Radon–Nikodym theorem

Radon–Nikodym derivative

The function  f  satisfying the above equality is uniquely defined up to a μ-null set, that is, if g is another function which satisfies the same property, then  f = g  μ-almost everywhere.  f  is commonly written d ν d μ and is called the Radon–Nikodym derivative. The choice of notation and the name of the function reflects the fact that the function is analogous to a derivative in calculus in the sense that it describes the rate of change of density of one measure with respect to another (the way the Jacobian determinant is used in multivariable integration). A similar theorem can be proven for signed and complex measures: namely, that if μ is a nonnegative σ-finite measure, and ν is a finite-valued signed or complex measure such that ν ≪ μ, i.e. ν is absolutely continuous with respect to μ, then there is a μ-integrable real- or complex-valued function g on X such that for every measurable set A,

Applications

The theorem is very important in extending the ideas of probability theory from probability masses and probability densities defined over real numbers to probability measures defined over arbitrary sets. It tells if and how it is possible to change from one probability measure to another. Specifically, the probability density function of a random variable is the Radon–Nikodym derivative of the induced measure with respect to some base measure (usually the Lebesgue measure for continuous random variables).

For example, it can be used to prove the existence of conditional expectation for probability measures. The latter itself is a key concept in probability theory, as conditional probability is just a special case of it.

Amongst other fields, financial mathematics uses the theorem extensively. Such changes of probability measure are the cornerstone of the rational pricing of derivatives and are used for converting actual probabilities into those of the risk neutral probabilities.

Properties

Information divergences

If μ and ν are measures over X, and μ ≪ ν

The assumption of σ-finiteness

The Radon–Nikodym theorem makes the assumption that the measure μ with respect to which one computes the rate of change of ν is σ-finite. Here is an example when μ is not σ-finite and the Radon–Nikodym theorem fails to hold.

Consider the Borel σ-algebra on the real line. Let the counting measure, μ, of a Borel set A be defined as the number of elements of A if A is finite, and ∞ otherwise. One can check that μ is indeed a measure. It is not σ-finite, as not every Borel set is at most a countable union of finite sets. Let ν be the usual Lebesgue measure on this Borel algebra. Then, ν is absolutely continuous with respect to μ, since for a set A one has μ(A) = 0 only if A is the empty set, and then ν(A) is also zero.

Assume that the Radon–Nikodym theorem holds, that is, for some measurable function  f  one has

for all Borel sets. Taking A to be a singleton set, A = {a}, and using the above equality, one finds

for all real numbers a. This implies that the function  f , and therefore the Lebesgue measure ν, is zero, which is a contradiction.

Proof

This section gives a measure-theoretic proof of the theorem. There is also a functional-analytic proof, using Hilbert space methods, that was first given by von Neumann.

For finite measures μ and ν, the idea is to consider functions  f  with  f dμ ≤ dν. The supremum of all such functions, along with the monotone convergence theorem, then furnishes the Radon–Nikodym derivative. The fact that the remaining part of μ is singular with respect to ν follows from a technical fact about finite measures. Once the result is established for finite measures, extending to σ-finite, signed, and complex measures can be done naturally. The details are given below.

For finite measures

First, suppose μ and ν are both finite-valued nonnegative measures. Let F be the set of those measurable functions  f  : X → [0, ∞) such that:

F ≠ ∅, since it contains at least the zero function. Now let  f₁,  f₂ ∈ F, and suppose A be an arbitrary measurable set, and define:

Then one has

and therefore, max{ f ₁,  f ₂} ∈ F.

Now, let { f_n } be a sequence of functions in F such that

By replacing  f_n  with the maximum of the first n functions, one can assume that the sequence { f_n } is increasing. Let g be an extended-valued function defined as

By Lebesgue's monotone convergence theorem, one has

for each A ∈ Σ, and hence, g ∈ F. Also, by the construction of g,

Now, since g ∈ F,

defines a nonnegative measure on Σ. Suppose ν₀ ≠ 0; then, since μ is finite, there is an ε > 0 such that ν₀(X) > ε μ(X). Let (P, N) be a Hahn decomposition for the signed measure ν₀ − ε μ. Note that for every A ∈ Σ one has ν₀(A ∩ P) ≥ ε μ(A ∩ P), and hence,

where 1_P is the indicator function of P. Also, note that μ(P) > 0; for if μ(P) = 0, then (since ν is absolutely continuous in relation to μ) ν₀(P) ≤ ν(P) = 0, so ν₀(P) = 0 and

contradicting the fact that ν₀(X) > εμ(X).

Then, since

g + ε 1_P ∈ F and satisfies

This is impossible, therefore, the initial assumption that ν₀ ≠ 0 must be false. So ν₀ = 0, as desired.

Now, since g is μ-integrable, the set {x ∈ X : g(x) = ∞} is μ-null. Therefore, if a  f  is defined as

then  f  has the desired properties.

As for the uniqueness, let  f, g : X → [0, ∞) be measurable functions satisfying

for every measurable set A. Then, g − f  is μ-integrable, and

In particular, for A = {x ∈ X : f(x) > g(x)}, or {x ∈ X : f(x) < g(x)}. It follows that

and so, that (g − f )⁺ = 0 μ-almost everywhere; the same is true for (g − f )⁻, and thus,  f  = g μ-almost everywhere, as desired.

For σ-finite positive measures

If μ and ν are σ-finite, then X can be written as the union of a sequence {B_n}_n of disjoint sets in Σ, each of which has finite measure under both μ and ν. For each n, by the finite case, there is a Σ-measurable function  f_n  : B_n → [0, ∞) such that

for each Σ-measurable subset A of B_n. The union  f  of those functions is then the required function.

As for the uniqueness, since each of the  f_n  is μ-almost everywhere unique, then so is  f .

For signed and complex measures

If ν is a σ-finite signed measure, then it can be Hahn–Jordan decomposed as ν = ν⁺ − ν⁻ where one of the measures is finite. Applying the previous result to those two measures, one obtains two functions, g, h : X → [0, ∞), satisfying the Radon–Nikodym theorem for ν⁺ and ν⁻ respectively, at least one of which is μ-integrable (i.e., its integral with respect to μ is finite). It is clear then that  f = g − h satisfies the required properties, including uniqueness, since both g and h are unique up to μ-almost everywhere equality.

If ν is a complex measure, it can be decomposed as ν = ν₁ + iν₂, where both ν₁ and ν₂ are finite-valued signed measures. Applying the above argument, one obtains two functions, g, h : X → [0, ∞), satisfying the required properties for ν₁ and ν₂, respectively. Clearly,  f  = g + ih is the required function.

Generalisation

If the condition that ν is absolutely continuous with respect to μ is dropped, then the following is true. Given a σ-finite measure μ on the measure space ( X , Σ ) and a σ-finite, signed measure ν on Σ , there exist unique signed measures ν a and ν s on Σ such that ν = ν a + ν s , ν a ≪ μ , and ν s ⊥ μ . Moreover, there exists some extended μ -integrable function f such that