Radon–Nikodym derivatives
Express one measure as weighted integration against another and identify the singular mass that no density can represent.
On this page
The question for today
What you’ll understand
- Check absolute continuity before seeking a measure density.
- Use a Radon–Nikodym derivative to transform integrals and event weights.
- Distinguish almost-everywhere uniqueness, reciprocal densities, and singular components.
A density is a comparison between two measures, not an intrinsic attribute of a distribution. A function described as a probability density usually means a Radon–Nikodym derivative relative to Lebesgue measure, but other reference measures can be more natural. The existence theorem says when such a comparison is possible. It also identifies the obstruction: if the new measure charges a set that the reference measure regards as null, no ordinary reference-measure density can account for that mass.
Absolute continuity is the null-set test
For positive measures ν and μ on the same measurable space, ν is absolutely continuous with respect to μ if every μ-null measurable set is also ν-null. The direction matters. Under the sigma-finite assumptions on both measures used here, the Radon–Nikodym theorem supplies a nonnegative measurable density h such that ν(A)=∫A h dμ for every measurable A. The density is unique only up to μ-null changes. It is a derivative of measures, not necessarily the ordinary derivative of a pointwise function.
The theorem's conclusion explains why the null-set condition is necessary: integrating any nonnegative measurable h over a μ-null set gives zero. Existence is the difficult converse. Sigma-finiteness is part of the standard theorem being invoked, not a decorative simplification; without appropriate hypotheses a general absolute-continuity claim does not automatically yield a density. For finite signed ν, one applies the theorem to its Jordan components when its variation is absolutely continuous with respect to a sigma-finite positive μ.
Reweight integrals, not coordinates
Once the identity holds for indicator functions, linearity extends it to simple functions and monotone convergence extends it to nonnegative measurable functions. Thus integrating f against ν is the same as integrating f times h against μ. For signed f, require integrability with respect to ν, equivalently finite ∫|f|h dμ. This is the measure-theoretic basis of importance weighting. It does not require a change of coordinates, and a density ratio should not be confused with a Jacobian unless a separate coordinate transformation establishes that interpretation.
For sigma-finite positive measures satisfying ν≪μ≪λ, the chain rule identifies dν/dλ with the product (dν/dμ)(dμ/dλ), interpreted with the usual almost-everywhere qualifications. A reciprocal rule requires mutual absolute continuity: then a positive finite density has reciprocal as the reverse derivative on the relevant full-measure set. If the first density vanishes on a region of positive μ measure, the reverse absolute continuity fails there. Blindly taking reciprocals can therefore lose exactly the region the weighting scheme cannot see.
A singular component is different from a large density
A measure ν is singular with respect to μ if it is carried by a μ-null measurable set. A point mass relative to Lebesgue measure is the simplest example. In the finite positive-measure setting with a sigma-finite reference μ, the Lebesgue decomposition writes ν uniquely as an absolutely continuous part plus a singular part. The first has a Radon–Nikodym density; the second cannot be represented by any ordinary μ-density. A narrow tall density can approximate a point mass weakly, but approximation does not turn the limiting atom into a Lebesgue density.
This distinction is practical when modeling mixed data. A response distribution might have a continuous spread and a separate probability of an exact zero. Writing only a continuous density omits the atom, even if that density integrates correctly over positive values. One can instead use a reference measure containing both Lebesgue measure and the relevant point mass, then describe everything with one density relative to that reference. The choice of measure changes the representation while leaving the underlying event probabilities unchanged.
Work it through
Reweighting the upper half of a unit interval
Let μ be length measure on [0,1] and define . Check normalization and compute ν((1/2,1]).
The density is nonnegative and its integral over [0,1] is [x³]₀¹=1, so ν is a probability measure absolutely continuous with respect to μ.
Use the event indicator in the change-of-measure identity: ν((1/2,1]) is the integral of 3x² from 1/2 to one.
The result is 1−(1/2)³=7/8. Although the same interval has μ mass 1/2, its ν mass is larger because the density emphasizes values near one.
The density changes event weights from 1/2 to 7/8 for the upper half. The reference measure and the density must both be specified.
The living notebook
Explore the model
Interactive lab · function
A quadratic used as a density
Set a=1 and vary the nonnegative endpoint x between zero and one in y=a x². Predict the area x³/3, then multiply that readout by three to obtain the event probability x³ for the lesson’s normalized density 3x². The coefficient control stops at two, so the normalization is a calculation rather than a selectable curve. The lab does not perform arbitrary changes of measure or represent atomic components.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA density value is a local weight per unit reference measure; probabilities come from integrating it over sets.
Pause & reason
Check your understanding
Your turn
Try it yourself
On [0,1], let ν=0.7m+0.3δ₀, where m is length measure and δ₀ is unit mass at zero. Identify its Lebesgue decomposition relative to m and decide whether ν has a single m-density.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Test absolute continuity on the singleton {0} before guessing a formula.
Compare with the solution
The absolutely continuous part is 0.7m, with m-density 0.7. The singular part is 0.3δ₀, carried by the m-null set {0}. The whole measure has no m-density because ν({0})=0.3 while m({0})=0. A density relative to the alternative reference m+δ₀ does exist: it can be 0.7 away from zero and 0.3 at zero. This represents the same probabilities using a reference measure that already sees the atom.
Keep the idea
Take it with you
- A Radon–Nikodym derivative compares measures and is unique only almost everywhere.
- Absolute continuity is directional and must precede density-ratio manipulations.
- Singular mass needs a separate component or a reference measure that can see it.
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