Radon–Nikodym theorem

The Radon–Nikodym theorem involves a measurable space $(X,\; \backslash Sigma)$ on which two σ-finite measures are defined, $\backslash mu$ and $\backslash nu.$

It states that, if $\backslash nu\; \backslash ll\; \backslash mu$ (that is, if $\backslash nu$ is absolutely continuous with respect to $\backslash mu$), then there exists a $\backslash Sigma$-measurable function $f\; :\; X\; \backslash to\; [0,\; \backslash infty),$ such that for any measurable set $A\; \backslash in\; \backslash Sigma,$

$$\backslash nu(A)\; =\; \backslash int\_A\; f\; \backslash ,\; d\backslash mu.$$

The function $f$ satisfying the above equality is {{em|uniquely defined up to a $\backslash mu$-null set}}, that is, if $g$ is another function which satisfies the same property, then $f\; =\; g$ {{nowrap|$\backslash mu$-almost everywhere}}. The function $f$ is commonly written \backslash frac\{d\backslash nu\}\{d\backslash mu\} and is called the {{visible anchor|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 $\backslash mu$ is a nonnegative σ-finite measure, and $\backslash nu$ is a finite-valued signed or complex measure such that $\backslash nu\; \backslash ll\; \backslash mu,$ that is, $\backslash nu$ is absolutely continuous with respect to $\backslash mu,$ then there is a $\backslash mu$-integrable real- or complex-valued function $g$ on $X$ such that for every measurable set $A,$

$$\backslash nu(A)\; =\; \backslash int\_A\; g\; \backslash ,\; d\backslash mu.$$

In the following examples, the set {{mvar|X}} is the real interval [0,1], and $\backslash Sigma$ is the Borel sigma-algebra on {{mvar|X}}.

1. $\backslash mu$ is the length measure on {{mvar|X}}. $\backslash nu$ assigns to each subset {{mvar|Y}} of {{mvar|X}}, twice the length of {{mvar|Y}}. Then, $\backslash frac\{d\backslash nu\}\{d\backslash mu\}\; =\; 2$.
2. $\backslash mu$ is the length measure on {{mvar|X}}. $\backslash nu$ assigns to each subset {{mvar|Y}} of {{mvar|X}}, the number of points from the set {0.1, …, 0.9} that are contained in {{mvar|Y}}. Then, $\backslash nu$ is not absolutely-continuous with respect to $\backslash mu$ since it assigns non-zero measure to zero-length points. Indeed, there is no derivative $\backslash frac\{d\backslash nu\}\{d\backslash mu\}$: there is no finite function that, when integrated e.g. from $(0.1\; -\; \backslash varepsilon)$ to $(0.1\; +\; \backslash varepsilon)$, gives $1$ for all $\backslash varepsilon\; >\; 0$.
3. $\backslash mu\; =\; \backslash nu\; +\; \backslash delta\_0$, where $\backslash nu$ is the length measure on {{mvar|X}} and $\backslash delta\_0$ 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, $\backslash nu$ is absolutely continuous with respect to $\backslash mu$, and $\backslash frac\{d\backslash nu\}\{d\backslash mu\}\; =\; 1\_\{X\backslash setminus\; \backslash \{0\backslash \}\}$ – the derivative is 0 at $x\; =\; 0$ and 1 at $x\; >\; 0$.{{cite web |title=Calculating Radon Nikodym derivative |date=April 7, 2018 |work=Stack Exchange |url=https://math.stackexchange.com/q/588442/29780 }}

