Let $\nu$ be a signed measure and $|\nu|$ be its total variation. Then does $\nu \text{ finite } \implies |\nu| \text{ finite} $? If so, why?
I can almost see this by appealing to the Jordan decomposition of $\nu$ and the definition of $|\nu|$, \begin{align*} \nu = \nu^+ - \nu^- \\ |\nu| = \nu^+ + \nu^- \end{align*} but I am not completely sure that there can't be a measurable set $A$ such that $\nu(A) =0$ but $\nu^+(A) = \nu^-(A) = \infty$.
For instance, suppose our measurable space is $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ denotes the Borel subsets of $\mathbb{R}$. Let $\nu^+$ and $\nu^-$ map a set to the number of integers it contains. Then $\nu^+, \nu^-$ seem to be measures, and $\nu$ would seem to be a signed measure. And if so, the implication of interest fails.
First of all: An expression of the form "$\infty - \infty$" is not well defined. So for a decomposition to be well defined such an expressions may not occur (that's why your example won't work because your so constructed $\nu$ is not well-defined)
Then: Yes, your implication holds, moreover for a signed measure $\nu$ it's true that $$\nu \text{ finite} \iff |\nu| \text{ finite}$$
To see that consider Hahn's decomposition theorem which states the existence of a positive set $P$ and a disjoint negative set $N$ s.t. $\Omega = P \cup N$.
From this it follows that $$\nu^+(A) = \nu(A \cap P), \nu^{-}(A) = \nu(A \cap N)$$ holds for all measurable sets $A$ and it's easy to see that this implies the equivalence above.