable measurable partitions in [L1] proves $\mathrm{\mid }\nu \mathrm{\mid }(E)<+\mathrm{\infty }$. Applying this with $E=X$ gives $\mathrm{\mid }\nu \mathrm{\mid }(X)<+\mathrm{\infty }$.