= Solution
For a <signed measure> define its <total variation measure> by
$$
|\mu|(E)=\sup\left\{\sum_{r=1}^m|\mu(E_r)|:E=\bigsqcup_{r=1}^m E_r,\ E_r\in\mathfrak M\right\}.
$$
The value at the empty set is zero. For disjoint <measurable sets> $A_j$ with union $A$, any finite partition $E_r$ of $A$ gives
$$
\sum_r|\mu(E_r)|\leq\sum_j\sum_r|\mu(E_r\cap A_j)|\leq\sum_j|\mu|(A_j).
$$
Taking the <supremum> proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus\bigcup_{j\leq k}A_j$ gives
$$
|\mu|(A)\geq\sum_{j=1}^k|\mu|(A_j)-\varepsilon.
$$
If one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\varepsilon\downarrow0$ and $k\to\infty$ proves \b[<countable additivity>], including infinite values. Thus $|\mu|$ is a <positive measure>. This uses only <countable additivity> of $\mu$, without assuming a <Hahn decomposition>.
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