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 $M:={\sup }_{g\in \mathcal{F}}\int gd\mu$, 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.

Choose measurable $\mu$-null carriers $N_1,N_2$ for ${\nu }_s,{\tilde{\nu }}_s$, respectively. These are just two existential witnesses, so no choice axiom is needed. Let $N=N_1\cup N_2$, again measurable and $\mu$-null. Both singular parts vanish on every measurable subset of $X\setminus N$, and both absolutely continuous parts vanish on every measurable subset of $N$.

For every measurable $E$, evaluate each given decomposition on $E\setminus N$. It gives $\nu (E\setminus N)={\nu }_a(E\setminus N)={\tilde{\nu }}_a(E\setminus N)$. As both absolutely continuous parts vanish on $E\cap N$, additivity yields ${\nu }_a(E)=\nu (E\setminus N)={\tilde{\nu }}_a(E)$. No difference of infinite values occurs.

Similarly, evaluating on $E\cap N$ gives $\nu (E\cap N)={\nu }_s(E\cap N)={\tilde{\nu }}_s(E\cap N)$. Both singular parts vanish on $E\setminus N$, so ${\nu }_s(E)=\nu (E\cap N)={\tilde{\nu }}_s(E)$. Since $E$ was arbitrary, both components are unique.

By [L1], write $\nu ={\nu }_a+{\nu }_s$, where ${\nu }_a\ll \mu$, ${\nu }_s\perp \mu$, and ${\nu }_a(E)={\int }_{E}fd\mu$ for every measurable $E$, with $f$ real-valued and ${\int }_{X_n}\mathrm{\mid }f\mathrm{\mid }d\mu <\mathrm{\infty }$.

Because $\nu \ll \mu$ and ${\nu }_a\ll \mu$, on each $\mu$-null set $E$ the decomposition gives $0=0+{\nu }_s(E)$, so ${\nu }_s\ll \mu$ without subtracting infinite values. 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 $n$. The formula on $X_n$ gives ${\int }_{X_n}gd\mu =\nu (X_n)\in \mathbb{R}$, so both positive and negative integrals of $g$ there are finite. Thus $f,g\in L^1(X_n)$, and [L3] permits subtraction of their integrals over every measurable subset of $X_n$. 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 $\mathrm{\mid }\nu \mathrm{\mid }(X)<\mathrm{\infty }$, use the representation on the measurable sets $P=\{f\ge 0\}$ and $N=\{f<0\}$. It gives $\int f^{+}d\mu =\nu (P)$ and $\int f^{-}d\mu =-\nu (N)$. Both are finite: $\mathrm{\mid }\nu (E)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(X)$ for every $E$, directly from the partition definition of variation. Hence $f\in L^1(\mu )$ by [L3].

Steps 2.1, 3.1, and 3.2 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. Under AC as in [L6], 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].

For each nonzero level set $E_j$, the hypothesis $\mathrm{\mid }\nu \mathrm{\mid }(E_j)<+\mathrm{\infty }$ makes $h{\mathbf{1}}_{E_j}$ integrable. In the signed case, apply step 1.1 to $E_j\cap \{h\ge 0\}$ and $E_j\cap \{h<0\}$: the positive and negative integrals are finite because the corresponding values of $\nu$ have absolute value at most $\mathrm{\mid }\nu \mathrm{\mid }(E_j)$. In the finite complex case, $h\in L^1(\mu )$ already by the definition of its derivative.

If $g={\sum }_{j=1}^m{c}_j{\chi }_{E_j}$ is canonical disjoint, then [L3] and step 1.1 give $\int gd\nu ={\sum }_{j=1}^m{c}_j\nu (E_j)={\sum }_{j=1}^m{c}_j{\int }_{E_j}hd\mu \mathrm{.}$ By step 2.1, every $h{\mathbf{1}}_{E_j}$ is in $L^1(\mu )$, so [L4] identifies this finite sum with $\int ghd\mu$.

The sum ${\nu }_1+{\nu }_2$ is finite-valued and countably additive, and vanishes on every $\mu$-null set, so is a finite signed or finite complex measure of the same type and is absolutely continuous. In the signed case [L3] gives finite total variation for both summands and the sum. Any increasing finite-measure exhaustion for $\mu$ is therefore common to these measures. Under AC apply [L1] to choose $h_1,h_2\in L^1(\mu )$ representing ${\nu }_1,{\nu }_2$, and to obtain the almost-everywhere unique density of their sum.

By [L2], $h_1+h_2$ is L1, and for every measurable $E$, $({\nu }_1+{\nu }_2)(E)={\int }_E{h}_{1}d\mu +{\int }_E{h}_{2}d\mu ={\int }_E(h_1+h_2)d\mu \mathrm{.}$ Thus it is a density of the sum. Uniqueness in [L1] gives the asserted equality of RN classes.

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.

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. Under AC as in [L3], 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.

In the finite signed case, [L3] gives an integrable real-valued representative $h$ of $d\nu \mathrm{/}d\mu$, and [L1] applied to that density yields exactly $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu (E\in \mathcal{A})\mathrm{.}$

In the finite complex case, [L3] gives an integrable complex representative $h$ of $d\nu \mathrm{/}d\mu$, and [L2] applied to that density yields the same formula $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu (E\in \mathcal{A})\mathrm{.}$