Proof of The Radon–Nikodym Theorem for a σ-Finite Measure, and Uniqueness of Densities

Throughout, arithmetic and order in $\mathbb{R}$ use the field and ordered-ring laws (The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii)) and the rules Rules of Arithmetic and Order in an Ordered Field (iv), Rules of Arithmetic and Order in an Ordered Field (vi) and Rules of Arithmetic and Order in an Ordered Field (viii). For a measure $\rho$ on $(X,\mathcal{F})$, a measurable $h:X\to\mathbb{R}$ with $h\ge0$ lies in $\mathcal{M}^{+}$, and for $A\in\mathcal{F}$ it is integrable on $A$ with respect to $\rho$ exactly when its integral over $A$ in $\mathcal{M}^{+}$ is finite, the two integrals then agreeing (Integrals over Measurable Subsets (iii)); and integrals over $A$ are integrals over $X$ of the product with $\mathbf{1}_{A}$ (Integrals over Measurable Subsets (i), Integrals over Measurable Subsets (ii)).

Uniqueness: a local statement. Let $B\in\mathcal{F}$ be a set on which $f$ and $g$ are integrable, with $\int_{A}f\,d\mu=\int_{A}g\,d\mu$ for the integrals as integrable functions for every $A\in\mathcal{F}$ with $A\subseteq B$, and let

$$C=B\cap\{x\in X:f(x)>g(x)\}.$$

We show $\mu(C)=0$. The set $C$ lies in $\mathcal{F}$, as $f-g$ is measurable (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v)). Since $f\mathbf{1}_{B}\in\mathcal{L}^{1}(\mu)$ and $|f\mathbf{1}_{C}|\le|f\mathbf{1}_{B}|$, also $f\mathbf{1}_{C}\in\mathcal{L}^{1}(\mu)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (v), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x)), and likewise $g\mathbf{1}_{C}\in\mathcal{L}^{1}(\mu)$. By linearity (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)),

$$\int_{X}(f-g)\mathbf{1}_{C}\,d\mu=\int_{C}f\,d\mu-\int_{C}g\,d\mu=0 .$$

The function $(f-g)\mathbf{1}_{C}$ is nonnegative, so its integral in $\mathcal{M}^{+}$ is $0$ as well (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)), and the set where it is positive, which is $C$, has measure $0$ (The Lebesgue Integral and Null Sets (iv)).

Uniqueness for integrable functions. If $f,g\in\mathcal{L}^{1}(\mu)$, both are integrable on $B=X$ (Integrals over Measurable Subsets (vi)), so $\mu(\{f>g\})=0$ by the local statement, and $\mu(\{g>f\})=0$ by the same argument with $f$ and $g$ exchanged. The set $\{f\neq g\}$ is the union of these two sets of $\mathcal{F}$, so it lies in $\mathcal{F}$ and has measure $0$ (Basic Properties of a Measure (vii)). This proves (i).

Uniqueness for nonnegative functions. Let $f,g\ge0$ be measurable, with equal integrals in $\mathcal{M}^{+}$ over every $A\in\mathcal{F}$. As $\mu$ is $\sigma$-finite, there is a sequence $(X_{k})$ in $\mathcal{F}$ with $\bigcup_{k}X_{k}=X$ and $\mu(X_{k})<\infty$ for every $k$; we fix one. Let $E=\{f>g\}\in\mathcal{F}$, and for $k,j\in\mathbb{N}$ let

$$B_{k,j}=X_{k}\cap\{x\in X:g(x)\le j\}\in\mathcal{F}.$$

Since $g\mathbf{1}_{B_{k,j}}\le j\mathbf{1}_{X_{k}}$, we get $\int_{B_{k,j}}g\,d\mu\le j\,\mu(X_{k})<\infty$ (Basic Properties of the Integral of Nonnegative Measurable Maps (iii), Basic Properties of the Integral of Nonnegative Measurable Maps (iv), Basic Properties of the Integral of Nonnegative Measurable Maps (ii)), and $\int_{A}f\,d\mu=\int_{A}g\,d\mu$ is finite for every $A\in\mathcal{F}$ with $A\subseteq B_{k,j}$ (Integrals over Measurable Subsets (viii)). So $f$ and $g$ are integrable on $B_{k,j}$ with equal integrals over its measurable subsets, and the local statement gives $\mu(B_{k,j}\cap E)=0$. For fixed $k$ the sets $B_{k,j}\cap E$ increase with $j$, and their union is $X_{k}\cap E$ because $g$ is real-valued; hence $\mu(X_{k}\cap E)=\sup_{j}\mu(B_{k,j}\cap E)=0$ (Basic Properties of a Measure (viii)). As $E=\bigcup_{k}(X_{k}\cap E)$, countable subadditivity gives $\mu(E)=0$ (Basic Properties of a Measure (vi)). Exchanging $f$ and $g$ and combining as before proves (ii). All exceptional sets here are explicit members of $\mathcal{F}$; no choice is used.

