nLab
Radon–Nikodym derivative

RadonNikodym derivatives

Radon–Nikodym derivatives

Idea

Given two measures μ,ν\mu, \nu on the same measurable space, their Radon–Nikodym derivative is essentially their ratio μ/ν\mu/\nu, although this is traditionally written dμ/dν\mathrm{d}\mu/\mathrm{d}\nu because of analogies with differentiation. This ratio or derivative is a measurable function which is defined up to equality almost everywhere with respect to the divisor ν\nu. It only exists iff μ\mu is absolutely continuous with respect to ν\nu.

Integration on a general measure space can be seen as the process of multiplying a measure by a function to get a measure. Then the Radon–Nikodym derivative is the reverse of this: dividing two measures to get a function.

The Radon–Nikodym theorem

Definition

Suppose XX is a set, Σ\Sigma is a σ-algebra of subsets of XX, μ:Σ[0,]\mu\colon\Sigma\to[0,\infty] is a countably additive measure, and ν:ΣR\nu\colon\Sigma\to\mathbf{R} is a finitely additive map. We say that ν\nu is absolutely continuous with respect to μ\mu if for any ϵ>0\epsilon\gt0 there is δ>0\delta\gt0 such that |νE|<ϵ|\nu E|\lt\epsilon for any EΣE\in\Sigma such that μ(E)<δ\mu(E)\lt\delta. We say that ν\nu is truly continuous with respect to μ\mu if for any ϵ>0\epsilon\gt0 we can find EΣE\in\Sigma and δ>0\delta\gt0 such that μ(E)\mu(E) is finite and |ν(F)|<ϵ|\nu(F)|\lt\epsilon whenever FΣF\in\Sigma and μ(EF)<δ\mu(E\cap F)\lt\delta.

Remark

If we assume ν\nu to be countably additive, then the definition of absolutely continuity simplifies as follows: ν\nu is absolutely continuous with respect to μ\mu if and only if for any EΣE\in\Sigma we have ν(E)=0\nu(E)=0 whenever μ(E)=0\mu(E)=0.

Remark

ν\nu is truly continuous with respect to μ\mu if and only if ν\nu it is countable additive, ν\nu absolutely continuous with respect to μ\mu, and for any EΣE\in\Sigma such that ν(E)>0\nu(E)\gt0 we can find FΣF\in\Sigma such that FEF\subset E, μ(F)\mu(F) is finite, and ν(F)>0\nu(F)\gt0.

Theorem

Suppose XX is a set, Σ\Sigma is a σ-algebra of subsets of XX, μ:Σ[0,]\mu\colon\Sigma\to[0,\infty] is a countably additive measure, and ν:ΣR\nu\colon\Sigma\to\mathbf{R} is a function. Then there is a μ\mu-integrable function ff such that ν(E)= Ef\nu(E)=\int_E f for all EΣE\in\Sigma if and only if ν\nu is finitely additive and truly continuous with respect to μ\mu.

Corollary

In the context of the above theorem, if (X,Σ,μ)(X,\Sigma,\mu) is σ-finite?, then there is a μ\mu-integrable function ff such that ν(E)= Ef\nu(E)=\int_E f for all EΣE\in\Sigma if and only if ν\nu is countably additive and absolutely continuous with respect to μ\mu. If μ(X)\mu(X) is finite, it suffices to require that ν\nu is finitely additive.

Example

The following example shows that the condition of being truly continuous is necessary in the non-σ-finite case. Consider an uncountable set XX, a σ-algebra Σ\Sigma of subsets of XX that are countable or have countable complement, the counting measure μ:Σ[0,]\mu\colon\Sigma\to[0,\infty], and the measure ν:Σ[0,1]\nu\colon\Sigma\to[0,1] that vanishes on countable sets and takes value 1 on all uncountable subsets. Then ν\nu is absolutely continuous with respect to μ\mu because μ\mu only vanishes on the empty set. However, ν\nu is not truly continuous with respect to μ\mu because μ\mu takes finite values only on finite sets, but for any such a set EE we have ν(XE)=1\nu(X\setminus E)=1, which yields a contradiction if ϵ<1\epsilon\lt1.

Example

The measure space in the last example is quite pathological: it is not localizable?. A localizable? example can be constructed if and only if real-valued-measurable cardinals? exist. In this case, take XX to be a real-valued-measurable cardinal, μ:2 X[0,]\mu\colon 2^X\to[0,\infty] to be the counting measure, and ν:2 X[0,1]\nu\colon 2^X\to[0,1] a probability measure that vanishes on all countable subsets of XX. Then ν\nu is absolutely continuous with respect to μ\mu because only the empty subset of XX has μ\mu-measure 0. However, ν\nu is not truly continuous with respect to μ\mu because μ\mu takes finite values only on finite sets, but for any such a set EE we have ν(XE)=1\nu(X\setminus E)=1, which yields a contradiction if ϵ<1\epsilon\lt1.

Remark

Truly continuous countably (or finitely) additive measures on a measure space (X,Σ,μ)(X,\Sigma,\mu) form a module over the complex *-algebra of measurable maps XRX\to\mathbf{R} modulo equality almost everywhere. The Radon–Nikodym theorem then says that this module is a free module of rank 1. Furthermore, generators of this module can be identified with truly continuous measures ν\nu such that μ\mu is absolutely continuous with respect to ν\nu.

