The Radon-Nikodym Theorem

If \(f\) is a nonnegative measurable function on a probability space (a measure space, more generally), say \((\Omega, \mathscr F, \mathop{\mathrm{\mathbf{P}}})\), then \(\mathbf Q (A) = \int_A f \dif \mathop{\mathrm{\mathbf{P}}}\) defines another measure on the same probability space. This means \(\mathbf Q\) has a density \(f\) with respect to \(\mathop{\mathrm{\mathbf{P}}}\). In other words, for all \(A \in \mathscr F\), \(\mathop{\mathrm{\mathbf{P}}}(A) = 0\) implies \(\mathbf Q(A) = 0\). The converse asks whether every measure that vanishes on the original null sets can be obtained by integrating a density. This is true if \(\mathop{\mathrm{\mathbf{P}}}\) and \(\mathbf Q\) are \(\sigma\)-finite. The proof constructs the largest density whose integral stays below the target measure, then shows that no residual mass remains.

We use a finite signed measure \(\nu\): a countably additive real set function on \(\mathscr F\) with finite total variation, that is, \(\sup_{(A_i)}\sum_i|\nu(A_i)|<\infty\), where the supremum is over finite measurable partitions of \(\Omega\). A difference of two finite positive measures has this property. Finite additivity alone would not justify the limits of sets used below.

Lemma 1. If \(A_n\uparrow A\) or \(A_n\downarrow A\), then \(\nu(A_n)\to\nu(A)\).

Proof. For an increasing sequence, write its union as the disjoint union of its successive differences and use countable additivity. The corresponding series converges absolutely by finite total variation. For a decreasing sequence, apply the increasing case to complements and subtract from \(\nu(\Omega)\).

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Hahn’s decomposition separates a signed measure into a region on which every measurable subset has nonnegative mass and a region on which every measurable subset has nonpositive mass.

Theorem 2. (Hahn). For a finite signed measure \(\nu\), there is a measurable partition \(\Omega=A^+\sqcup A^-\) such that every measurable subset of \(A^+\) has nonnegative measure and every measurable subset of \(A^-\) has nonpositive measure.

Proof.

A set of maximal signed mass gives the decomposition: removing a negative subset or adding a positive subset of its complement would increase its mass. It remains to construct a maximiser.

Let \(s=\sup_{A\in\mathscr F}\nu(A)<\infty\) and choose \(A_n\) with \(\nu(A_n)>s-2^{-n}\). Since \(\nu(B\cap A_n)\leq s\),

\[\nu(B\cup A_n)=\nu(B)+\nu(A_n)-\nu(B\cap A_n)\geq\nu(B)-2^{-n}.\]

Induction gives \(\nu(\bigcup_{n=m}^NA_n)\geq s-\sum_{n=m}^N2^{-n}\). With \(U_m=\bigcup_{n\geq m}A_n\), continuity from below yields \(\nu(U_m)\geq s-2^{1-m}\). The sets \(U_m\) decrease, so continuity from above gives \(\nu(A^+)=s\) for \(A^+=\bigcap_mU_m\). The maximality argument in the first paragraph proves the sign claims for \(A^+\) and \(A^-=\Omega\setminus A^+\).

Explicitly, if \(B\subseteq A^+\) is measurable and \(\nu(B)<0\), then \(\nu(A^+\setminus B)=s-\nu(B)>s\), contradicting the definition of \(s\). If \(B\subseteq A^-\) and \(\nu(B)>0\), then disjoint additivity gives \(\nu(A^+\cup B)=s+\nu(B)>s\), the same contradiction. These arguments concern every measurable subset, which is stronger than merely requiring \(\nu(A^+)\geq0\) and \(\nu(A^-)\leq0\).

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By way of nomenclature, defining

\[\begin{aligned} \nu^+ (A) &\triangleq \nu(A \cap A^+) ~, \\ \nu^- (A) &\triangleq -\nu(A \cap A^-) ~, \\ \nu(A) &\triangleq \nu^+ (A) - \nu^- (A) ~, \end{aligned}\tag{1}\]

Here \(\nu^+\) and \(\nu^-\) are finite positive measures, called the positive and negative variations. Their difference \(\nu\) is a signed measure. This is the Jordan decomposition.

Two measures \(P,Q\) are mutually singular if there is a measurable \(S\) with \(P(S)=0\) and \(Q(S^c)=0\). They need not have disjoint topological supports. We write \(Q\ll P\) if \(P(A)=0\) implies \(Q(A)=0\) for every measurable \(A\). The construction below decomposes \(Q\) into a part with a density and a singular remainder; absolute continuity then forces the remainder to vanish.

Theorem 3 (Radon–Nikodym). Let \(P,Q\) be sigma-finite measures on the same measurable space, with \(Q\ll P\). There is a nonnegative measurable \(f\) such that \(Q(A)=\int_A f\,dP\) for every measurable \(A\). Any two such densities agree \(P\)-almost everywhere.

