Let $m$ be a signed measure. For any set $A$, define the upper upper variation $m^+(A)$ as $\sup \{m(B) : B\subset A\}$ and the lower variation $\sup \{-m(B) : B\subset A\}$. We’re of course restricting ourselves to measurable subsets $B$ of $A$ here.
Let’s check that $m^+$ is additive. If $X$ and $Y$ are two disjoint measurable sets with subsets $B_X, B_Y$, then $m^+(X\cup Y) \geq m(B_X \cup B_Y) = m(B_X) + m(B_Y)$. Taking the supremum over the rightmost sum gives us $m^+(X\cup Y) \geq m^+(X) + m^+(Y)$. On the other hand, $m^+(X) + m^+(Y) \geq m(B_X) + m(B_Y) = m(B_X \cup B_Y)$. Taking the supremum again yields $m^+(X) + m^+(Y) \geq m^+(X\cup Y)$. It’s easy to extend this argument to $\sigma$-additivity as well, and obviously it follows that $m^-$ is also $\sigma$-additive by similar reasoning.
Observe, then, that $m^+$ and $m^-$ are both (unsigned) measures, since $\emptyset$ is measurable with signed measure 0.
Let’s try to understand the behavior of signed measures and their upper/lower variations.
Theorem A. You can’t have both $m(A) = \infty$ and $m(B) = -\infty$. This would break additivity: $m(A) = m(A \setminus B) + m(B)$, so $m(A \setminus B)$ is undefined.
Theorem B. $m(B) = \infty, B \subset A$ implies $m(A) = \infty$. Proof is obvious given the above theorem.
Theorem C. Upward and downward continuity of signed measure. Proof is exactly the same as the case of unsigned measures.
Theorem D. $m^+(A) = \infty$ implies $m(A) = \infty$.
Proof. We basically just carve out disjoint chunks of $A$ so that the union of them is arbitrarily large. Let $A_1 = A$ and find $B_1 \subset A_1$ with $m(B_1) > 1$. If $m(B_1) = \infty$, we’re done. Otherwise, subtract this out of $A_1$; call the remainder $A_2$. Its measure is still infinite, so keep going in this way; we end up with disjoint $B_k$ whose measures are all greater than 1. Let $B = \cup B_k$; clearly $m(B) = \infty$, so $m(A) = \infty$. $\blacksquare$
A set is totally positive if each of its subsets has non-negative measure. Similarly for totally negative. (A set is a null set if each of its subsets has zero measure.) The existence of such sets is not, in fact, obvious.
Theorem E. Suppose $-\infty < m(A) < 0$. Then $m(B) \leq m(A)$ for some totally negative $B \subset A$.
Proof. Set $A_0 = A$. We can, by definition of the supremum, find some $B_1 \subset A_0$ with $m(B_1) \geq m^+(A_0)/2 \geq 0$. Subtract this out of $A_0$ and call the result $A_1$. Keep doing this infinitely and let $A^* = \cap A_i = A \setminus \cup B_i$. The $B_i$’s are disjoint, so $m(A^*) = m(A) – \sum m(B_i) \leq m(A)$. If some $C \subset A^*$ remains with $m(C) > 0$, then for each $i$ we have $m^+(A_i) \geq m(C)$, hence $m(B_i) \geq m(C)/2$ and the sum diverges.
Theorem F. Suppose $\infty > m(A) > 0$. Then $m(B) \geq m(A)$ for some totally positive $B \subset A$. This is an obvious corollary of Theorem E.
Theorem G. If $m^+(A) = c < \infty$, then this is actually achieved: $m(B) = c$ for some totally positive $B \subset A$. Similarly for $m^-(A)$.
Proof. Assume WLOG $c > 0$. We can find a sequence $B_k \subset A$ with $m(B_k) \geq m^+(A) – 1/k$. By Theorem F, we may assume these are all totally positive. But then $m(\cup B_k) \geq m(B_i)$ for each $i$ (since $m$ is an unsigned measure on the sigma-algebra generated by the $B_i$’s), so $m(\cup B_k) \geq m^+(A)$. The reverse inequality is also true by definition of $m^+$. $\blacksquare$
Theorem H. Theorems E and F still hold dropping the assumption of finitude.
Proof. We deal with the negative case: $m(A) = -\infty$. Then $m^+(A) < \infty$, so by theorem G, we can find some totally positive $B \subset A$ with $m(B) = m^+(A)$. Then clearly $C = A \setminus B$ is totally negative; otherwise it has a positive subset disjoint from $B$, which would mean $m^+(A) > m(B)$.
Theorem I (Hahn Decomposition). On a signed measure space over $A$, we can write $A = B \cup C$ where $B$ is totally positive and $C$ is totally negative. Moreover, any other such decomposition differs only by a null set.
Proof. Let $c = m^+(A)$. We can then find a totally positive subset $B$ with $m(B) = c$. Suppose $C = A \setminus B$ contained a further positive subset $D$. Then $B$ and $D$ are disjoint, so $m(B \cup D) = m(B) + m(D) > c = m^+(A)$, which is a contradiction. So $C$ is negative.
Now suppose also $A = B’ \cup C’$ with corresponding total positivity/negativity. Since $B$ and $B’$ both have measure $c$, $m(B \setminus B’) = m(B) – m(B’) = 0$ and so forth; the symmetric differences are obviously null sets. $\blacksquare$