Radon–Nikodym theorem

Radon–Nikodym theorem The Radon–Nikodym theorem involves a measurable space (X, \Sigma) on which two σ-finite measures are defined, \mu and \nu. It states that, if \nu \ll \mu (that is, if \nu is absolutely continuous with respect to \mu), then there exists a \Sigma-measurable function f : X \to [0, \infty), such that for any measurable set A \in \Sigma, \nu(A) = \int_A f \, d\mu. Radon–Nikodym derivative The function f satisfying the above equality is , that is, if g is another function which satisfies the same property, then f = g . The function f is commonly written d\nu/d\mu and is called the ''''''. 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). Extension to signed or complex measures A similar theorem can be proven for signed and complex measures: namely, that if \mu is a nonnegative σ-finite measure, and \nu is a finite-valued signed or complex measure such that \nu \ll \mu, that is, \nu is absolutely continuous with respect to \mu, then there is a \mu-integrable real- or complex-valued function g on X such that for every measurable set A, \nu(A) = \int_A g \, d\mu. == Examples ==

In the following examples, the set X is the real interval [0,1], and \Sigma is the Borel sigma-algebra on X. • Let \mu be the length measure on X, and let \nu assign to each subset Y of X twice the length of Y. Then \frac{d\nu}{d\mu} = 2. • Let \mu be the length measure on X, and let \nu assign to each subset Y of X the number of points from the set \{0.1,\dots,0.9\} that are contained in Y. Then \nu is not absolutely continuous with respect to \mu since it assigns non-zero measure to zero-length points. Indeed, there is no derivative \frac{d\nu}{d\mu}: there is no finite function that, when integrated e.g. from (0.1 - \varepsilon) to (0.1 + \varepsilon), gives 1 for all \varepsilon > 0. • \mu = \nu + \delta_0, where \nu is the length measure on X and \delta_0 is the Dirac measure on 0 (it assigns a measure of 1 to any set containing 0 and a measure of 0 to any other set). Then, \nu is absolutely continuous with respect to \mu, and \frac{d\nu}{d\mu} = 1_{X\setminus \{0\}} – the derivative is 0 at x = 0 and 1 at x > 0. ==Properties==

• Let ν, μ, and λ be σ-finite measures on the same measurable space. If ν ≪ λ and μ ≪ λ (ν and μ are both absolutely continuous with respect to λ), then \frac{d(\nu+\mu)}{d\lambda} = \frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda} \quad \lambda\text{-almost everywhere}. • If ν ≪ μ ≪ λ, then \frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}\quad\lambda\text{-almost everywhere}. • In particular, if μ ≪ ν and ν ≪ μ, then \frac{d\mu}{d\nu}=\left(\frac{d\nu}{d\mu}\right)^{-1}\quad\nu\text{-almost everywhere}. • If μ ≪ λ and is a μ-integrable function, then \int_X g\,d\mu = \int_X g\frac{d\mu}{d\lambda}\,d\lambda. • If ν is a finite signed or complex measure, then {d|\nu|\over d\mu} = \left|{d\nu\over d\mu}\right|. ==Applications==

Probability theory 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. Financial mathematics Amongst other fields, financial mathematics uses the theorem extensively, in particular via the Girsanov theorem. 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. Information divergences If μ and ν are measures over , and μ ≪ ν • The Kullback–Leibler divergence from ν to μ is defined to be D_\text{KL}(\mu \parallel \nu) = \int_X \log \left( \frac{d \mu}{d \nu} \right) \; d\mu. • For α > 0, α ≠ 1 the Rényi divergence of order α from ν to μ is defined to be D_\alpha(\mu \parallel \nu) = \frac{1}{\alpha - 1} \log\left(\int_X\left(\frac{d\mu}{d\nu}\right)^{\alpha-1}\; d\mu\right). ==The assumption of σ-finiteness==