Remark 4. Note that a density \(f\) is integrable with respect to \(\mathop{\mathrm{\mathbf{P}}}\) if and only if \(\mathbf Q\) is finite. But, note that \(\mathbf Q\) is assumed to be \(\sigma\)-finite; this means \(f < \infty\) everywhere except on a set of measure zero. Therefore, we can restrict to analyzing the case where \(f\) is finite everywhere. This integral can be expressed as

\[\begin{aligned} \int_A h \dif \mathbf Q = \int_A h f \dif \mathop{\mathrm{\mathbf{P}}}\quad \text{for any } A \in \mathscr F ~, \end{aligned}\]

For this change-of-measure identity, \(h\) may be nonnegative measurable, or integrable with respect to \(Q\). The density \(f=dQ/dP\) is the Radon–Nikodym derivative.

Proof.

First suppose \(P,Q\) are finite. Let \(\mathcal E=\{g\geq0:\ g\text{ measurable},\ \int_Ag\,dP\leq Q(A)\text{ for all }A\in\mathscr F\}\). This class contains zero. For \(g,h\in\mathcal E\), split an arbitrary measurable \(A\) into the two regions where each function is larger:

\[\begin{aligned} \int_A\max(g,h)\,dP &=\int_{A\cap\{g\geq h\}}g\,dP+\int_{A\cap\{g<h\}}h\,dP\\ &\leq Q(A\cap\{g\geq h\})+Q(A\cap\{g<h\})=Q(A). \end{aligned}\]

Thus \(\mathcal E\) is closed under finite maxima. If \(g_n\uparrow g\) with each \(g_n\in\mathcal E\), monotone convergence on each \(A\) gives \(\int_Ag\,dP=\lim_n\int_Ag_n\,dP\leq Q(A)\), so it is also closed under increasing limits.

Put \(\alpha=\sup_{g\in\mathcal E}\int g\,dP\leq Q(\Omega)\). Choose \(g_n\in\mathcal E\) with \(\int g_n\,dP\to\alpha\), and set \(f_n=\max(g_1,\ldots,g_n)\), \(f=\lim_nf_n\). Then \(f\in\mathcal E\) and \(\int f\,dP=\alpha\). Define the nonnegative finite residual measure

\[R(A)=Q(A)-\int_A f\,dP.\]

For each positive integer \(m\), apply Hahn’s theorem to \(R-m^{-1}P\) and let \(H_m\) be its positive set. Every measurable \(E\subseteq H_m\) satisfies \(R(E)\geq m^{-1}P(E)\). Consequently \(f+m^{-1}1_{H_m}\in\mathcal E\). Maximality of \(f\) forces \(P(H_m)=0\). Put \(N=\bigcup_mH_m\). Then \(P(N)=0\), while the negative-set property gives \(R(N^c)\leq m^{-1}P(N^c)\) for every \(m\). Hence \(R(N^c)=0\): the residual is singular. Under \(Q\ll P\), we also have \(R\ll P\), so \(R(N)=0\) and \(R=0\).

For sigma-finite \(P,Q\), choose measurable covers \(\Omega=\bigcup_r B_r=\bigcup_s C_s\) with \(P(B_r)<\infty\) and \(Q(C_s)<\infty\). Enumerate the countably many intersections \(B_r\cap C_s\) as \(D_1,D_2,\ldots\), and disjointify them by \(E_j=D_j\setminus\bigcup_{i<j}D_i\). Both measures are finite on every \(E_j\). Apply the finite case to \(P_j(A)=P(A\cap E_j)\) and \(Q_j(A)=Q(A\cap E_j)\), obtaining densities \(f_j\). The measurable function \(f=\sum_jf_j\mathbf1_{E_j}\) then satisfies, by monotone convergence,

\[\int_Af\,dP=\sum_j\int_{A\cap E_j}f_j\,dP =\sum_jQ(A\cap E_j)=Q(A).\]

Finally, if \(f,g\) are two densities, on each \(E_j\) their integrals are finite. Integrating \(f-g\) over \(E_j\cap\{f>g\}\) gives zero, hence \(P(E_j\cap\{f>g\})=0\). Interchange \(f,g\) and take the countable union to obtain \(P(f\ne g)=0\).

To justify the finiteness and integration assertions in Remark 4, let \((C_j)\) cover \(\Omega\) with \(Q(C_j)<\infty\). If \(P(C_j\cap\{f=\infty\})>0\), then \(\int_{C_j}f\,dP=\infty\), a contradiction. Thus \(P(f=\infty)=0\); changing \(f\) to zero there gives a finite-valued version. The identity \(\int_Ah\,dQ=\int_Ahf\,dP\) holds first for indicators by the density formula, then for nonnegative simple functions by linearity, and for nonnegative measurable functions by monotone convergence. For integrable signed \(h\), apply it to the positive and negative parts; both integrals are finite, so subtraction is valid.

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