Radon–Nikodym theorem

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In mathematics, the Radon–Nikodym theorem is a result in functional analysis that states that, given a measure space (X,Σ), if a measure ν is absolutely continuous with respect to another measure μ which is sigma-finite, then there is a measurable function f on X and taking values in [0,∞), such that

\nu(A) = \int_A f \, d\mu

for any measurable set A.

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.

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[edit] 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. It is commonly written dν/dμ 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. 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 |\nu| \ll \mu, there is μ-integrable real- or complex-valued function g on X such that

\nu(A) = \int_A g \, d\mu,

for any measurable set A.

[edit] Applications

The theorem is very important in probability theory. It tells if and how it is possible to change from one probability measure to another.

For example, it is necessary when proving existence of conditional expectation. 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 derivative securities.

[edit] Properties

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

[edit] The assumption of sigma-finiteness

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

Consider the Borel sigma-algebra on the real line. Let the 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 sigma-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 in 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

\nu(A) = \int_A f \, d\mu

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

0 = f(a)

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

[edit] Proof of the theorem

[edit] For finite measures

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

\int_A f\,d\mu\leq\nu(A)

for every A ∈ Σ (this set is not empty, for it contains at least the zero function). Let f1, f2 ∈ F; let A be an arbitrary measurable set, A1 = {x ∈ A | f1(x) > f2(x)}, and A2 = {x ∈ A | f2(x) ≥ f1(x)}. Then one has

\int_A\max\{f_1,f_2\}\,d\mu = \int_{A_1} f_1\,d\mu+\int_{A_2} f_2\,d\mu \leq \nu(A_1)+\nu(A_2)=\nu(A),

and therefore, max{f1,f2} ∈ F.

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

\lim_{n\to\infty}\int_X f_n\,d\mu=\sup_{f\in F} \int_X f\,d\mu.

By replacing fn with the maximum of the first n functions, one can assume that the sequence {fn} is increasing. Let g be a function defined as

g(x):=\lim_{n\to\infty}f_n(x).

By Lebesgue's monotone convergence theorem, one has

\int_A g\,d\mu=\lim_{n\to\infty} \int_A f_n\,d\mu \leq \nu(A)

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

\int_X g\,d\mu=\sup_{f\in F}\int_X f\,d\mu.

Now, since g ∈ F,

\nu_0(A):=\nu(A)-\int_A g\,d\mu

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,

\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(g+\varepsilon1_P)\,d\mu.

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

\nu_0(X)-\varepsilon\mu(X)=(\nu_0-\varepsilon\mu)(N)\leq 0,

contradicting the fact that ν0(X) > ε μ (X).

Then, since

\int_X g\,d\mu \leq \nu(X) < +\infty,

g + ε 1P ∈ F and satisfies

\int_X (g+\varepsilon 1_P)\,d\mu>\int_X g\,d\mu=\sup_{f\in F}\int_X f\,d\mu.

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

f(x)=\begin{cases} g(x)&\mbox{if }g(x) < \infty\\0&\mbox{otherwise,}\end{cases}

then f has the desired properties.

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

\nu(A)=\int_A f\,d\mu=\int_A g\,d\mu

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

\int_A (g-f)\,d\mu=0.

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

\int_X (g-f)^+\,d\mu=0=\int_X (g-f)^-\,d\mu,

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.

[edit] For σ-finite positive measures

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. For each n, there is a Σ-measurable function fn : Bn→[0,+∞) such that

\nu(A)=\int_A f_n\,d\mu

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

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

[edit] For signed and complex measures

If ν is a 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+i ν2, 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+i h is the required function.

Q.E.D.

[edit] References

  • Shilov, G. E., and Gurevich, B. L., 1978. Integral, Measure, and Derivative: A Unified Approach, Richard A. Silverman, trans. Dover Publications. ISBN 0486635198.

This article incorporates material from Radon-Nikodym theorem on PlanetMath, which is licensed under the GFDL.

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