The Radon–Nikodym theorem above makes the assumption that the measure μ with respect to which one computes the rate of change of ν is σ-finite. Negative example 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, , of a Borel set be defined as the number of elements of if is finite, and otherwise. One can check that is indeed a measure. It is not -finite, as not every Borel set is at most the union of countably many finite sets. Let be the usual Lebesgue measure on this Borel algebra. Then, is absolutely continuous with respect to , since for a set one has only if is the empty set, and then is also zero. Assume that the Radon–Nikodym theorem holds, that is, for some measurable function one has :\nu(A) = \int_A f \,d\mu for all Borel sets. Taking to be a singleton set, {{math|1=A = {a}}}, and using the above equality, one finds : 0 = f(a) for all real numbers . This implies that the function , and therefore the Lebesgue measure , is zero, which is a contradiction. Positive result Assuming \nu\ll\mu, the Radon–Nikodym theorem also holds if \mu is localizable and \nu is accessible with respect to \mu, i.e., \nu(A)=\sup\{\nu(B):B\in{\cal P}(A)\cap\mu^\operatorname{pre}(\R_{\ge0})\} for all A\in\Sigma. ==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 and , the idea is to consider functions with . 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 is singular with respect to follows from a technical fact about finite measures. Once the result is established for finite measures, extending to -finite, signed, and complex measures can be done naturally. The details are given below. For finite measures Constructing an extended-valued candidate First, suppose and are both finite-valued nonnegative measures. Let be the set of those extended-value measurable functions such that: :\forall A \in \Sigma:\qquad \int_A f\,d\mu \leq \nu(A) , since it contains at least the zero function. Now let , and suppose is an arbitrary measurable set, and define: :\begin{align} A_1 &= \left\{ x \in A : f_1(x) > f_2(x) \right\}, \\ A_2 &= \left\{ x \in A : f_2(x) \geq f_1(x) \right\}. \end{align} Then one has :\int_A\max\left\{f_1, f_2\right\}\,d\mu = \int_{A_1} f_1\,d\mu + \int_{A_2} f_2\,d\mu \leq \nu\left(A_1\right) + \nu\left(A_2\right) = \nu(A), and therefore, {{math|max{ f 1, f 2} ∈ F}}. Now, let be a sequence of functions in such that :\lim_{n\to\infty}\int_X f_n\,d\mu = \sup_{f\in F} \int_X f\,d\mu. By replacing with the maximum of the first functions, one can assume that the sequence is increasing. Let be an extended-valued function defined as :g(x) := \lim_{n\to\infty}f_n(x). By Lebesgue's monotone convergence theorem, one has :\lim_{n\to\infty} \int_A f_n\,d\mu = \int_A \lim_{n\to\infty} f_n(x)\,d\mu(x) = \int_A g\,d\mu \leq \nu(A) for each , and hence, . Also, by the construction of , :\int_X g\,d\mu = \sup_{f\in F}\int_X f\,d\mu. Proving equality Now, since , :\nu_0(A) := \nu(A) - \int_A g\,d\mu defines a nonnegative measure on . To prove equality, we show that . Suppose ; then, since is finite, there is an such that . To derive a contradiction from , we look for a positive set for the signed measure (i.e. a measurable set , all of whose measurable subsets have non-negative measure), where also has positive -measure. Conceptually, we're looking for a set , where in every part of . A convenient approach is to use the Hahn decomposition for the signed measure . Note then that for every one has , and hence, :\begin{align} \nu(A) &= \int_A g\,d\mu + \nu_0(A) \\ &\geq \int_A g\,d\mu + \nu_0(A\cap P)\\ &\geq \int_A g\,d\mu + \varepsilon\mu(A\cap P) = \int_A\left(g + \varepsilon 1_P\right)\,d\mu, \end{align} where is the indicator function of . Also, note that as desired; for if , then (since is absolutely continuous in relation to ) , so and :\nu_0(X) - \varepsilon\mu(X) = \left(\nu_0 - \varepsilon\mu\right)(N) \leq 0, contradicting the fact that . Then, since also :\int_X\left(g + \varepsilon1_P\right)\,d\mu \leq \nu(X) and satisfies :\int_X\left(g + \varepsilon 1_P\right)\,d\mu > \int_X g\,d\mu = \sup_{f\in F}\int_X f\,d\mu. This is impossible because it violates the definition of a supremum; therefore, the initial assumption that must be false. Hence, , as desired. Restricting to finite values Now, since is -integrable, the set is -null. Therefore, if a is defined as :f(x) = \begin{cases} g(x) & \text{if }g(x) then has the desired properties. Uniqueness As for the uniqueness, let be measurable functions satisfying :\nu(A) = \int_A f\,d\mu = \int_A g\,d\mu for every measurable set . Then, is -integrable, and :\int_A(g - f)\,d\mu = 0. (Recall that we can split the integral into two as long as they are measurable and non-negative) In particular, for {{math|1=A = {x ∈ X : f(x) > g(x)},}} or . It follows that :\int_X(g - f)^+\,d\mu = 0 = \int_X(g - f)^-\,d\mu, and so, that -almost everywhere; the same is true for , and thus, -almost everywhere, as desired. For -finite positive measures If and are -finite, then can be written as the union of a sequence {{math|{Bn}n}} of disjoint sets in , each of which has finite measure under both and . For each , by the finite case, there is a -measurable function such that :\nu_n(A) = \int_A f_n\,d\mu for each -measurable subset of . The sum \left(\sum_n f_n 1_{B_n}\right) := f of those functions is then the required function such that \nu(A) = \int_A f \, d\mu. As for the uniqueness, since each of the is -almost everywhere unique, so is . For signed and complex measures If is a -finite signed measure, then it can be Hahn–Jordan decomposed as where one of the measures is finite. Applying the previous result to those two measures, one obtains two functions, , satisfying the Radon–Nikodym theorem for and respectively, at least one of which is -integrable (i.e., its integral with respect to is finite). It is clear then that satisfies the required properties, including uniqueness, since both and are unique up to -almost everywhere equality. If is a complex measure, it can be decomposed as , where both and are finite-valued signed measures. Applying the above argument, one obtains two functions, , satisfying the required properties for and , respectively. Clearly, is the required function. ==The Lebesgue decomposition theorem==

Lebesgue's decomposition theorem shows that the assumptions of the Radon–Nikodym theorem can be found even in a situation which is seemingly more general. Consider a σ-finite positive measure \mu on the measure space (X,\Sigma) and a σ-finite signed measure \nu on \Sigma, without assuming any absolute continuity. Then there exist unique signed measures \nu_a and \nu_s on \Sigma such that \nu=\nu_a+\nu_s, \nu_a\ll\mu, and \nu_s\perp\mu. The Radon–Nikodym theorem can then be applied to the pair \nu_a,\mu. ==See also==