Examples of a measure include area and volume, where the subsets are sets of points; or the probability of an event, which is a subset of possible outcomes within a wider probability space.
The integral is with respect to an existing measure μ, which may often be the canonical Lebesgue measure on the real line R or the n-dimensional Euclidean space Rn (corresponding to our standard notions of length, area and volume).
For example, if f represented mass density and μ was the Lebesgue measure in three-dimensional space R3, then ν(A) would equal the total mass in a spatial region A.
The Radon–Nikodym theorem essentially states that, under certain conditions, any measure ν can be expressed in this way with respect to another measure μ on the same space.
The theorem is named after Johann Radon, who proved the theorem for the special case where the underlying space is Rn in 1913, and for Otto Nikodym who proved the general case in 1930.
[3] A Banach space Y is said to have the Radon–Nikodym property if the generalization of the Radon–Nikodym theorem also holds, mutatis mutandis, for functions with values in Y.
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
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.
Here is an example when μ is not σ-finite and the Radon–Nikodym theorem fails to hold.
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.
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.
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.
Constructing an extended-valued candidate First, suppose μ and ν are both finite-valued nonnegative measures.
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, Proving equality Now, since g ∈ F, defines a nonnegative measure on Σ.
Conceptually, we're looking for a set P, where ν0 ≥ ε μ in every part of P. A convenient approach is to use the Hahn decomposition (P, N) for the signed measure ν0 − ε μ.
Then, since also g + ε 1P ∈ F and satisfies This is impossible because it violates the definition of a supremum; therefore, the initial assumption that ν0 ≠ 0 must be false.
Restricting to finite values Now, since g is μ-integrable, the set {x ∈ X : g(x) = ∞} is μ-null.
If μ and ν are σ-finite, then X can be written as the union of a sequence {Bn}n of disjoint sets in Σ, each of which has finite measure under both μ and ν.
As for the uniqueness, since each of the fn is μ-almost everywhere unique, so is f. 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).
Applying the above argument, one obtains two functions, g, h : X → [0, ∞), satisfying the required properties for ν1 and ν2, respectively.