Radon-Nikodym derivatives satisfy the chain rule along nu << mu << lambda

Apply [L1] under AC to choose finite-valued $k$ representing $d\mu \mathrm{/}d\lambda$. It may be made nonnegative everywhere: on $E_{n,m}=X_n\cap \{k\le -1\mathrm{/}m\}$, finite-piece integrability gives $0\le \mu (E_{n,m})={\int }_{E_{n,m}}kd\lambda \le -\lambda (E_{n,m})\mathrm{/}m$, so each set is null. Their countable union is $\{k<0\}$. Replacing $k$ there by zero leaves its integrals unchanged by [L4]. Thus $\mu =kd\lambda$ with finite nonnegative $k$.

If $\nu$ is signed, [L1] gives a finite real density $h$ for $\nu$ with respect to $\mu$. Its integral on $X$ is defined, so at least one of $\int h^{+}d\mu$ and $\int h^{-}d\mu$ is finite. Since $(hk{)}^{\pm }=h^{\pm }k$, [L2] gives ${\int }_E(hk{)}^{\pm }d\lambda ={\int }_E{h}^{\pm }d\mu$ for every measurable $E$. The same finite-sign bound permits subtraction, proving ${\int }_{E}hkd\lambda =\nu (E)$. By [L5], $\nu \ll \lambda$ and the given exhaustion permits [L1] for this pair. Its uniqueness clause identifies $hk=d\nu \mathrm{/}d\lambda$ almost everywhere.

If $\nu$ is finite complex, [L1] gives $h=u+iv\in L^1(\mu )$. By [L2], $\int \mathrm{\mid }hk\mathrm{\mid }d\lambda =\int \mathrm{\mid }h\mathrm{\mid }d\mu <\mathrm{\infty }$. Apply [L2] to the four nonnegative parts of $u,v$ and combine their finite integrals by [L3]; this yields ${\int }_{E}hkd\lambda =\nu (E)$ for every $E$. Transitivity [L5] and the complex uniqueness clause [L1] again give $hk=d\nu \mathrm{/}d\lambda$ almost everywhere.

The product is independent of finite-valued representatives. If $h^{\mathrm{\prime }}$ agrees with $h$ outside a measurable $\mu$-null set $Z$, then ${\int }_{Z}kd\lambda =\mu (Z)=0$, so [L4] gives $k=0$ $\lambda$-almost everywhere on $Z$. Hence $h^{\mathrm{\prime }}k=hk$ $\lambda$-almost everywhere. Changing $k$ on a $\lambda$-null set also leaves the product class unchanged. Thus the preceding identities give the stated identity of RN classes in both cases.