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

Each result cited below is universally quantified over the data in its own statement.

Two conventions are used throughout. First, if $\lambda$ is a measure on $(X,\mathcal{F})$ and $u:X\to\mathbb{R}$ is measurable, integrable with respect to $\lambda$ and satisfies $0\le u(x)$ for every $x$, then its integral in the sense of the integrable case equals its integral as a $[0,\infty]$-valued map, because by that definition its positive part is $u$ and its negative part is the zero function; this identification is used without further comment. Second, (B0): if $\lambda$ is a finite measure on $(X,\mathcal{F})$ and $u:X\to\mathbb{R}$ is measurable with $|u(x)|\le M$ for every $x$, for a real $M\ge0$, then $|u|$ is measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and by claim 1 of Linearity and Monotonicity of the Lebesgue Integral together with The Integral of an Indicator Function is the Measure of the Set we have $\int_X|u|\,d\lambda\le\int_X M\mathbf{1}_X\,d\lambda=M\lambda(X)<\infty$, so $u$ is integrable with respect to $\lambda$ by the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §integral.

Claim 1 (Uniqueness). Let $h_1,h_2$ be densities of $\nu$ with respect to $\mu$.

Step 1. For $i\in\{1,2\}$, taking $A=X$ in the defining identity gives $\int_X h_i\,d\mu=\int_X\mathbf{1}_Xh_i\,d\mu=\nu(X)$, a real number because $\nu$ is finite. Since $0\le h_i$, we have $|h_i|=h_i$, so $h_i$ is integrable by Measure Spaces and the Lebesgue Integral: Standing Notation §integral, with integral $\nu(X)$.

Step 2. Let $E=\{x\in X:h_2(x)<h_1(x)\}$. The difference $h_1-h_2$ is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, so $E=\{x:0<(h_1-h_2)(x)\}$ belongs to $\mathcal{F}$ by claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line. The function $u=\mathbf{1}_E(h_1-h_2)$ is measurable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; it is positive on $E$ and $0$ off $E$. Thus $\mathbf{1}_Eh_1=\mathbf{1}_Eh_2+u$ with all three functions nonnegative and measurable, and claim 1 of Linearity and Monotonicity of the Lebesgue Integral gives

Both integrals $\int_X\mathbf{1}_Eh_i\,d\mu$ equal $\nu(E)$, which is real since $\nu(E)\le\nu(X)$ by claim 2 of Basic Properties of a Measure. If $\int_Xu\,d\mu$ were $\infty$ the right-hand side would be $\infty$; so it is real, and then it equals $0$. By The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing, $u(x)=0$ for almost every $x$; the set where $u\ne0$ is $E$, so $E$ is null: $E\subseteq B$ for some $B\in\mathcal{F}$ with $\mu(B)=0$. As $E\in\mathcal{F}$, claim 2 of Basic Properties of a Measure gives $\mu(E)\le\mu(B)=0$.

Step 3. Interchanging $h_1$ and $h_2$, the set $E'=\{x:h_1(x)<h_2(x)\}$ belongs to $\mathcal{F}$ and $\mu(E')=0$. The set $\{x:h_1(x)\ne h_2(x)\}$ is the union of the disjoint sets $E$ and $E'$, so it belongs to $\mathcal{F}$ and has measure $\mu(E)+\mu(E')=0$ by claim 1 of Basic Properties of a Measure.

Claim 2 (Existence). Suppose $\mu$ is $\sigma$-finite and $\nu(A)=0$ for every $A\in\mathcal{F}$ with $\mu(A)=0$.

