Radon-Nikodym and Chain Rule Computation Workshop

A Practical View of Radon–Nikodym Computations

A Radon–Nikodym derivative is a density relative to a chosen reference measure. The same measure can have different densities relative to different references, and a measure with a singular component cannot in general be represented by one density relative to the reference measure. Computations are most reliable when they keep these distinctions explicit: first identify absolute continuity, then find the density, separate any singular part, and finally check the result by integrating over sets.

Unless specified otherwise, the measures below are finite positive measures on the stated measurable space. A finite signed measure is a countably additive real-valued set function with finite total variation. For a signed measure \(\nu\), the notation \(\nu\ll\mu\) means \(|\nu|\ll\mu\). A statement that a density is unique is understood up to equality \(\mu\)-almost everywhere.

The Radon–Nikodym theorem and the Lebesgue decomposition theorem are standard results established earlier in the course. In the finite positive-measure setting, the first says that \(\nu\ll\mu\) has a nonnegative measurable density. The second says that any finite positive \(\nu\) can be written uniquely as \(\nu=\nu_{\mathrm{ac}}+\nu_{\mathrm{s}}\), where \(\nu_{\mathrm{ac}}\ll\mu\) and \(\nu_{\mathrm{s}}\perp\mu\). For a finite signed measure, apply these results to its positive and negative parts. The computational task is to determine which component a proposed formula actually describes.

Finite-Space and Signed-Measure Calculations

This finite calculation is the general method on a countable space when the reference measure assigns positive mass to each point: divide the target mass at each point by the reference mass there. If a point has zero reference mass but nonzero target mass, absolute continuity fails, and no Radon–Nikodym density with respect to that reference measure can represent the target measure.

Proof. Put \(h=f-g\), which is integrable. The assumption gives \(\int_A h\,d\mu=0\) for every measurable \(A\). For each positive integer \(n\), let \(E_n=\{h>1/n\}\). If \(\mu(E_n)>0\), then \(\int_{E_n}h\,d\mu\ge\mu(E_n)/n>0\), contradicting the assumption. Hence every \(E_n\) is null. Since \(\{h>0\}=\bigcup_{n\ge1}E_n\), the set \(\{h>0\}\) is null. Applying the same argument to \(-h\) shows that \(\{h<0\}\) is null. Therefore \(h=0\), and \(f=g\), \(\mu\)-almost everywhere. \(\square\)

This gives a direct check for a proposed density: show the integral identity on measurable sets. It also explains why a density is not generally a pointwise-unique formula; values on a \(\mu\)-null set do not affect any of those integrals.

Separating an Absolutely Continuous Part from an Atom

A singular measure cannot be represented by a density with respect to the reference measure unless it is the zero measure. Indeed, suppose \(\rho\perp\mu\) and \(\rho(A)=\int_A f\,d\mu\) for a nonnegative measurable \(f\). Choose \(N\) with \(\mu(N)=0\) and \(\rho(X\setminus N)=0\). Then \(\rho(N)=\int_N f\,d\mu=0\), and \(\rho(X\setminus N)=0\) by singularity. Hence \(\rho(X)=0\), so \(\rho\) is zero. The zero singular measure is represented by the density \(f=0\); it is only a nonzero singular measure that has no such representation.

Changing Reference Measures: The Chain Rule

Proof. Absolute continuity is transitive: if \(\gamma(A)=0\), then \(\beta(A)=0\) because \(\beta\ll\gamma\), and then \(\alpha(A)=0\) because \(\alpha\ll\beta\). Thus \(\alpha\ll\gamma\). The density identity for \(\beta\) gives \(\int q\,d\beta=\int qg\,d\gamma\) for every nonnegative measurable \(q\). This identity follows first for indicators from \(\beta(A)=\int_Ag\,d\gamma\), then for nonnegative simple functions by finite additivity of the integral, and then for arbitrary nonnegative measurable functions by Monotone Convergence. Apply it with \(q=h\chi_A\). For every \(A\in\mathcal{M}\),

Thus \(hg\) is a nonnegative measurable density of \(\alpha\) with respect to \(\gamma\). Uniqueness of Radon–Nikodym derivatives gives \(d\alpha/d\gamma=hg\) \(\gamma\)-almost everywhere. The densities can be chosen finite outside null sets because they are integrable for finite measures; their values on exceptional null sets do not affect this identity. \(\square\)

A Common Error and a Reliable Verification Routine

A common mistake is to use the absolutely continuous density as if it represented an entire mixed measure. In the preceding example, the incorrect calculation is \(\nu(\{0\})=\int_{\{0\}}(1+x)\,dm=0\), followed by the claim that this equals the measure of \(\{0\}\). The integral is indeed \(0\), but it computes only \(\nu_{\mathrm{ac}}(\{0\})\); the full measure is \(\nu(\{0\})=\nu_{\mathrm{ac}}(\{0\})+\nu_{\mathrm{s}}(\{0\})=0+3=3\). The correction is to identify any singular component before asserting a density for the whole measure.

For a candidate derivative \(f\), the decisive check is \(\nu(A)=\int_A f\,d\mu\) on measurable sets. On a finite space, it suffices to check every singleton. For a density on an interval, test sets that expose endpoints, zeros, or atoms as well as the whole interval. A chain-rule computation should also check that the measures are absolutely continuous in the required order: the formula multiplies derivatives only when \(d\alpha/d\beta\) and \(d\beta/d\gamma\) both exist. Equalities between derivatives are almost-everywhere statements with respect to the final reference measure, not necessarily pointwise identities everywhere.