# Satz von Radon-Nikodym

Radon–Nikodym theorem In mathematics, the Radon–Nikodym theorem is a result in measure theory that expresses the relationship between two measures defined on the same measurable space. A measure is a set function that assigns a consistent magnitude to the measurable subsets of a measurable space. 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.

One way to derive a new measure from one already given is to assign a density to each point of the space, then integrate over the measurable subset of interest. This can be expressed as {Anzeigestil Nr (EIN)=int _{EIN}f,dmu ,} where ν is the new measure being defined for any measurable subset A and the function f is the density at a given point. 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). Zum Beispiel, if f represented mass density and μ was the Lebesgue measure in three-dimensional space R3, then ν(EIN) would equal the total mass in a spatial region A.

The Radon–Nikodym theorem essentially states that, unter bestimmten Bedingungen, any measure ν can be expressed in this way with respect to another measure μ on the same space. The function  f  is then called the Radon–Nikodym derivative and is denoted by {Anzeigestil {tfrac {dnu }{dmu }}} .[1] An important application is in probability theory, leading to the probability density function of a random variable.

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.[2] Im 1936 Hans Freudenthal generalized the Radon–Nikodym theorem by proving the Freudenthal spectral theorem, a result in Riesz space theory; this contains the Radon–Nikodym theorem as a special case.[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. All Hilbert spaces have the Radon–Nikodym property.

Inhalt 1 Formal description 1.1 Satz von Radon-Nikodym 1.2 Radon–Nikodym derivative 1.3 Extension to signed or complex measures 2 Beispiele 3 Eigenschaften 4 Anwendungen 4.1 Probability theory 4.2 Financial mathematics 4.3 Information divergences 5 The assumption of σ-finiteness 5.1 Negative example 5.2 Positive result 6 Nachweisen 6.1 For finite measures 6.2 For σ-finite positive measures 6.3 For signed and complex measures 7 The Lebesgue decomposition theorem 8 Siehe auch 9 Anmerkungen 10 References Formal description Radon–Nikodym theorem The Radon–Nikodym theorem involves a measurable space {Anzeigestil (X,Sigma )} on which two σ-finite measures are defined, {zeige ihn an } und {Anzeigestil Nr .} Es sagt, dass, wenn {displaystyle nu ll mu } (das ist, wenn {Anzeigestil Nr } is absolutely continuous with respect to {zeige ihn an } ), then there exists a {displaystyle Sigma } -measurable function {Anzeigestil f:Xto [0,unendlich ),} such that for any measurable set {displaystyle Asubseteq X,} {Anzeigestil Nr (EIN)=int _{EIN}f,dmu .} Radon–Nikodym derivative The function {Anzeigestil f} satisfying the above equality is uniquely defined up to a {zeige ihn an } -null set, das ist, wenn {Anzeigestil g} is another function which satisfies the same property, dann {displaystyle f=g} {zeige ihn an } -almost everywhere. Function {Anzeigestil f} is commonly written {frac {dnu }{dmu }} 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).

Extension to signed or complex measures A similar theorem can be proven for signed and complex measures: nämlich, that if {zeige ihn an } is a nonnegative σ-finite measure, und {Anzeigestil Nr } is a finite-valued signed or complex measure such that {displaystyle nu ll mu ,} das ist, {Anzeigestil Nr } is absolutely continuous with respect to {zeige ihn an ,} then there is a {zeige ihn an } -integrable real- or complex-valued function {Anzeigestil g} an {Anzeigestil X} such that for every measurable set {Anzeigestil A,} {Anzeigestil Nr (EIN)=int _{EIN}g,dmu .} Examples In the following examples, the set X is the real interval [0,1], und {displaystyle Sigma } is the Borel sigma-algebra on X.