• 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 $$\backslash frac\{d(\backslash nu+\backslash mu)\}\{d\backslash lambda\}\; =\; \backslash frac\{d\backslash nu\}\{d\backslash lambda\}+\backslash frac\{d\backslash mu\}\{d\backslash lambda\}\; \backslash quad\; \backslash lambda\backslash text\{-almost\; everywhere\}.$$
• If ν ≪ μ ≪ λ, then $$\backslash frac\{d\backslash nu\}\{d\backslash lambda\}=\backslash frac\{d\backslash nu\}\{d\backslash mu\}\backslash frac\{d\backslash mu\}\{d\backslash lambda\}\backslash quad\backslash lambda\backslash text\{-almost\; everywhere\}.$$
• In particular, if μ ≪ ν and ν ≪ μ, then $$\backslash frac\{d\backslash mu\}\{d\backslash nu\}=\backslash left(\backslash frac\{d\backslash nu\}\{d\backslash mu\}\backslash right)^\{-1\}\backslash quad\backslash nu\backslash text\{-almost\; everywhere\}.$$
• If μ ≪ λ and {{mvar|g}} is a μ-integrable function, then $$\backslash int\_X\; g\backslash ,d\backslash mu\; =\; \backslash int\_X\; g\backslash frac\{d\backslash mu\}\{d\backslash lambda\}\backslash ,d\backslash lambda.$$
• If ν is a finite signed or complex measure, then $$\{d|\backslash nu|\backslash over\; d\backslash mu\}\; =\; \backslash left|\{d\backslash nu\backslash over\; d\backslash mu\}\backslash right|.$$

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, 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.

If μ and ν are measures over {{mvar|X}}, and μ ≪ ν

• The Kullback–Leibler divergence from ν to μ is defined to be $$D\_\backslash text\{KL\}(\backslash mu\; \backslash parallel\; \backslash nu)\; =\; \backslash int\_X\; \backslash log\; \backslash left(\; \backslash frac\{d\; \backslash mu\}\{d\; \backslash nu\}\; \backslash right)\; \backslash ;\; d\backslash mu.$$
• For α > 0, α ≠ 1 the Rényi divergence of order α from ν to μ is defined to be $$D\_\backslash alpha(\backslash mu\; \backslash parallel\; \backslash nu)\; =\; \backslash frac\{1\}\{\backslash alpha\; -\; 1\}\; \backslash log\backslash left(\backslash int\_X\backslash left(\backslash frac\{d\backslash mu\}\{d\backslash nu\}\backslash right)^\{\backslash alpha-1\}\backslash ;\; d\backslash mu\backslash right).$$

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.

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, {{mvar|μ}}, of a Borel set {{mvar|A}} be defined as the number of elements of {{mvar|A}} if {{mvar|A}} is finite, and {{math|∞}} otherwise. One can check that {{mvar|μ}} is indeed a measure. It is not {{mvar|σ}}-finite, as not every Borel set is at most a countable union of finite sets. Let {{mvar|ν}} be the usual Lebesgue measure on this Borel algebra. Then, {{mvar|ν}} is absolutely continuous with respect to {{mvar|μ}}, since for a set {{mvar|A}} one has {{math|1=μ(A) = 0}} only if {{mvar|A}} is the empty set, and then {{math|ν(A)}} is also zero.

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

:$\backslash nu(A)\; =\; \backslash int\_A\; f\; \backslash ,d\backslash mu$

for all Borel sets. Taking {{mvar|A}} to be a singleton set, {{math|1=A = {a}}}, and using the above equality, one finds

:$0\; =\; f(a)$

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

Assuming $\backslash nu\backslash ll\backslash mu,$ the Radon–Nikodym theorem also holds if $\backslash mu$ is localizable and $\backslash nu$ is accessible with respect to $\backslash mu$,{{cite book |last1=Brown |first1=Arlen |last2=Pearcy |first2=Carl |title=Introduction to Operator Theory I: Elements of Functional Analysis |date=1977 |isbn=978-1461299288}}{{rp|at=p. 189, Exercise 9O}} i.e., $\backslash nu(A)=\backslash sup\backslash \{\backslash nu(B):B\backslash in\{\backslash cal\; P\}(A)\backslash cap\backslash mu^\backslash operatorname\{pre\}(\backslash R\_\{\backslash ge0\})\backslash \}$ for all $A\backslash in\backslash Sigma.${{cite book |last1=Fonseca |first1=Irene |last2=Leoni |first2=Giovanni |title=Modern Methods in the Calculus of Variations: L^p Spaces |publisher=Springer |isbn=978-0-387-35784-3 |page=68}}{{rp|at=Theorem 1.111 (Radon–Nikodym, II)}}{{rp|at=p. 190, Exercise 9T(ii)}}

