Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure

First assume $\eta$ and $\mu$ are finite positive measures. Let $\mathcal{F}$ be the set of measurable $g\ge 0$ such that ${\int }_{A}gd\mu \le \eta (A)$ for every measurable $A$. Then $0\in \mathcal{F}$, every $g\in \mathcal{F}$ satisfies $\int gd\mu \le \eta (X)$, and if $g,h\in \mathcal{F}$ then $\max \{g,h\}\in \mathcal{F}$ because on $B:=\{g\ge h\}$ one has $\max \{g,h\}=g{\chi }_B+h{\chi }_{X\setminus B}$ and the defining inequality splits over $A\cap B$ and $A\setminus B$.

Let $M:={\sup }_{g\in \mathcal{F}}\int gd\mu$, use AC to choose $g_k\in \mathcal{F}$ with $\int g_{k}d\mu \to M$, and put $f_n:={\max }_{k\le n}{g}_k$. By step 1.1 each $f_n$ lies in $\mathcal{F}$ and $f_n\uparrow f:={\sup }_n{f}_n$. Monotone convergence gives $\int fd\mu =M$ and ${\int }_{A}fd\mu ={\lim }_n{\int }_A{f}_{n}d\mu \le \eta (A)$ for every measurable $A$. Because [L2] makes $A\mapsto {\int }_{A}fd\mu$ a measure, the difference $\rho (A):=\eta (A)-{\int }_{A}fd\mu$ defines a finite positive measure with $\eta =fd\mu +\rho$.

The residual $\rho$ is singular to $\mu$. Assume not. For each $m\ge 1$, use AC to choose a Hahn decomposition $X=P_m\bigsqcup N_m$ of the signed measure $\rho -\mu \mathrm{/}m$ using [L6]. If every $\mu (P_m)$ were $0$, then $P:={\bigcup }_m{P}_m$ would be $\mu$-null. For every measurable $A\subseteq X\setminus P$, one has $A\subseteq N_m$ for every $m$, so $(\rho -\mu \mathrm{/}m)(A)\le 0$ and therefore $\rho (A)\le \mu (A)\mathrm{/}m$. Because $\mu$ is finite, this forces $\rho (A)=0$. Thus $\rho$ would be concentrated on the $\mu$-null set $P$, contradicting the assumption that $\rho$ is not singular to $\mu$. Hence $\mu (P_m)>0$ for some $m$; fix such an $m$ and write $\epsilon :=1\mathrm{/}m$, $P:=P_m$. Every measurable $A\subseteq P$ then satisfies $\rho (A)\ge \epsilon \mu (A)$, so for every measurable $A$, $\eta (A)={\int }_{A}fd\mu +\rho (A)\ge {\int }_A(f+\epsilon {\chi }_P)d\mu \mathrm{.}$ Thus $f+\epsilon {\chi }_P\in \mathcal{F}$, while $\int (f+\epsilon {\chi }_P)d\mu =\int fd\mu +\epsilon \mu (P)>M,$ contradicting step 2.1. Therefore $\rho \perp \mu$.

In the finite construction, $Z=\{f=\mathrm{\infty }\}$ is measurable and $\mu$-null by [L8]. Replace $f$ by $0$ on $Z$. This leaves every integral unchanged by [L8], so the density may be taken finite-valued everywhere.

For a positive $\eta$ satisfying the exhaustion hypothesis, put $X_0=\mathrm{\varnothing }$ and $D_n=X_n\setminus X_{n-1}$. Apply the finite construction to the restrictions of $\mu ,\eta$ to each $D_n$. AC selects these decompositions and their singular carriers simultaneously. Write $\eta (E\cap D_n)={\int }_{E\cap D_n}{g}_{n}d\mu +{\sigma }_n(E)$ with $g_n$ nonnegative and finite-valued, and choose a carrier $N_n\subseteq D_n$ with $\mu (N_n)=0$. This last condition follows by intersecting a carrier null for ${\mu }_{D_n}$ with $D_n$; also ${\sigma }_n$ vanishes outside $D_n$ since it is dominated by ${\eta }_{D_n}$. Set $g=g_n$ on $D_n$, ${\eta }_a=gd\mu$ and ${\eta }_s(E)={\sum }_n{\sigma }_n(E)$. The function $g$ is measurable and finite-valued. The sum is a measure: for disjoint $E_j$, exchange the two nonnegative sums in ${\sum }_n{\sum }_j{\sigma }_n(E_j)$, which is legitimate because each is the supremum of finite subsums. Monotone convergence gives ${\eta }_a(E)={\sum }_n{\int }_{E\cap D_n}{g}_{n}d\mu$, hence $\eta ={\eta }_a+{\eta }_s$. The measure ${\eta }_a$ vanishes on $\mu$-null sets by [L8], and ${\eta }_s$ is carried by the $\mu$-null set ${\bigcup }_n{N}_n$. Both components are dominated by $\eta$.

Write $\nu ={\nu }^{+}-{\nu }^{-}$ by [L7]. By [L3], both parts are finite on every $X_n$. Apply step 5.1 to obtain ${\nu }^{\pm }={\alpha }_{\pm }+{\sigma }_{\pm }$ with ${\alpha }_{\pm }=g_{\pm }d\mu$, where $g_{\pm }$ are nonnegative finite-valued, and ${\sigma }_{\pm }\perp \mu$. At least one Jordan part is finite on $X$: on a Hahn partition $P,N$ its total masses are $\nu (P)$ and $-\nu (N)$; if both were infinite, $\nu$ would take both infinite signs, contrary to [L9]. Domination therefore makes at least one sign finite for each of ${\alpha }_{+}-{\alpha }_{-}$ and ${\sigma }_{+}-{\sigma }_{-}$. These differences are signed measures: countable additivity follows by subtracting the nonnegative series, one of which has finite total. Define ${\nu }_a={\alpha }_{+}-{\alpha }_{-}$ and ${\nu }_s={\sigma }_{+}-{\sigma }_{-}$. The same finite-sign bound makes their sum defined and gives $\nu ={\nu }_a+{\nu }_s$. Null-set vanishing gives ${\nu }_a\ll \mu$; the union of carriers of ${\sigma }_{+},{\sigma }_{-}$ gives ${\nu }_s\perp \mu$.

Set $f=g_{+}-g_{-}$. It is measurable and real-valued. To justify its integral, put $h=\min (g_{+},g_{-})$, so $g_{+}=f^{+}+h$ and $g_{-}=f^{-}+h$. At least one of $\int g_{+}d\mu ,\int g_{-}d\mu$ is finite by step 6.1; thus $\int hd\mu$ and at least one of $\int f^{+}d\mu ,\int f^{-}d\mu$ are finite. Additivity of the nonnegative integral now permits cancellation of the finite common term ${\int }_{E}hd\mu$, giving ${\int }_{E}fd\mu ={\alpha }_{+}(E)-{\alpha }_{-}(E)={\nu }_a(E)$ for every $E$. Furthermore $\mathrm{\mid }f\mathrm{\mid }\le g_{+}+g_{-}$ gives ${\int }_{X_n}\mathrm{\mid }f\mathrm{\mid }d\mu \le {\nu }^{+}(X_n)+{\nu }^{-}(X_n)=\mathrm{\mid }\nu \mathrm{\mid }(X_n)<\mathrm{\infty }$. This proves all assertions.