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 $M:={\sup }_{g\in \mathcal{F}}\int gd\mu$, 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 $m\ge 1$, 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 $\rho \perp \mu$.

Now assume $\eta$ is a positive measure satisfying the same finite-exhaustion hypothesis as $\nu$. For each $n$, [L1] makes ${\mu }_{X_n}$ a finite positive measure, and ${\eta }_{X_n}(A):=\eta (A\cap X_n)$ is finite because $\eta (X_n)<+\mathrm{\infty }$. Apply steps 1.1-3.1 to $({\mu }_{X_n},{\eta }_{X_n})$. This yields ${\eta }_{X_n}={\alpha }_n+{\sigma }_n,{\alpha }_n=g_{n}d{\mu }_{X_n},{\alpha }_n\ll {\mu }_{X_n},{\sigma }_n\perp {\mu }_{X_n},$ with $g_n\ge 0$ measurable. If $m<n$, restricting the $n$th decomposition to $X_m$ gives another absolutely-continuous/singular decomposition of ${\eta }_{X_m}$ relative to ${\mu }_{X_m}$. Subtracting the two decompositions and applying [L8] to the common difference shows ${\alpha }_n(E\cap X_m)={\alpha }_m(E),{\sigma }_n(E\cap X_m)={\sigma }_m(E)(E\in \mathcal{A})\mathrm{.}$ For each $n\ge 1$, replacing $g_n$ on $X_{n-1}$ by $g_{n-1}$ does not change ${\alpha }_n$, so we may assume $g_n=g_{n-1}$ pointwise on $X_{n-1}$. Define a measurable nonnegative function $f$ by $f=g_n$ on $X_n$.

Let ${\eta }_a:=fd\mu$. By [L2], this is a positive measure, and for every measurable $E$ and every $n$ the overlap construction gives ${\eta }_a(E\cap X_n)={\alpha }_n(E)\mathrm{.}$ The compatibility in step 4.1 makes the sequence ${\sigma }_n(E)$ increasing, so define ${\eta }_s(E):={\lim }_{n\to \mathrm{\infty }}{\sigma }_n(E)\mathrm{.}$ An increasing compatible limit of the measures ${\sigma }_n$ is a positive measure: countable additivity follows by applying it to each ${\sigma }_n$ and then using monotone convergence for the resulting nonnegative double sequence. Taking limits in $\eta (E\cap X_n)={\alpha }_n(E)+{\sigma }_n(E)$ and using [L5] yields $\eta (E)={\eta }_a(E)+{\eta }_s(E)$ without any subtraction of infinite values. Choose $\mu$-null sets $N_n$ on which each ${\sigma }_n$ is concentrated and put $N:={\bigcup }_n{N}_n$. Then $\mu (N)=0$, and if $A\subseteq X\setminus N$ is measurable, every ${\sigma }_n(A)=0$, hence ${\eta }_s(A)=0$. Thus ${\eta }_s\perp \mu$, and every sigma-finite positive $\eta$ admits a decomposition $\eta ={\eta }_a+{\eta }_s,{\eta }_a=fd\mu ,{\eta }_a\ll \mu ,{\eta }_s\perp \mu \mathrm{.}$

Return to the given signed measure $\nu$. By [L7], write $\nu ={\nu }^{+}-{\nu }^{-}$. The formula in [L3] and the hypothesis $\mathrm{\mid }\nu \mathrm{\mid }(X_n)<+\mathrm{\infty }$ show that ${\nu }^{+}(X_n)$ and ${\nu }^{-}(X_n)$ are finite for every $n$, so both Jordan parts satisfy the same finite-exhaustion hypothesis as $\nu$. Apply step 5.1 to ${\nu }^{+}$ and ${\nu }^{-}$ to obtain ${\nu }^{\pm }={\alpha }^{\pm }+{\sigma }^{\pm },{\alpha }^{\pm }=f^{\pm }d\mu ,{\alpha }^{\pm }\ll \mu ,{\sigma }^{\pm }\perp \mu ,$ with $f^{\pm }\ge 0$ measurable. Put $f:=f^{+}-f^{-},{\nu }_a:={\alpha }^{+}-{\alpha }^{-},{\nu }_s:={\sigma }^{+}-{\sigma }^{-}\mathrm{.}$ Then $\nu ={\nu }_a+{\nu }_s$, the difference of two absolutely continuous measures is absolutely continuous, and if $N^{\pm }$ are $\mu$-null sets on which ${\sigma }^{\pm }$ are concentrated, then ${\nu }_s$ is concentrated on $N^{+}\cup N^{-}$ and hence singular with respect to $\mu$. Finally, ${\nu }_a(E)={\int }_E{f}^{+}d\mu -{\int }_E{f}^{-}d\mu ={\int }_{E}fd\mu$ whenever the integral is defined. This is exactly the claimed Lebesgue decomposition with density.