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 {{mvar|μ}} and {{mvar|ν}}, the idea is to consider functions {{math| f }} with {{math|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 {{mvar|μ}} is singular with respect to {{mvar|ν}} follows from a technical fact about finite measures. Once the result is established for finite measures, extending to {{mvar|σ}}-finite, signed, and complex measures can be done naturally. The details are given below.

Constructing an extended-valued candidate First, suppose {{mvar|μ}} and {{mvar|ν}} are both finite-valued nonnegative measures. Let {{mvar|F}} be the set of those extended-value measurable functions {{math|f  : X → [0, ∞]}} such that:

:$\backslash forall\; A\; \backslash in\; \backslash Sigma:\backslash qquad\; \backslash int\_A\; f\backslash ,d\backslash mu\; \backslash leq\; \backslash nu(A)$

{{math|F ≠ ∅}}, since it contains at least the zero function. Now let {{math|f₁,  f₂ ∈ F}}, and suppose {{mvar|A}} is an arbitrary measurable set, and define:

:$\backslash begin\{align\}$

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

:$\backslash int\_A\backslash max\backslash left\backslash \{f\_1,\; f\_2\backslash right\backslash \}\backslash ,d\backslash mu\; =\; \backslash int\_\{A\_1\}\; f\_1\backslash ,d\backslash mu\; +\; \backslash int\_\{A\_2\}\; f\_2\backslash ,d\backslash mu\; \backslash leq\; \backslash nu\backslash left(A\_1\backslash right)\; +\; \backslash nu\backslash left(A\_2\backslash right)\; =\; \backslash nu(A),$

and therefore, {{math|max{ f ₁,  f ₂} ∈ F}}.

Now, let {{math|{{mset| f_n }}}} be a sequence of functions in {{mvar|F}} such that

:$\backslash lim\_\{n\backslash to\backslash infty\}\backslash int\_X\; f\_n\backslash ,d\backslash mu\; =\; \backslash sup\_\{f\backslash in\; F\}\; \backslash int\_X\; f\backslash ,d\backslash mu.$

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

:$g(x)\; :=\; \backslash lim\_\{n\backslash to\backslash infty\}f\_n(x).$

By Lebesgue's monotone convergence theorem, one has

:$\backslash lim\_\{n\backslash to\backslash infty\}\; \backslash int\_A\; f\_n\backslash ,d\backslash mu\; =\; \backslash int\_A\; \backslash lim\_\{n\backslash to\backslash infty\}\; f\_n(x)\backslash ,d\backslash mu(x)\; =\; \backslash int\_A\; g\backslash ,d\backslash mu\; \backslash leq\; \backslash nu(A)$

for each {{math|A ∈ Σ}}, and hence, {{math|g ∈ F}}. Also, by the construction of {{mvar|g}},

:$\backslash int\_X\; g\backslash ,d\backslash mu\; =\; \backslash sup\_\{f\backslash in\; F\}\backslash int\_X\; f\backslash ,d\backslash mu.$

Proving equality Now, since {{math|g ∈ F}},

:$\backslash nu\_0(A)\; :=\; \backslash nu(A)\; -\; \backslash int\_A\; g\backslash ,d\backslash mu$

defines a nonnegative measure on {{math|Σ}}. To prove equality, we show that {{math|1=ν₀ = 0}}.

