The Radon Nikodym Theorem and Lebesgue Decomposition

Choose measurable sets $P,N$ as in [L1]. Then every measurable subset of $P$ has $\nu$-value $0$.

Let $A\in \mathcal{A}$. Because $A\cap N\subseteq N$, [L1] gives $\mu (A\cap N)=0$, so [L2] yields $\nu (A\cap N)=0$; because $A\cap P\subseteq P$, [L1] also gives $\nu (A\cap P)=0$.

The partition $A=(A\cap P)\bigsqcup (A\cap N)$ therefore gives $\nu (A)=0$. Since $A$ was arbitrary, $\nu =0$.

Suppose first that $\nu$ is signed. If $\nu \ll \mu$ and $\mu (E)=0$, then every measurable $F\subseteq E$ is $\mu$-null, hence $\nu (F)=0$; [L1] therefore gives ${\nu }^{+}(E)=0$ and ${\nu }^{-}(E)=0$. Conversely, if ${\nu }^{+}\ll \mu$ and ${\nu }^{-}\ll \mu$, then $\nu (E)={\nu }^{+}(E)-{\nu }^{-}(E)=0$ on every $\mu$-null set, so $\nu \ll \mu$. Thus $\nu \ll \mu$ iff ${\nu }^{+}\ll \mu$ and ${\nu }^{-}\ll \mu$.

Suppose now that $\nu$ is complex. If $\nu \ll \mu$ and $\mu (E)=0$, then $\nu (E)=0$, so both $\mathrm{Re}\nu (E)$ and $\mathrm{Im}\nu (E)$ vanish; hence $\mathrm{Re}\nu \ll \mu$ and $\mathrm{Im}\nu \ll \mu$. Conversely, if both real and imaginary parts are absolutely continuous, then [L3] gives $\nu (E)=0$ on every $\mu$-null set, so $\nu \ll \mu$.

Still in the signed case, if $\nu \ll \mu$, then step 1.1 and [L1] give $\mathrm{\mid }\nu \mathrm{\mid }(E)={\nu }^{+}(E)+{\nu }^{-}(E)=0$ on every $\mu$-null set, so $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$. Conversely, if $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$, then [L2] gives $\mathrm{\mid }\nu (E)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(E)=0$ on every $\mu$-null set, hence $\nu \ll \mu$. This completes clause 1.

For a complex measure, $\nu \ll \mu$ implies $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$ because if $\mu (E)=0$, then every piece of every measurable partition of $E$ is $\mu$-null, so every partition sum in [L2] is $0$ and therefore $\mathrm{\mid }\nu \mathrm{\mid }(E)=0$. Conversely, if $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$, then [L2] again gives $\mathrm{\mid }\nu (E)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(E)=0$ on every $\mu$-null set, hence $\nu \ll \mu$. Together with step 1.2 this proves clause 2.

Steps 1.1 and 2.1 prove clause 1, and steps 1.2 and 2.2 prove clause 2.

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.

Subtract the two decompositions to obtain ${\nu }_a-{\tilde{\nu }}_a={\tilde{\nu }}_s-{\nu }_s\mathrm{.}$ The left-hand side is absolutely continuous with respect to $\mu$, because differences of absolutely continuous measures are again absolutely continuous.

Choose $\mu$-null sets $N$ and $\tilde{N}$ on which ${\nu }_s$ and ${\tilde{\nu }}_s$ are concentrated. Then ${\tilde{\nu }}_s-{\nu }_s$ is concentrated on $N\cup \tilde{N}$, which is still $\mu$-null, so [L1] makes the right-hand side singular with respect to $\mu$.

The common difference in steps 1.1 and 1.2 is therefore both absolutely continuous and singular with respect to $\mu$, so [L2] forces ${\nu }_a-{\tilde{\nu }}_a=0$. Substituting back into the decomposition identity gives ${\nu }_s-{\tilde{\nu }}_s=0$ as well.

By [L1], write $\nu ={\nu }_a+{\nu }_s$, where ${\nu }_a\ll \mu$, ${\nu }_s\perp \mu$, and ${\nu }_a(E)={\int }_{E}fd\mu$ whenever the integral is defined.

Because $\nu \ll \mu$ and ${\nu }_a\ll \mu$, the difference ${\nu }_s=\nu -{\nu }_a$ is also absolutely continuous with respect to $\mu$. Step 1.1 already gives ${\nu }_s\perp \mu$, so [L2] forces ${\nu }_s=0$. Hence $\nu ={\nu }_a$, and the displayed integral formula for $f$ represents $\nu$ on every measurable set.