Existence when the measure is finite. We first prove: if $\rho$ is a finite measure on $(X,\mathcal{F})$ and $\nu$ a finite measure with $\nu\ll\rho$, there is a measurable $f:X\to\mathbb{R}$ with $f\ge0$ and $\nu(A)=\int_{A}f\,d\rho$ in $\mathcal{M}^{+}$ for every $A\in\mathcal{F}$. Here $\nu(A)\le\nu(X)<\infty$ for every $A$ (Basic Properties of a Measure (iii)). Let $\mathcal{K}$ be the set of measurable $g:X\to\mathbb{R}$ with $g\ge0$ and

$$\int_{A}g\,d\rho\le\nu(A)\qquad(A\in\mathcal{F}).$$

The zero function lies in $\mathcal{K}$ (Basic Properties of the Integral of Nonnegative Measurable Maps (v)), and $\int_{X}g\,d\rho\le\nu(X)$ for $g\in\mathcal{K}$; let $\alpha=\sup\{\int_{X}g\,d\rho:g\in\mathcal{K}\}$, a real number with $0\le\alpha\le\nu(X)$.

(a) $\mathcal{K}$ is closed under maxima. Let $g,g'\in\mathcal{K}$ and $A\in\mathcal{F}$. The function $\max(g,g')$ is measurable (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)), and $A_{1}=A\cap\{g\ge g'\}$ and $A_{2}=A\cap\{g<g'\}$ are disjoint sets of $\mathcal{F}$ with union $A$, on which $\max(g,g')$ equals $g$ and $g'$ respectively. By Integrals over Measurable Subsets (vii) and the countable additivity of $\nu$ (Basic Properties of a Measure (ii)),

$$\int_{A}\max(g,g')\,d\rho=\int_{A_{1}}g\,d\rho+\int_{A_{2}}g'\,d\rho\le\nu(A_{1})+\nu(A_{2})=\nu(A),$$

so $\max(g,g')\in\mathcal{K}$.

(b) A maximizing sequence (countable choice). For $n\in\mathbb{N}$ the set $\mathcal{K}_{n}=\{g\in\mathcal{K}:\int_{X}g\,d\rho>\alpha-1/n\}$ is nonempty, by the definition of $\alpha$. Under the axiom of dependent choice, countable choice (Countable Choice Follows from Dependent Choice (i), with $b$ the set of measurable functions $X\to\mathbb{R}$) gives a sequence $(g_{n})$ with $g_{n}\in\mathcal{K}_{n}$ for every $n$. Put $f_{1}=g_{1}$ and $f_{n+1}=\max(f_{n},g_{n+1})$. By (a) and induction, $f_{n}\in\mathcal{K}$; the sequence $(f_{n})$ is increasing; and $\alpha-1/n<\int_{X}g_{n}\,d\rho\le\int_{X}f_{n}\,d\rho\le\alpha$. Let $\bar f:X\to[0,\infty]$ be the function $\bar f(x)=\sup_{n}f_{n}(x)$. By the monotone convergence theorem, applied to the increasing sequences $(f_{n}\mathbf{1}_{A})$ with supremum $\bar f\mathbf{1}_{A}$, $\bar f\in\mathcal{M}^{+}$ and

$$\int_{A}\bar f\,d\rho=\sup_{n}\int_{A}f_{n}\,d\rho\le\nu(A)\qquad(A\in\mathcal{F}),\qquad\int_{X}\bar f\,d\rho=\alpha$$

(Monotone Convergence Theorem (i), Monotone Convergence Theorem (ii)).

(c) A real-valued maximal element. As $\int_{X}\bar f\,d\rho=\alpha<\infty$, the set $N=\{\bar f=\infty\}$ lies in $\mathcal{F}$ and $\rho(N)=0$ (The Lebesgue Integral and Null Sets (v)). Let $f=\bar f\mathbf{1}_{X\setminus N}$; it is measurable (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (iii)), real-valued and nonnegative, hence measurable as a function into $\mathbb{R}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)). For $A\in\mathcal{F}$ the functions $f\mathbf{1}_{A}$ and $\bar f\mathbf{1}_{A}$ agree outside the null set $N$, so $\int_{A}f\,d\rho=\int_{A}\bar f\,d\rho$ (The Lebesgue Integral and Null Sets (ii)). Hence $f\in\mathcal{K}$ and $\int_{X}f\,d\rho=\alpha$.