{zeige ihn an } is the length measure on X. {Anzeigestil Nr } assigns to each subset Y of X, twice the length of Y. Dann, {textstyle {frac {dnu }{dmu }}=2} . {zeige ihn an } is the length measure on X. {Anzeigestil Nr } assigns to each subset Y of X, the number of points from the set {0.1, …, 0.9} that are contained in Y. Dann, {Anzeigestil Nr } is not absolutely-continuous with respect to {zeige ihn an } since it assigns non-zero measure to zero-length points. In der Tat, there is no derivative {textstyle {frac {dnu }{dmu }}} : there is no finite function that, when integrated e.g. aus {Anzeigestil (0.1-varepsilon )} zu {Anzeigestil (0.1+varepsilon )} , gibt {Anzeigestil 1} für alle {displaystyle varepsilon >0} . {displaystyle mu =nu +delta _{0}} , wo {Anzeigestil Nr } is the length measure on X and {displaystyle 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). Dann, {Anzeigestil Nr } is absolutely continuous with respect to {zeige ihn an } , und {textstyle {frac {dnu }{dmu }}=1_{Xsetminus {0}}} – the derivative is 0 bei {displaystyle x=0} und 1 bei {displaystyle x>0} .[4] Properties Let ν, m, 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), dann {Anzeigestil {frac {d(nu +mu )}{dlambda }}={frac {dnu }{dlambda }}+{frac {dmu }{dlambda }}quad lambda {Text{-almost everywhere}}.} If ν ≪ μ ≪ λ, dann {Anzeigestil {frac {dnu }{dlambda }}={frac {dnu }{dmu }}{frac {dmu }{dlambda }}quad lambda {Text{-almost everywhere}}.} Im Speziellen, if μ ≪ ν and ν ≪ μ, dann {Anzeigestil {frac {dmu }{dnu }}=links({frac {dnu }{dmu }}Rechts)^{-1}quad nu {Text{-almost everywhere}}.} If μ ≪ λ and g is a μ-integrable function, dann {Anzeigestil int _{X}g,Blut = du _{X}g{frac {dmu }{dlambda }},dlambda .} If ν is a finite signed or complex measure, dann {Anzeigestil {d|nicht | over dmu }=links|{dnu over dmu }Rechts|.} 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. Speziell, 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).

Zum Beispiel, 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 X, and μ ≪ ν The Kullback–Leibler divergence from ν to μ is defined to be {displaystyle D_{Text{KL}}(mu parallel nu )=int _{X}Protokoll übrig({frac {dmu }{dnu }}Rechts);dmu .} For α > 0, α ≠ 1 the Rényi divergence of order α from ν to μ is defined to be {displaystyle D_{Alpha }(mu parallel nu )={frac {1}{Alpha -1}}Protokoll übrig(int _{X}links({frac {dmu }{dnu }}Rechts)^{Alpha -1};dmu stimmt).} 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, m, 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. Dann, ν is absolutely continuous with respect to μ, since for a set A one has μ(EIN) = 0 only if A is the empty set, and then ν(EIN) is also zero.

Assume that the Radon–Nikodym theorem holds, das ist, for some measurable function f one has {Anzeigestil Nr (EIN)=int _{EIN}f,dmu } for all Borel sets. Taking A to be a singleton set, A = {a}, and using the above equality, one finds {displaystyle 0=f(a)} for all real numbers a. This implies that the function  f , and therefore the Lebesgue measure ν, ist Null, which is a contradiction.

Positive result Assuming {displaystyle nu ll mu ,} the Radon-Nikodym theorem also holds if {zeige ihn an } is localizable and {Anzeigestil Nr } is accessible with respect to {zeige ihn an } ,[5]:p. 189, Exercise 9O  i.e., {Anzeigestil Nr (EIN)=sup{nicht (B):Bin {Kal {P}}(EIN)cap mu ^{Name des Bedieners {pre} }(mathbb {R} _{geq 0})}} für alle {displaystyle Ain Sigma .} [6]: Theorem 1.111 (Radon-Nikodym, II)[5]:p. 190, Exercise 9T(ii)  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 Constructing an extended-valued candidate First, suppose μ and ν are both finite-valued nonnegative measures. Let F be the set of those extended-value measurable functions f  : X → [0, ∞] so dass: {displaystyle forall Ain Sigma :qquad int _{EIN}f,dmu leq nu (EIN)} F ≠ ∅, since it contains at least the zero function. Now let f1,  f2 ∈ F, and suppose A is an arbitrary measurable set, und definieren: {Anzeigestil {Start{ausgerichtet}EIN_{1}&=left{xin A:f_{1}(x)>f_{2}(x)Rechts},\EIN_{2}&=left{xin A:f_{2}(x)geq f_{1}(x)Rechts},Ende{ausgerichtet}}} Then one has {Anzeigestil int _{EIN}max left{f_{1},f_{2}Rechts},Blut = du _{EIN_{1}}f_{1},dmu +int _{EIN_{2}}f_{2},dmu leq nu left(EIN_{1}Rechts)+nu left(EIN_{2}Rechts)=nu (EIN),} und deshalb, max{ f 1,  f 2} ∈ F.