Suppose $g$ is another measurable real-valued function with $\nu (E)={\int }_{E}gd\mu$ for every measurable $E$. Fix $n$. For each $m\ge 1$, let $P_{n,m}:=X_n\cap \{f-g\ge 1\mathrm{/}m\},N_{n,m}:=X_n\cap \{g-f\ge 1\mathrm{/}m\}\mathrm{.}$ Applying the two representation formulas to $P_{n,m}$ gives $0={\int }_{P_{n,m}}(f-g)d\mu \ge \mu (P_{n,m})\mathrm{/}m,$ so $\mu (P_{n,m})=0$; the same argument with $N_{n,m}$ gives $\mu (N_{n,m})=0$. Therefore $X_n\cap \{f\mathrm{\ne }g\}={\bigcup }_{m\ge 1}{P}_{n,m}\cup {\bigcup }_{m\ge 1}{N}_{n,m}$ is $\mu$-null. Since $X={\bigcup }_n{X}_n$, the functions $f$ and $g$ agree $\mu$-almost everywhere on $X$.

If $\mathrm{\mid }\nu \mathrm{\mid }(X)<+\mathrm{\infty }$, let $\nu ={\nu }^{+}-{\nu }^{-}$ be the Jordan decomposition from [L4]. Then [L3] makes both ${\nu }^{+}$ and ${\nu }^{-}$ absolutely continuous with respect to $\mu$. Apply the positive-measure part of [L1] to obtain nonnegative measurable functions $f^{+},f^{-}$ with ${\nu }^{+}(E)={\int }_E{f}^{+}d\mu ,{\nu }^{-}(E)={\int }_E{f}^{-}d\mu (E\in \mathcal{A})\mathrm{.}$ Then $f^{+}-f^{-}$ is another representative of $\nu$, so step 3.1 gives $f=f^{+}-f^{-}$ $\mu$-almost everywhere. Hence $\mathrm{\mid }f\mathrm{\mid }\le f^{+}+f^{-}$ almost everywhere, and $\int \mathrm{\mid }f\mathrm{\mid }d\mu \le \int f^{+}d\mu +\int f^{-}d\mu ={\nu }^{+}(X)+{\nu }^{-}(X)=\mathrm{\mid }\nu \mathrm{\mid }(X)<+\mathrm{\infty }\mathrm{.}$ Therefore $f\in L^1(\mu )$ by the definition of integrability.

Steps 2.1, 3.1, and 4.1 prove existence, almost-everywhere uniqueness, and the finite-measure integrability clause.

By [L1] and [L2], the signed measures $\mathrm{Re}\nu$ and $\mathrm{Im}\nu$ are absolutely continuous with respect to $\mu$. By [L5], partition sums for their total variations are bounded by those of $\mathrm{\mid }\nu \mathrm{\mid }$, so both have finite total variation. Choose an increasing exhaustion $(X_n)$ with $\mu (X_n)<\mathrm{\infty }$; it is then common to $\mu$ and both signed measures. Applying [L3], choose real-valued $u,v\in L^1(\mu )$ such that $\mathrm{Re}\nu (E)={\int }_{E}ud\mu ,\mathrm{Im}\nu (E)={\int }_{E}vd\mu (E\in \mathcal{A})\mathrm{.}$

Put $h:=u+iv$. Then [L4] gives $h\in L^1(\mu )$, and for every measurable $E$ one has $\nu (E)=\mathrm{Re}\nu (E)+i\mathrm{Im}\nu (E)={\int }_{E}ud\mu +i{\int }_{E}vd\mu ={\int }_{E}hd\mu \mathrm{.}$ If $h^{\mathrm{\prime }}$ is another such density, then its real and imaginary parts give alternative signed densities for $\mathrm{Re}\nu$ and $\mathrm{Im}\nu$, so the uniqueness part of [L3] makes $h^{\mathrm{\prime }}=h$ $\mu$-almost everywhere.

The measurable-set identity $\nu (E)={\int }_{E}hd\mu$ is exactly [L1].

If $g={\sum }_{j=1}^m{c}_j{\chi }_{E_j}$ is canonical disjoint and each $\mathrm{\mid }\nu \mathrm{\mid }(E_j)<+\mathrm{\infty }$, then [L3] and step 1.1 give $\int gd\nu ={\sum }_{j=1}^m{c}_j\nu (E_j)={\sum }_{j=1}^m{c}_j{\int }_{E_j}hd\mu \mathrm{.}$ If, in addition, the Lebesgue integral of $gh$ is defined, then [L4] identifies the same finite sum with $\int ghd\mu$.