(d) $f$ is a density (Hahn decomposition). We show $\nu(X)=\alpha$. Suppose instead that $\delta=\nu(X)-\alpha>0$, and put

$$\varepsilon=\frac{\delta}{2(\rho(X)+1)}>0,\qquad\text{so that}\qquad\varepsilon\,\rho(X)<\delta .$$

The function $f+\varepsilon$ lies in $\mathcal{M}^{+}$ with $\int_{X}(f+\varepsilon)\,d\rho=\alpha+\varepsilon\rho(X)<\infty$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i), Basic Properties of the Integral of Nonnegative Measurable Maps (iv), Basic Properties of the Integral of Nonnegative Measurable Maps (ii)), so the measure $(f+\varepsilon)\rho$ with density $f+\varepsilon$ is a finite measure (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (v)), and

$$\lambda(A)=\nu(A)-\int_{A}(f+\varepsilon)\,d\rho\qquad(A\in\mathcal{F})$$

defines a finite signed measure $\lambda$ (Finite Signed Measures from Finite Measures and from Integrals (i)). Under the axiom of dependent choice, $\lambda$ has a Hahn set $P$ (Hahn and Jordan Decompositions of a Finite Signed Measure (i)). For $A\in\mathcal{F}$ we have $A\cap P\subseteq P$, so $\lambda(A\cap P)\ge0$, that is, $\int_{A\cap P}(f+\varepsilon)\,d\rho\le\nu(A\cap P)$. Let $g=f+\varepsilon\mathbf{1}_{P}$, a nonnegative measurable function equal to $f+\varepsilon$ on $P$ and to $f$ off $P$. Since $f\in\mathcal{K}$,

$$\int_{A}g\,d\rho=\int_{A\cap P}(f+\varepsilon)\,d\rho+\int_{A\setminus P}f\,d\rho\le\nu(A\cap P)+\nu(A\setminus P)=\nu(A)$$

(Integrals over Measurable Subsets (vii), Basic Properties of a Measure (ii)). Thus $g\in\mathcal{K}$, and $\alpha\ge\int_{X}g\,d\rho=\alpha+\varepsilon\rho(P)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i), Basic Properties of the Integral of Nonnegative Measurable Maps (iv), Basic Properties of the Integral of Nonnegative Measurable Maps (ii)), whence $\rho(P)=0$ and therefore $\nu(P)=0$, as $\nu\ll\rho$. On the other hand, the Hahn property applied to $X\setminus P$ itself gives $\lambda(X\setminus P)\le0$, that is, $\nu(X\setminus P)\le\int_{X\setminus P}f\,d\rho+\varepsilon\rho(X\setminus P)\le\alpha+\varepsilon\rho(X)$ (Integrals over Measurable Subsets (viii), Integrals over Measurable Subsets (v), Basic Properties of a Measure (iii)). Then

$$\nu(X)=\nu(P)+\nu(X\setminus P)\le\alpha+\varepsilon\rho(X)<\alpha+\delta=\nu(X)$$

(Basic Properties of a Measure (ii)), a contradiction. Hence $\nu(X)=\alpha$. Finally, for $A\in\mathcal{F}$, $\int_{A}f\,d\rho\le\nu(A)$ and $\int_{X\setminus A}f\,d\rho\le\nu(X\setminus A)$, while

$$\int_{A}f\,d\rho+\int_{X\setminus A}f\,d\rho=\alpha=\nu(X)=\nu(A)+\nu(X\setminus A)$$

(Integrals over Measurable Subsets (vii), Integrals over Measurable Subsets (v), Basic Properties of a Measure (ii)). Both inequalities between real numbers are therefore equalities, and $\nu(A)=\int_{A}f\,d\rho$.

