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Radon-Nikodym Theorem and Its Applications

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Radon-Nikodym Theorem and Its Applications Radon-Nikodym Theorem And Its Applications Li Tao PB13001106 Abstraction: Radon-Nikodym theorem is one of the most important result in real analy- sis. In this paper, I will introduce the background of this problem and some basic definitions at first. Then I will give the proof, as well as several applications of Radon-Nikodym theorem, including conditional expectation, the dual space of Lp and change of measure in stochastic analysis. Keywords: Radon-Nikodym theorem, conditional expectation, dual space of Lp, change of measure 1 Introduction and notations Generally speaking, Radon-Nikodym theorem gives the connection between two measures. 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. In 1936 Hans Freudenthal further generalized the Radon-Nikodym theorem by proving the Freudenthal spectral theorem, a result in Riesz space theory, which contains the Radon-Nikodym theorem as a special case. However, before we state the theorem, we need some basic definitions. In this lecture, we always assume the background mesurable space is (Ω; F). Definition(Measure) (Ω; F) is a measurable space, a set function µ : F! [0; 1] is a 1 P1 measure if µ(;) = 0 and µ([i=1Ai) = i=1 µ(Ai) for disjoint Ai. Definition(Finite Measure) A measure µ on (Ω; F) is finite if µ(Ω) < 1. Definition(σ-Finite Measure) A measure µ on (Ω; F) is σ-finite if 9An 2 F; n ≥ 1 such that [nAn = Ω and µ(An) < 1. As we all know, Lebesgue measure m is a measure on (R; B). In probability theory, we often deal with probability measure P on (Ω; F),where Ω is called the sample space and F is called the events field. Lebesgue measure on (R; B) is not a finite measure because m(R) = 1 while probability measure must be finite: P (Ω) = 1. However, Lebesgue measure is σ-finite because R = [n[−n; n] and m([−n; n]) < 1. Definition(Signed Measure) (Ω; F) is a measurable space, a set map ν : F! (−∞; +1] 1 P1 is a signed measure if ν(;) = 0 and ν([i=1Ai) = i=1 ν(Ai) for disjoint Ai. 1 Loosely speaking, a signed measure possesses all the properties of a measure, except that it may take positive or negative values. Let's see an example of signed measure. Consider we have a function f on (Ω; F; µ) which is \integral" in the extent sense: R f +dµ ≤ +1 while R − R f dµ < +1. Then for 8E 2 F, let ν(E) = E fdµ, we claim that ν is a signed measure. Given two measures defined on a common σ-algebra, we can describe some special rela- tions between them.More concretely, consider two measures µ and ν on (Ω; F), two extreme scenarios are • µ and ν are \supported" on separate parts of F. • The \support" of ν is an essential part of the \support" of µ. Here, the meaning of \support" is similar to the meaning of support in function theory. We say µ is supported on A if µ(E) = µ(E \ A); 8E 2 F. That is to say, the non-zero part of µ is on A. To formally describe the relations above, we have the following definitions. Definition(Mutually Singular) Two signed measures µ and ν on (Ω; F) are mutually singular if 9A; B 2 F;A \ B = ; such that µ(E) = µ(E \ A) and ν(E) = ν(E \ B) for all E 2 F. We denote this by µ ? ν. Definition(Absolutely Continuous) ν is a signed measure on (Ω; F) and µ is a positive measure on (Ω; F), if µ(A) = 0 implies ν(A) = 0 then we say ν is absolutely continuous with respect to µ. We denote this by ν µ. Just as what we do in Lebesgue integration theory, we can decompose a signed measure ν into two parts: positive and negative. However, the reason is not obvious. Here I state the theorem without proof. Theorem(Jordan Decomposition) Let ν be a signed measure on (Ω; F). There exists a unique pair (ν+; ν−) of mutually singular measures, one of which is finite, such that ν = ν+ − ν−. R Consider the example ν(E) = E fdµ we discussed above. It's easily to see that ν µ. The Radon-Nikodym theorem indicates the converse is also true. Theorem(Radon-Nikodym Theorem) Suppose µ is a σ-finite positive measure on (Ω; F) and ν is a σ-finite signed measure on (Ω; F). Then there exists unique signed measures νa and νs such that ν = νa + νs, νa µ, νs ? µ. In addition, νa takes the form dνa = fdµ. That is to say, for some extended µ-integrable function f, Z νa(E) = fdµ E dνa f here is said to be the Radon-Nikodym derivative dµ . 