The Lebesgue decomposition

Theorem

Suppose XX is a set, Σ\Sigma is a σ-algebra of subsets of XX, μ:Σ[0,]\mu\colon\Sigma\to[0,\infty] is a countably additive measure, and ν:ΣR\nu\colon\Sigma\to\mathbf{R} is a countably additive function. Then there is a unique decomposition

ν=ν s+ν t+ν e,\nu=\nu_s+\nu_t+\nu_e,

where ν t\nu_t is truly continuous with respect to μ\mu, ν e\nu_e is absolutely continuous with respect to μ\mu and vanishes on every measurable set of finite μ\mu-measure, and ν s\nu_s is singular with respect to μ\mu, meaning there is a set FΣF\in\Sigma such that μ(F)=0\mu(F)=0 and ν(E)=0\nu(E)=0 for all EΣE\in\Sigma such that EXFE\subset X\setminus F. In particular, setting ν a=ν t+ν e\nu_a=\nu_t+\nu_e yields a unique decomposition ν=ν s+ν a\nu=\nu_s+\nu_a of ν\nu as a sum of a singular and absolutely continuous measure.

Definitions

Let XX be a measurable space (so XX consists of a set |X|{|X|} and a σ\sigma-algebra X\mathcal{M}_X), and let μ\mu and ν\nu be measures on XX, valued in the real numbers (and possibly taking infinite values) or in the complex numbers (and taking only finite values). Let ff be a measurable function ff (with real or complex values) on XX.

Definition

The function ff is a Radon–Nikodym derivative of μ\mu with respect to ν\nu if, given any measurable subset AA of XX, the μ\mu-measure of AA equals the integral of ff on AA with respect to ν\nu:

μ(A)= Afν= xAf(x)dν(x). \mu(A) = \int_A f \nu = \int_{x \in A} f(x) \mathrm{d}\nu(x) .

(The latter two expressions in this equation are different notations for the same thing.)

Properties

These properties are basic to the concept; the notation is as in the definition above.

Theorem

Let ff be a Radon–Nikodym derivative of μ\mu with respect to ν\nu, and let gg be a measurable function on XX. Then gg is a Radon–Nikodym derivative of μ\mu with respect to ν\nu if and only if ff and gg are equal almost everywhere with respect to ν\nu.

Theorem

If a Radon–Nikodym derivative of μ\mu with respect to ν\nu exists, then μ\mu is absolutely continuous with respect to ν\nu.

Theorem

If μ\mu is absolutely continuous with respect to ν\nu and both μ\mu and ν\nu are σ\sigma-finite, then a Radon–Nikodym derivative of μ\mu with respect to ν\nu exists.

Proofs

For fairly elementary proofs, see Bartels (2003).

(This last theorem is not as general as it could be.)

Note the repetition of ‘with respect to ν\nu’ in various guises; let us fix ν\nu (assumed to be σ\sigma-finite) and take everything with respect to it. Then it is convenient to treat all measurable functions up to equality almost everywhere; and given any absolutely continuous μ\mu (also assumed to be σ\sigma-finite), we speak of the Radon–Nikodym derivative of μ\mu.

Notation

See also the discussion of notation at measure space.

Using the simplest notation for integrals, the definition of Radon–Nikodym derivative reads

μ(A)= Afν, \mu(A) = \int_A f \nu ,

or equivalently

Aμ= Afν. \int_A \mu = \int_A f \nu .

In other words, the measure μ\mu is the product of the function ff and the measure ν\nu:

μ=fν; \mu = f \nu ;

and so ff is the ratio of μ\mu to ν\nu:

f=μ/ν. f = \mu/\nu .

So this is the simplest notation for the Radon–Nikodym derivative.

However, this notation for integrals is uncommon; one is more likely to see

Adμ= Afdν, \int_A \mathrm{d}\mu = \int_A f \,\mathrm{d}\nu ,

which leads to

f=dμ/dν f = \mathrm{d}\mu/\mathrm{d}\nu

for the Radon–Nikodym derivative. But none of these ‘d\mathrm{d}’s are really necessary.

We can also use a fuller notation with a dummy variable as the object of the symbol ‘d\mathrm{d}’:

xAμ(dx)= xAf(x)ν(dx); \int_{x \in A} \mu(\mathrm{d}x) = \int_{x \in A} f(x) \,\nu(\mathrm{d}x) ;

this leads to

f(x)=μ(dx)/ν(dx), f(x) = \mu(\mathrm{d}x)/\nu(\mathrm{d}x) ,

which does not give a symbol for ff directly. If instead of μ(dx)\mu(\mathrm{d}x) one unwisely writes dμ(x)\mathrm{d}\mu(x), then this gives the previous notation for the Radon–Nikodym derivative.

Now let ν\nu be Lebesgue measure on the real line and let FF be an upper semicontinuous function on the real line, so that FF defines a Borel measure μ\mu generated by

μ(],a])F(a). \mu({]-\infty,a]}) \coloneqq F(a) .

Then FF is absolutely continuous if and only if μ\mu is absolutely continuous, in which case the derivative FF' exists almost everywhere and is a Radon–Nikodym derivative of μ\mu. That is,

μ/ν=F=dF/dt. \mu/\nu = F' = \mathrm{d}F/\mathrm{d}t .

The presence of ‘d\mathrm{d}’ on the right-hand side inspires people to put it on the left-hand side as well; but this is spurious, since we really want to write

μ=dF \mu = \mathrm{d}F

and

ν=dt, \nu = \mathrm{d}t ,

where tt is the identity function on the real line.

References

A comprehensive treatment can be found in Chapter 23 of

Some fairly elementary proofs prepared for a substitute lecture in John Baez's introductory measure theory course are here:

The strategy there is based on:

  • Richard Bradley (1989): An Elementary Treatment of the Radon-Nikodym Derivative, American Mathematical Monthly 96(5), 437–440.

Last revised on June 12, 2020 at 23:19:58. See the history of this page for a list of all contributions to it.