Choose representatives $h_j$ of $d{\nu }_j\mathrm{/}d\mu$ for $j=1,2$. For every measurable set $E$, [L1] gives $({\nu }_1+{\nu }_2)(E)={\nu }_1(E)+{\nu }_2(E)={\int }_E{h}_{1}d\mu +{\int }_E{h}_{2}d\mu ={\int }_E(h_1+h_2)d\mu \mathrm{.}$

The function $h_1+h_2$ is therefore a density for ${\nu }_1+{\nu }_2$, so [L2] yields $\frac{d({\nu }_1+{\nu }_2)}{d\mu }=h_1+h_2=\frac{d{\nu }_1}{d\mu }+\frac{d{\nu }_2}{d\mu }\mu \text{-almost everywhere.}$

First assume that $\nu$ is a positive measure. Choose the nonnegative representatives $h$ of $d\nu \mathrm{/}d\mu$ and $k$ of $d\mu \mathrm{/}d\lambda$ furnished by the positive-measure case of the Radon-Nikodym theorem. For every measurable set $E$, that theorem gives $\nu (E)={\int }_{E}hd\mu ,\mu (E)={\int }_{E}kd\lambda \mathrm{.}$ Applying [L1] to the nonnegative density $h$ therefore yields $\nu (E)={\int }_{E}hkd\lambda (E\in \mathcal{A})\mathrm{.}$ Hence [L2] gives $d\nu \mathrm{/}d\lambda =hk$ $\lambda$-almost everywhere in the positive case.

Now assume that $\nu$ is a signed measure. By [L4], write $\nu ={\nu }^{+}-{\nu }^{-}$. Because $\nu \ll \mu$, [L3] gives ${\nu }^{\pm }\ll \mu$. Apply step 1.1 to ${\nu }^{+}$ and ${\nu }^{-}$ separately to obtain nonnegative representatives $h^{+},h^{-}$ with $\frac{d{\nu }^{+}}{d\lambda }=h^{+}k,\frac{d{\nu }^{-}}{d\lambda }=h^{-}k\lambda \text{-almost everywhere.}$ Then $h:=h^{+}-h^{-}$ represents $d\nu \mathrm{/}d\mu$, while $(h^{+}-h^{-})k$ represents $d\nu \mathrm{/}d\lambda$. Uniqueness from [L2] therefore gives $\frac{d\nu }{d\lambda }=hk=\frac{d\nu }{d\mu }\frac{d\mu }{d\lambda }\lambda \text{-almost everywhere.}$

Finally assume that $\nu$ is a finite complex measure, and choose a representative $h=u+iv$ of $d\nu \mathrm{/}d\mu$. By [L5], the finite signed measures $\mathrm{Re}\nu$ and $\mathrm{Im}\nu$ are both absolutely continuous with respect to $\mu$, and the measurable-set identity for $h$ shows that $u$ and $v$ represent $d(\mathrm{Re}\nu )\mathrm{/}d\mu$ and $d(\mathrm{Im}\nu )\mathrm{/}d\mu$. Applying step 2.1 to those signed measures gives $\frac{d(\mathrm{Re}\nu )}{d\lambda }=u\frac{d\mu }{d\lambda },\frac{d(\mathrm{Im}\nu )}{d\lambda }=v\frac{d\mu }{d\lambda }\lambda \text{-almost everywhere.}$ Therefore, for every measurable set $E$, $\nu (E)=\mathrm{Re}\nu (E)+i\mathrm{Im}\nu (E)={\int }_{E}u\frac{d\mu }{d\lambda }d\lambda +i{\int }_{E}v\frac{d\mu }{d\lambda }d\lambda ={\int }_{E}h\frac{d\mu }{d\lambda }d\lambda \mathrm{.}$ So $h(d\mu \mathrm{/}d\lambda )$ represents $d\nu \mathrm{/}d\lambda$, and [L2] yields $\frac{d\nu }{d\lambda }=h\frac{d\mu }{d\lambda }=\frac{d\nu }{d\mu }\frac{d\mu }{d\lambda }\lambda \text{-almost everywhere.}$

Choose increasing finite-measure exhaustions $(A_n)$ for $\mu$ and $(B_n)$ for $\nu$, and put $X_n:=A_n\cap B_n$. After replacing both exhaustions by finite unions, $(X_n)$ is increasing, covers $X$, and is finite for both measures. Apply [L1] to the chain $\mu \ll \nu \ll \mu$ on this common exhaustion. Then $\frac{d\mu }{d\mu }=\frac{d\mu }{d\nu }\frac{d\nu }{d\mu }\mu \text{-almost everywhere.}$ By [L2], $d\mu \mathrm{/}d\mu =1$ almost everywhere, so $\frac{d\mu }{d\nu }\frac{d\nu }{d\mu }=1\mu \text{-almost everywhere.}$

Interchanging the roles of $\mu$ and $\nu$ gives $\frac{d\nu }{d\mu }\frac{d\mu }{d\nu }=1\nu \text{-almost everywhere.}$ Because $\mu$ and $\nu$ have the same null sets, the two almost-everywhere conclusions are equivalent.