Step 1 (a finite measure with the null sets of $\mu$). By the definition of $\sigma$-finiteness there is a sequence $(X_m)_{m\in\mathbb{N}}$ in $\mathcal{F}$ with $X=\bigcup_mX_m$ and $\mu(X_m)<\infty$ for every $m$. Put $D_1=X_1$ and $D_m=X_m\setminus(X_1\cup\dots\cup X_{m-1})$ for $m\ge2$; these belong to $\mathcal{F}$, are pairwise disjoint, and each $x\in X$ lies in exactly one of them, namely $D_m$ for the least $m$ with $x\in X_m$. Put $c_m=2^{-m}(1+\mu(X_m))^{-1}$, a positive real number by claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field, and define $w:X\to\mathbb{R}$ by $w(x)=c_m$ for $x\in D_m$. Then $0<w(x)$ for every $x$. For real $a$, the set $\{x:a<w(x)\}$ is $\bigcup_mE_m$ with $E_m=D_m$ if $a<c_m$ and $E_m=\varnothing$ otherwise, a member of $\mathcal{F}$; so $w$ is measurable by claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line. For $n\in\mathbb{N}$ let $w_n=\sum_{m=1}^{n}c_m\mathbf{1}_{D_m}$, measurable by claims 1 and 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. For each $x$, $w_n(x)$ is $0$ for $n$ below the index $m$ with $x\in D_m$ and $w(x)$ from then on; so $0\le w_n\le w_{n+1}$ and $w(x)$ is the least upper bound of $\{w_n(x):n\in\mathbb{N}\}$. By claim 1 of Linearity and Monotonicity of the Lebesgue Integral, The Integral of an Indicator Function is the Measure of the Set, the bound $\mu(D_m)\le\mu(X_m)$ from claim 2 of Basic Properties of a Measure, the elementary bound $\mu(X_m)(1+\mu(X_m))^{-1}\le1$, and the formula of Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric with $r=\tfrac12$,

By Monotone Convergence Theorem, $\int_Xw\,d\mu\le1$. By claim 3 of Image Measures, Measures with Densities, and Change of Variables, $\mu'(A)=\int_X\mathbf{1}_Aw\,d\mu$ ($A\in\mathcal{F}$) defines a measure $\mu'$ on $(X,\mathcal{F})$, the measure with density $w$ with respect to $\mu$, and $\mu'(X)=\int_Xw\,d\mu\le1$, so $\mu'$ is finite.

For $A\in\mathcal{F}$ we have $\mu(A)=0$ if and only if $\mu'(A)=0$. If $\mu(A)=0$ then $A$ is null and $\mathbf{1}_Aw$ vanishes off $A$, so $\mu'(A)=0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral. If $\mu'(A)=0$ then $\mathbf{1}_Aw=0$ almost everywhere by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing; since $w>0$, the set where $\mathbf{1}_Aw\ne0$ is $A$, so $A$ is null, and as $A\in\mathcal{F}$ we get $\mu(A)=0$ as in Step 2 of Claim 1.

Step 2 (the sum measure). For $A\in\mathcal{F}$ put $\rho(A)=\nu(A)+\mu'(A)$, a real number since $\nu(A)\le\nu(X)$ and $\mu'(A)\le\mu'(X)$ by claim 2 of Basic Properties of a Measure. Clearly $\rho(\varnothing)=0$. Let $(A_m)_{m\in\mathbb{N}}$ be pairwise disjoint members of $\mathcal{F}$, and let $s_n=\sum_{m=1}^{n}\nu(A_m)$ and $t_n=\sum_{m=1}^{n}\mu'(A_m)$. By claims 1 and 2 of Basic Properties of a Measure, $s_n=\nu(\bigcup_{m\le n}A_m)\le\nu(X)$ and likewise $t_n\le\mu'(X)$, so both sequences are nondecreasing and bounded above; by countable additivity and the definition of the sum of a sequence in that definition, $(s_n)$ converges to $\nu(\bigcup_mA_m)$ and $(t_n)$ to $\mu'(\bigcup_mA_m)$. Hence, by claim 1 of Arithmetic of Limits of Real Sequences, the partial sums $s_n+t_n$ of $\sum_m\rho(A_m)$, which are bounded above by $\rho(X)$, converge to $\rho(\bigcup_mA_m)$, and by the same definition this limit is $\sum_m\rho(A_m)$. So $\rho$ is a finite measure on $(X,\mathcal{F})$ with $\rho(X)=\nu(X)+\mu'(X)$. If $\rho(A)=0$ then $\nu(A)=\mu'(A)=0$, being nonnegative reals with sum $0$.