Jetzt, Lassen { fn } be a sequence of functions in F such that {Anzeigestil lim _{nto infty }int _{X}f_{n},dmu =sup _{fin F}int _{X}f,dmu .} By replacing  fn  with the maximum of the first n functions, one can assume that the sequence { fn } is increasing. Let g be an extended-valued function defined as {Anzeigestil g(x):=lim _{nto infty }f_{n}(x).} By Lebesgue's monotone convergence theorem, hat man {Anzeigestil lim _{nto infty }int _{EIN}f_{n},Blut = du _{EIN}lim _{nto infty }f_{n}(x),dmu (x)=int _{EIN}g,dmu leq nu (EIN)} for each A ∈ Σ, und daher, g ∈ F. Ebenfalls, by the construction of g, {Anzeigestil int _{X}g,dmu =sup _{fin F}int _{X}f,dmu .} Proving equality Now, since g ∈ F, {Anzeigestil Nr. _{0}(EIN):=nu (EIN)-int _{EIN}g,dmu } defines a nonnegative measure on Σ. To prove equality, we show that ν0 = 0.

Suppose ν0 ≠ 0; dann, since μ is finite, there is an ε > 0 such that ν0(X) > ε μ(X). To derive a contradiction from ν0 ≠ 0, we look for a positive set P ∈ Σ for the signed measure ν0 − ε μ (d.h. a measurable set P, all of whose measurable subsets have non-negative ν0−ε μ measure), where also P has positive μ-measure. 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.

Note then that for every A ∈ Σ one has ν0(A ∩ P) ≥ ε μ(A ∩ P), und daher, {Anzeigestil {Start{ausgerichtet}nicht (EIN)&=int _{EIN}g,dmu +nu _{0}(EIN)\&geq int _{EIN}g,dmu +nu _{0}(Acap P)\&geq int _{EIN}g,dmu +varepsilon mu (Acap P)=int _{EIN}links(g+varepsilon 1_{P}Rechts),dmu ,Ende{ausgerichtet}}} where 1P is the indicator function of P. Ebenfalls, note that μ(P) > 0 wie gewünscht; for if μ(P) = 0, dann (since ν is absolutely continuous in relation to μ) ν0(P) ≤ ν(P) = 0, so ν0(P) = 0 und {Anzeigestil Nr. _{0}(X)-varepsilon mu (X)=links(nicht _{0}-varepsilon mu right)(N)leq 0,} contradicting the fact that ν0(X) > εμ(X).

Dann, since also {Anzeigestil int _{X}links(g+varepsilon 1_{P}Rechts),dmu leq nu (X)<+infty ,} g + ε 1P ∈ F and satisfies {displaystyle int _{X}left(g+varepsilon 1_{P}right),dmu >int _{X}g,dmu =sup _{fin F}int _{X}f,dmu .} This is impossible because it violates the definition of a supremum; deshalb, the initial assumption that ν0 ≠ 0 must be false. Somit, ν0 = 0, wie gewünscht.

Restricting to finite values Now, since g is μ-integrable, der Satz {x ∈ X : g(x) = ∞} is μ-null. Deswegen, if a  f  is defined as {Anzeigestil f(x)={Start{Fälle}g(x)&{Text{wenn }}g(x)

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