Suppose {{math|ν₀ ≠ 0}}; then, since {{mvar|μ}} is finite, there is an {{math|ε > 0}} such that {{math|ν₀(X) > ε μ(X)}}. To derive a contradiction from {{math|ν₀ ≠ 0}}, we look for a positive set {{math|P ∈ Σ}} for the signed measure {{math|ν₀ − ε μ}} (i.e. a measurable set {{mvar|P}}, all of whose measurable subsets have non-negative {{math|ν₀ − εμ}} measure), where also {{mvar|P}} has positive {{mvar|μ}}-measure. Conceptually, we're looking for a set {{mvar|P}}, where {{math|ν₀ ≥ ε μ}} in every part of {{mvar|P}}. A convenient approach is to use the Hahn decomposition {{math|(P, N)}} for the signed measure {{math|ν₀ − ε μ}}.

Note then that for every {{math|A ∈ Σ}} one has {{math|ν₀(A ∩ P) ≥ ε μ(A ∩ P)}}, and hence,

:$\backslash begin\{align\}$

\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 {{math|1_P}} is the indicator function of {{mvar|P}}. Also, note that {{math|μ(P) > 0}} as desired; for if {{math|1=μ(P) = 0}}, then (since {{mvar|ν}} is absolutely continuous in relation to {{mvar|μ}}) {{math|1=ν₀(P) ≤ ν(P) = 0}}, so {{math|1=ν₀(P) = 0}} and

:$\backslash nu\_0(X)\; -\; \backslash varepsilon\backslash mu(X)\; =\; \backslash left(\backslash nu\_0\; -\; \backslash varepsilon\backslash mu\backslash right)(N)\; \backslash leq\; 0,$

contradicting the fact that {{math|ν₀(X) > εμ(X)}}.

Then, since also

:

{{math|g + ε 1_P ∈ F}} and satisfies

:$\backslash int\_X\backslash left(g\; +\; \backslash varepsilon\; 1\_P\backslash right)\backslash ,d\backslash mu\; >\; \backslash int\_X\; g\backslash ,d\backslash mu\; =\; \backslash sup\_\{f\backslash in\; F\}\backslash int\_X\; f\backslash ,d\backslash mu.$

This is impossible because it violates the definition of a supremum; therefore, the initial assumption that {{math|ν₀ ≠ 0}} must be false. Hence, {{math|1=ν₀ = 0}}, as desired.

Restricting to finite values Now, since {{mvar|g}} is {{mvar|μ}}-integrable, the set {{math|{{mset|1=x ∈ X : g(x) = ∞}}}} is {{mvar|μ}}-null. Therefore, if a {{math| f }} is defined as

:$f(x)\; =\; \backslash begin\{cases\}$

g(x) & \text{if }g(x) < \infty \\

0 & \text{otherwise,}

\end{cases}

then {{math|f}} has the desired properties.

Uniqueness As for the uniqueness, let {{math| f, g : X → [0, ∞)}} be measurable functions satisfying

:$\backslash nu(A)\; =\; \backslash int\_A\; f\backslash ,d\backslash mu\; =\; \backslash int\_A\; g\backslash ,d\backslash mu$

for every measurable set {{mvar|A}}. Then, {{math|g − f }} is {{mvar|μ}}-integrable, and

:$\backslash int\_A(g\; -\; f)\backslash ,d\backslash mu\; =\; 0.$ (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 {{math|{{mset|x ∈ X : f(x) < g(x)}}}}. It follows that

:$\backslash int\_X(g\; -\; f)^+\backslash ,d\backslash mu\; =\; 0\; =\; \backslash int\_X(g\; -\; f)^-\backslash ,d\backslash mu,$

and so, that {{math|1=(g − f )⁺ = 0}} {{mvar|μ}}-almost everywhere; the same is true for {{math|(g − f )⁻}}, and thus, {{math|1=f = g}} {{mvar|μ}}-almost everywhere, as desired.