2 Note that the theorem above has two parts: decomposition and representation. The de- composition part tells us that a σ-finite signed measure can always be decomposed to two mutually singular measures. The representation part gives us an integration representation of a σ-finite signed measure which is absolutely continuous with respect to a σ-finite positive measure. Two direct corollaries of this theorem are as following: Corollary Suppose µ is a σ-finite positive measure on (Ω; F) and ν is a σ-finite signed measure on (Ω; F), then there exists unique signed measures νa and νs such that ν = νa + νs, νa µ, νs ? µ. Corollary Suppose µ is a σ-finite positive measure on (Ω; F) and ν is a σ-finite signed measure on (Ω; F), and ν µ. Then there exists some extended µ-integrable function f such R that ν(E) = E fdµ. Sometimes we call the first corollary the Lebesgue decomposition theorem and the second corollary the Radon-Nikodym theorem too. 2 Proof of Radon-Nikodym theorem This proof, due to Von Neumann, has the virtue that it exploits elegantly the application of a simple Hilbert space idea. The main references for this part are [1] and [6]. Proof: First, we assume that both µ and ν are positive and finite. Let ρ = µ + ν, and consider the Hilbert space L2(Ω; ρ). Define a mapping on L2(Ω; ρ) by Z 2 l : L (Ω; ρ) ! R; f 7! l(f) = f(x)dν(x) Ω We claim this mapping is a linear functional, since Z Z Z jl(f)j ≤ jf(x)jdν(x) ≤ jf(x)jdρ(x) ≤ ( f 2(x)dρ(x))1=2(ρ(Ω))1=2 Ω Ω Ω where the last inequality holds for Cauchy-Schwarz inequality. By the assumption that µ and ν are finite, we know ρ is finite, thus l is bounded, which implies l is continuous. The linearity of l is obvious, so l is indeed a linear functional on L2(Ω; ρ). By Risez representation theorem, we know there exists g 2 L2(Ω; ρ) such that Z Z l(f) = f(x)dν(x) = f(x)g(x)dρ(x); 8f 2 L2(Ω; ρ) Ω Ω R For E 2 F and ρ(E) > 0, let f(x) = χE, we get ν(E) = E g(x)dρ(x). But we have assumed ν and µ are positive, thus ν(E) ≤ ρ(E). We get the following 1 Z 0 ≤ g(x)dρ(x) ≤ 1 ρ(E) E 3 From the inequality above, we find Z Z gdρ ≥ 0; (1 − g)dρ ≥ 0; 8E 2 F E E This indicates 0 ≤ g(x) ≤ 1 almost everywhere with respect to measure ρ. Note that ρ = µ+ν, we can rewrite the Risez representation as Z Z f(1 − g)dν = fgdµ Ω Ω Now consider two sets A = fx 2 Ω; 0 ≤ g(x) < 1g;B = fx 2 Ω; g(x) = 1g and define two measures on F by νa(E) = ν(A \ E); νs(E) = ν(B \ E) R R In the equality Ω f(1 − g)dν = Ω fgdµ, let f = χB, we get Z Z 0 = (1 − g)dν = gdµ = µ(B) B B Thus, the support of νs is separate from the support of µ, νs ? µ. Pn n Finally, let f = χE i=0 g , we get n Z Z X (1 − gn+1)dν = g gndµ E E i=0 Note that if x 2 A, then (1 − gn+1)(x) ! 1; if x 2 B, then (1 − gn+1)(x) ! 0. When x 2 A, Pn n g g i=0 g ! 1−g . Let n ! 1, by dominated convergence theorem we get Z Z n Z n+1 X n g νa(E) = ν(E \ A) = lim (1 − g )dν = lim g g dµ = dµ n!1 n!1 1 − g E E i=0 E That is Z g νa(E) = fdµ, where f = E 1 − g 1 At this time, f 2 L (Ω; µ) since νa(Ω) ≤ ν(Ω) < 1. 4 Now consider the condition that µ and ν are σ-finite positive measures. Obviously, we can find Ej 2 F, [jEj = Ω and µ(Ej) < 1; ν(Ej) < 1; 8j We then define finite and positive measures by µj(E) = µ(E \ Ej); νj(E) = ν(E \ Ej) Now, we have reduced the problem to the case we have talked above. By the result above, R we claim νj = νj;a + νj;s, νj;s ? µj, νj;a µj and νj;a = fjdµj for some fj.
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