(B1) For every measurable $f:X\to[0,\infty]$, $\int_Xf\,d\rho=\int_Xf\,d\nu+\int_Xf\,d\mu'$ in $[0,\infty]$; in particular $\int_Xf\,d\nu\le\int_Xf\,d\rho$. Indeed, for a nonnegative simple function $s$ with standard representation $\sum_{i=1}^{r}c_i\mathbf{1}_{A_i}$, the definition of its integral in Simple Function and Its Integral gives $\int_Xs\,d\rho=\sum_ic_i\rho(A_i)=\sum_ic_i\nu(A_i)+\sum_ic_i\mu'(A_i)=\int_Xs\,d\nu+\int_Xs\,d\mu'$, all real. By Approximation of Measurable Functions by Simple Functions §nonnegative there are nonnegative simple functions $s_n\le s_{n+1}$ whose pointwise least upper bound is $f$. Put $\alpha_n=\int_Xs_n\,d\nu$ and $\beta_n=\int_Xs_n\,d\mu'$, so that $\int_Xs_n\,d\rho=\alpha_n+\beta_n$. By Monotone Convergence Theorem for $\nu$, $\mu'$ and $\rho$, these nondecreasing sequences have least upper bounds $\int_Xf\,d\nu$, $\int_Xf\,d\mu'$ and $\int_Xf\,d\rho$ in $[0,\infty]$, and converge to them. If $\int_Xf\,d\nu$ and $\int_Xf\,d\mu'$ are both real, claim 1 of Arithmetic of Limits of Real Sequences shows that $(\alpha_n+\beta_n)$ converges to their sum, which is therefore $\int_Xf\,d\rho$. If one of them, say $\int_Xf\,d\nu$, is $\infty$, then $(\alpha_n)$ is not bounded above, hence neither is $(\alpha_n+\beta_n)$ since $\beta_n\ge0$, and both sides equal $\infty$.

(B2) If $f:X\to\mathbb{R}$ is measurable and bounded, then $f$ is integrable with respect to $\nu$, $\mu'$ and $\rho$ by (B0), and $\int_Xf\,d\rho=\int_Xf\,d\nu+\int_Xf\,d\mu'$: apply (B1) to the bounded nonnegative positive and negative parts $f^{+},f^{-}$, whose integrals are all real, and subtract, using the definition of the integral in Integrable Function and the Lebesgue Integral.

Step 3 (the Riesz representation). Let $\mathcal{L}^2$ be the set of $2$-integrable functions on $(X,\mathcal{F},\rho)$ and $L^2=L^2(X,\mathcal{F},\rho)$ the Lebesgue space; by The Lebesgue Space of Square-Integrable Functions is a Real Hilbert Space §hilbert it is a real Hilbert space with the inner product of The Lebesgue Space of Square-Integrable Functions is a Real Hilbert Space §inner-product, whose norm is $[g]\mapsto\lVert g\rVert_2$.

(i) Every measurable $g:X\to\mathbb{R}$ with $|g(x)|\le M$ for all $x$ lies in $\mathcal{L}^2$: by Properties of Real Powers of Nonnegative Real Numbers §monotone, $|g(x)|^2\le M^2$, so $\int_X|g|^2\,d\rho\le M^2\rho(X)<\infty$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral and The Integral of an Indicator Function is the Measure of the Set. In particular the constant function $1$ lies in $\mathcal{L}^2$ with $\lVert1\rVert_2=\rho(X)^{1/2}$, since $1^2=1$ by Properties of Real Powers of Nonnegative Real Numbers §agreement.

(ii) Let $g\in\mathcal{L}^2$. The numbers $2,2$ are conjugate exponents in the sense of Conjugate Exponents and Young's Inequality §conjugate, since $\tfrac12+\tfrac12=1$, so Hoelder's Inequality, for Two and for Finitely Many Factors §holder applied to $g$ and $1$ gives $\int_X|g|\,d\rho\le\rho(X)^{1/2}\lVert g\rVert_2$. By (B1) applied to $|g|$, $\int_X|g|\,d\nu\le\int_X|g|\,d\rho<\infty$, so $g$ is integrable with respect to $\nu$ by Measure Spaces and the Lebesgue Integral: Standing Notation §integral.

