For the difference of two (signed) measures to be well-defined, we need one of the two to be finite. Otherwise, the measure of the difference on the base set involves subtracting $\infty$ from $\infty$.

A finite signed measure is one for which the measure of every set is finite. This condition is equivalent to the base set having finite measure (see below).

A signed measure is finite if and only if it is finite on the base set.

Let $(X, \mathcal{A} )$ be a measurable space. Let $\mu : \mathcal{A} \to \eri$ be a signed measure.

($\Rightarrow$) If $\mu $ is finite, then $\mu (X)$ is finite since $X \in \mathcal{A} $.

($\Leftarrow)$ Next, suppose $\mu (X)$ is finite. Let $A \in \mathcal{A} $. Then $X = A \union (X - A)$, with these sets disjoint, so by countable additivity of $\mu $, $\mu (X) = \mu (A) + \mu (X - A)$. Since $\mu (X)$ finite, $\mu (A)$ and $\mu (X - A)$ are both finite.