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

:$\backslash nu\_n(A)\; =\; \backslash int\_A\; f\_n\backslash ,d\backslash mu$

for each {{math|Σ}}-measurable subset {{mvar|A}} of {{math|B_n}}. The sum $\backslash left(\backslash sum\_n\; f\_n\; 1\_\{B\_n\}\backslash right)\; :=\; f$ of those functions is then the required function such that $\backslash nu(A)\; =\; \backslash int\_A\; f\; \backslash ,\; d\backslash mu$.

As for the uniqueness, since each of the {{math|f_n}} is {{mvar|μ}}-almost everywhere unique, so is {{math|f}}.

If {{mvar|ν}} is a {{mvar|σ}}-finite signed measure, then it can be Hahn–Jordan decomposed as {{math|1=ν = ν⁺ − ν⁻}} where one of the measures is finite. Applying the previous result to those two measures, one obtains two functions, {{math|g, h : X → [0, ∞)}}, satisfying the Radon–Nikodym theorem for {{math|ν⁺}} and {{math|ν⁻}} respectively, at least one of which is {{mvar|μ}}-integrable (i.e., its integral with respect to {{mvar|μ}} is finite). It is clear then that {{math|1=f = g − h}} satisfies the required properties, including uniqueness, since both {{mvar|g}} and {{mvar|h}} are unique up to {{mvar|μ}}-almost everywhere equality.

If {{mvar|ν}} is a complex measure, it can be decomposed as {{math|1=ν = ν₁ + iν₂}}, where both {{math|ν₁}} and {{math|ν₂}} are finite-valued signed measures. Applying the above argument, one obtains two functions, {{math|g, h : X → [0, ∞)}}, satisfying the required properties for {{math|ν₁}} and {{math|ν₂}}, respectively. Clearly, {{math|1=f = g + ih}} is the required function.

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 $\backslash mu$ on the measure space $(X,\backslash Sigma)$ and a σ-finite signed measure $\backslash nu$ on $\backslash Sigma$, without assuming any absolute continuity. Then there exist unique signed measures $\backslash nu\_a$ and $\backslash nu\_s$ on $\backslash Sigma$ such that $\backslash nu=\backslash nu\_a+\backslash nu\_s$, $\backslash nu\_a\backslash ll\backslash mu$, and $\backslash nu\_s\backslash perp\backslash mu$. The Radon–Nikodym theorem can then be applied to the pair $\backslash nu\_a,\backslash mu$.

• Girsanov theorem
• Radon–Nikodym set

{{Reflist}}

• {{cite book |last=Lang |first=Serge |year=1969 |title=Analysis II: Real analysis |author-link= Serge Lang |publisher=Addison-Wesley }} Contains a proof for vector measures assuming values in a Banach space.
• {{cite book |last1=Royden |first1=H. L. | last2= Fitzpatrick | first2 = P. M.| author1-link= Halsey Royden |year=2010 |title=Real Analysis |edition= 4th |publisher= Pearson }} Contains a lucid proof in case the measure ν is not σ-finite.
• {{cite book |last=Shilov |first=G. E. |last2=Gurevich |first2=B. L. |year=1978 |title=Integral, Measure, and Derivative: A Unified Approach |others=Richard A. Silverman, trans. |publisher=Dover Publications |isbn=0-486-63519-8 }}
• {{Cite book| publisher = Princeton University Press| isbn = 978-0-691-11386-9| last1 = Stein| first1 = Elias M.| last2 = Shakarchi| first2 = Rami| title = Real analysis: measure theory, integration, and Hilbert spaces| location = Princeton, N.J| series = Princeton lectures in analysis| date = 2005}} Contains a proof of the generalisation.
• {{cite web |last=Teschl |first=Gerald |author-link=Gerald Teschl |title=Topics in Real and Functional Analysis |url=https://www.mat.univie.ac.at/~gerald/ftp/book-fa/index.html |others=(lecture notes) }}

{{PlanetMath attribution|id=3998|title=Radon–Nikodym theorem}}

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