(iii) For $g\in\mathcal{L}^2$ put $\ell([g])=\int_Xg\,d\nu$. This is well defined: if $[g]=[g']$ then, by The Lebesgue Space of Power-Integrable Functions §equivalence, the set where $g\ne g'$ is contained in some $B\in\mathcal{F}$ with $\rho(B)=0$, hence $\nu(B)=0$ by Step 2, so $g=g'$ $\nu$-almost everywhere and $\int_Xg\,d\nu=\int_Xg'\,d\nu$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison (for the measure $\nu$). By the operations of The Lebesgue Space of Power-Integrable Functions §space and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $\ell$ is linear. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral and (ii), $|\ell([g])|\le\int_X|g|\,d\nu\le\rho(X)^{1/2}\lVert g\rVert_2$, and $\lVert g\rVert_2$ is the norm of $[g]$ by The Lebesgue Space of Square-Integrable Functions is a Real Hilbert Space §inner-product and The Lebesgue Space of Power-Integrable Functions §norm. Thus $\ell$ is a bounded linear functional in the sense of Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §functional, and The Riesz Representation Theorem for a Real Hilbert Space §existence yields $z\in L^2$ with $\ell(x)=\langle x,z\rangle$ for every $x\in L^2$. Let $k_0\in\mathcal{L}^2$ be a representative of $z$. By The Lebesgue Space of Square-Integrable Functions is a Real Hilbert Space §inner-product, for every $g\in\mathcal{L}^2$ the product $gk_0$ is $\rho$-integrable and

Step 4 (a representative with values in $[0,1]$). For $A\in\mathcal{F}$, $\mathbf{1}_A\in\mathcal{L}^2$ by (i), so (B3) and The Integral of an Indicator Function is the Measure of the Set give $\nu(A)=\int_X\mathbf{1}_Ak_0\,d\rho$. Let $P=\{x:0<-k_0(x)\}$ and $Q=\{x:0<k_0(x)-1\}$; they lie in $\mathcal{F}$ by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line, and they are disjoint. The function $u=-\mathbf{1}_Pk_0$ is nonnegative, measurable and $\rho$-integrable with $\int_Xu\,d\rho=-\nu(P)\le0$ by claim 2 of Linearity and Monotonicity of the Lebesgue Integral and (B3) with $g=\mathbf{1}_P$; as an integral of a nonnegative function it is also $\ge0$, so it is $0$, and The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing shows that $u=0$ $\rho$-almost everywhere. Since $u>0$ on $P$, $P$ is $\rho$-null, and $\rho(P)=0$ as in Step 2 of Claim 1. Likewise $u'=\mathbf{1}_Qk_0-\mathbf{1}_Q$ is nonnegative, measurable and $\rho$-integrable ($\mathbf{1}_Q$ by (B0)), with $\int_Xu'\,d\rho=\nu(Q)-\rho(Q)\le0$, by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, (B3) with $g=\mathbf{1}_Q$, and The Integral of an Indicator Function is the Measure of the Set for $\int_X\mathbf{1}_Q\,d\rho=\rho(Q)$, the inequality holding because $\nu(Q)\le\rho(Q)$; hence $\rho(Q)=0$. Put $k=\min(1,\max(0,k_0))$, measurable by claims 1 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, with $0\le k(x)\le1$ for every $x$ and $k=k_0$ off $P\cup Q$, where $\rho(P\cup Q)=0$ by claim 1 of Basic Properties of a Measure. For $g\in\mathcal{L}^2$ the functions $gk$ and $gk_0$ therefore agree $\rho$-almost everywhere, so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison and (B3) $gk$ is $\rho$-integrable and

Step 5 (splitting $\rho$). Let $g:X\to\mathbb{R}$ be measurable and bounded. Then $g\in\mathcal{L}^2$ by (i), and $gk$ is measurable and bounded, so (B4) and (B2) give $\int_Xg\,d\nu=\int_Xgk\,d\nu+\int_Xgk\,d\mu'$, all integrands being integrable by (B0). By claim 2 of Linearity and Monotonicity of the Lebesgue Integral,

Let $N=\{x:k(x)=1\}=X\setminus\{x:0<1-k(x)\}\in\mathcal{F}$. With $g=\mathbf{1}_N$ the left side of (B5) is the integral of the zero function, which is $0$, while $gk=\mathbf{1}_N$; so $\mu'(N)=0$ by The Integral of an Indicator Function is the Measure of the Set. By Step 1, $\mu(N)=0$, and by hypothesis $\nu(N)=0$. For $x\notin N$ we have $0\le k(x)<1$.

