Assuming countable choice, every measure is the sum of its semifinite part and a zero-infinity-valued measure

A measurable subset of a sigma-finite set is sigma-finite. Under [L3], a countable union of sigma-finite measurable sets is sigma-finite: select a finite-measure cover for each member and flatten the resulting $\mathbb{N}\times \mathbb{N}$ family to one countable cover.

If $E$ is sigma-finite for $\mu$, restriction of the measure axioms makes $\mu {\mathrm{\mid }}_E$ a measure, and it is semifinite: a positive-measure subset must meet one member of a finite-measure cover in positive measure, since otherwise [L6] would make it null. The semifinite part of this restriction, evaluated at $E$, is exactly the supremum in [L7]. Thus [L1] applied to $\mu {\mathrm{\mid }}_E$ gives ${\mu }_{\mathrm{s}\mathrm{f}}(E)=\mu (E)$.

For nonuniqueness, take counting measure $\mathrm{\#}$ on the uncountable set $\mathbb{R}$. It is semifinite: every set of positive counting measure is nonempty, so one of its points gives a singleton subset of finite positive measure $1$ by [L8]. Hence [L1] makes its semifinite part itself. Besides the zero measure, define $\eta (E)=0$ for countable $E$ and $+\mathrm{\infty }$ for uncountable $E$. One has $\eta (\mathrm{\varnothing })=0$; for a disjoint sequence, [L4] makes the union countable exactly when every member is countable, so countable additivity has both sides $0$ in that case and both sides $+\mathrm{\infty }$ otherwise. Thus $\eta$ is a measure, and both $0$ and $\eta$ satisfy $\mathrm{\#}={\mathrm{\#}}_{\mathrm{s}\mathrm{f}}+0={\mathrm{\#}}_{\mathrm{s}\mathrm{f}}+\eta$.

Therefore, for a disjoint sequence $(E_k)$, the union is sigma-finite exactly when every $E_k$ is sigma-finite. Hence $\nu ({\bigcup }_k{E}_k)=0$ exactly when every $\nu (E_k)=0$, and otherwise both sides of $\nu ({\bigcup }_k{E}_k)={\sum }_k\nu (E_k)$ are $+\mathrm{\infty }$; also $\nu (\mathrm{\varnothing })=0$.

If $E$ is sigma-finite, step 1.2 gives $({\mu }_{\mathrm{s}\mathrm{f}}+\nu )(E)=\mu (E)+0$; if $E$ is not sigma-finite, then $\mu (E)=+\mathrm{\infty }$ and $({\mu }_{\mathrm{s}\mathrm{f}}+\nu )(E)={\mu }_{\mathrm{s}\mathrm{f}}(E)+(+\mathrm{\infty })=+\mathrm{\infty }$.

Step 2.1 proves that $\nu$ is a zero-infinity-valued measure.

Step 2.2 proves $\mu ={\mu }_{\mathrm{s}\mathrm{f}}+\nu$, and step 1.3 supplies two distinct zero-infinity summands for the same semifinite part, proving the final assertion.