Existence when $\mu$ is $\sigma$-finite. Fix a sequence $(X_{m})$ in $\mathcal{F}$ with $\bigcup_{m}X_{m}=X$ and $\mu(X_{m})<\infty$, and let $(B_{m})$ be its disjointing sequence: the $B_{m}$ lie in $\mathcal{F}$, are pairwise disjoint, have union $X$, and $B_{m}\subseteq X_{m}$, so $\mu(B_{m})<\infty$ (The Disjointing Sequence of a Sequence of Sets (v), The Disjointing Sequence of a Sequence of Sets (iii), The Disjointing Sequence of a Sequence of Sets (iv), The Disjointing Sequence of a Sequence of Sets (ii)). For $m\in\mathbb{N}$ let $\mu_{m}=\mathbf{1}_{B_{m}}\mu$ be the measure with density $\mathbf{1}_{B_{m}}$, a measure (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (v)) with

$$\mu_{m}(A)=\int_{X}\mathbf{1}_{B_{m}}\mathbf{1}_{A}\,d\mu=\mu(A\cap B_{m})\qquad(A\in\mathcal{F})$$

(Basic Properties of the Integral of Nonnegative Measurable Maps (ii)), so finite; and let $\nu_{m}(A)=\nu(A\cap B_{m})$, a finite measure (Restricting a Measure to a Measurable Set (iii), Restricting a Measure to a Measurable Set (iv)). If $\mu_{m}(A)=0$ then $\mu(A\cap B_{m})=0$, so $\nu_{m}(A)=0$: thus $\nu_{m}\ll\mu_{m}$. By the finite case, the set $\mathcal{D}_{m}$ of measurable $h:X\to\mathbb{R}$ with $h\ge0$ and $\nu_{m}(A)=\int_{A}h\,d\mu_{m}$ for every $A\in\mathcal{F}$ is nonempty. Under the axiom of dependent choice, countable choice (Countable Choice Follows from Dependent Choice (i)) gives a sequence $(h_{m})$ with $h_{m}\in\mathcal{D}_{m}$ for every $m$. For $A\in\mathcal{F}$ and $m\in\mathbb{N}$, integrating against the density (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (vi)),

$$\nu(A\cap B_{m})=\int_{X}h_{m}\mathbf{1}_{A}\,d\mu_{m}=\int_{X}h_{m}\mathbf{1}_{B_{m}}\mathbf{1}_{A}\,d\mu .$$

Let $f=\sum_{m}h_{m}\mathbf{1}_{B_{m}}$, pointwise in $[0,\infty]$. Each $x\in X$ lies in exactly one $B_{k}$, where only the $k$-th term is nonzero, so $f(x)=h_{k}(x)$: $f$ is real-valued and nonnegative. By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ii), applied to the sequences $(h_{m}\mathbf{1}_{B_{m}}\mathbf{1}_{A})_{m}$ with sum $f\mathbf{1}_{A}$, $f\in\mathcal{M}^{+}$ and

$$\int_{A}f\,d\mu=\sum_{m\in\mathbb{N}}\int_{X}h_{m}\mathbf{1}_{B_{m}}\mathbf{1}_{A}\,d\mu=\sum_{m\in\mathbb{N}}\nu(A\cap B_{m})=\nu(A),$$

the last equality by the countable additivity of $\nu$, the sets $A\cap B_{m}$ being pairwise disjoint with union $A$ (Measure, Measure Space, and Probability Measure (i)). So $f$ is measurable as a function into $\mathbb{R}$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)) with $\int_{X}f\,d\mu=\nu(X)<\infty$, hence $f\in\mathcal{L}^{1}(\mu)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)), and $\nu(A)=\int_{A}f\,d\mu$ for the integral over $A$ as an integrable function. This proves (iii).

Finite signed measures. Under the axiom of dependent choice, $\nu$ has a Hahn set $P$ (Hahn and Jordan Decompositions of a Finite Signed Measure (i)), and the measures $\nu^{+}$ and $\nu^{-}$ of the Jordan decomposition are finite, absolutely continuous with respect to $\mu$, and satisfy $\nu=\nu^{+}-\nu^{-}$ on $\mathcal{F}$ (Hahn and Jordan Decompositions of a Finite Signed Measure (ii)). By (iii) there are $f_{1},f_{2}\in\mathcal{L}^{1}(\mu)$ with $\nu^{+}(A)=\int_{A}f_{1}\,d\mu$ and $\nu^{-}(A)=\int_{A}f_{2}\,d\mu$ for every $A\in\mathcal{F}$. Then $f=f_{1}-f_{2}\in\mathcal{L}^{1}(\mu)$ and, by linearity (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)),

$$\int_{A}f\,d\mu=\int_{X}f_{1}\mathbf{1}_{A}\,d\mu-\int_{X}f_{2}\mathbf{1}_{A}\,d\mu=\nu^{+}(A)-\nu^{-}(A)=\nu(A)\qquad(A\in\mathcal{F}),$$

which proves (iv).

Uses of dependent choice. Dependent choice enters only through countable choice in (b) and the $\sigma$-finite step and through the Hahn decomposition in (d) and the signed case, so (i) and (ii) use no choice.