Step 6 (the density). Fix $A\in\mathcal{F}$. For $j\in\mathbb{N}$ let $p_j=\sum_{i=0}^{j-1}k^i$ (with $k^0=1$) and $q_j=\sum_{i=1}^{j}k^i$; they are measurable by claims 1 to 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $0\le p_j\le j$, $0\le q_j\le j$ since $0\le k^i\le1$. By telescoping, $(1-k)p_j=1-k^j$ and $kp_j=q_j$. Applying (B5) to $g=\mathbf{1}_Ap_j$ gives

Left side: for $x\in N$, $\mathbf{1}_A(x)(1-k(x)^j)=0=\mathbf{1}_{A\setminus N}(x)$ for every $j$; for $x\notin N$ the sequence $(k(x)^j)_j$ converges to $0$ by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric, so $\mathbf{1}_A(x)(1-k(x)^j)$ converges to $\mathbf{1}_A(x)=\mathbf{1}_{A\setminus N}(x)$ by claims 1 and 3 of Arithmetic of Limits of Real Sequences. Since $|\mathbf{1}_A(1-k^j)|\le1$ and the constant $1$ is $\nu$-integrable by (B0), claim 3 of Dominated Convergence Theorem shows that the left side of (B6) converges to $\int_X\mathbf{1}_{A\setminus N}\,d\nu=\nu(A\setminus N)$ (The Integral of an Indicator Function is the Measure of the Set), and $\nu(A\setminus N)=\nu(A)$ because $\nu(A)=\nu(A\setminus N)+\nu(A\cap N)$ and $\nu(A\cap N)\le\nu(N)=0$ by claims 1 and 2 of Basic Properties of a Measure.

Right side: let $r_j=\mathbf{1}_{A\setminus N}q_j$. Since $\mathbf{1}_Aq_j=r_j$ off $N$ and $\mu'(N)=0$, The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison (for $\mu'$) gives $\int_X\mathbf{1}_Aq_j\,d\mu'=\int_Xr_j\,d\mu'=:a_j$. The functions $r_j$ are nonnegative, measurable and nondecreasing in $j$, so $(a_j)$ is nondecreasing by claim 1 of Linearity and Monotonicity of the Lebesgue Integral. Define $h'(x)=k(x)(1-k(x))^{-1}$ for $x\notin N$ and $h'(x)=0$ for $x\in N$. For $x\notin N$, $q_j(x)$ are the partial sums of the geometric series $\sum_ik(x)^i$, which by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric converges with sum $h'(x)$, and this sum is the least upper bound of the partial sums by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion. Hence at every $x$ the sequence $(\mathbf{1}_{X\setminus N}(x)q_j(x))_j$ converges to $h'(x)$, so $h'$ is measurable by claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $\mathbf{1}_Ah'$ (measurable by claim 3 there) is the pointwise least upper bound of $(r_j)$. By Monotone Convergence Theorem, $\int_X\mathbf{1}_Ah'\,d\mu'$ is the least upper bound of $\{a_j:j\in\mathbb{N}\}$. By (B6) and the previous paragraph, $(a_j)$ converges to $\nu(A)$. For each $j$ the sequence $(a_{\max(i,j)})_{i\in\mathbb{N}}$ also converges to $\nu(A)$ and dominates the constant $a_j$, so $a_j\le\nu(A)$ by claim 1 of Order Properties of Limits of Real Sequences; and if $a_j\le U$ for all $j$ then $\nu(A)\le U$ by the same claim. So $\nu(A)$ is the least upper bound of the $a_j$, that is,

Finally, by claim 3 of Image Measures, Measures with Densities, and Change of Variables for the measure $\mu'$ with density $w$, applied to the measurable $\mathbf{1}_Ah':X\to[0,\infty)$, $\int_X\mathbf{1}_Ah'\,d\mu'=\int_X\mathbf{1}_Ah'w\,d\mu$. Let $h=h'w$; it is measurable by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, real-valued and nonnegative, and $\nu(A)=\int_X\mathbf{1}_Ah\,d\mu$ for every $A\in\mathcal{F}$. So $h$ is a density of $\nu$ with respect to $\mu$.