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 analysis. 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, ∞] is a ∞ P∞ 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) < ∞.

Deﬁnition(σ-Finite Measure) A measure µ on (Ω, F) is σ-ﬁnite if ∃An ∈ F, n ≥ 1 such that ∪nAn = Ω and µ(An) < ∞.

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) = ∞ while probability measure must be finite: P (Ω) = 1. However, Lebesgue measure is σ-finite because R = ∪n[−n, n] and m([−n, n]) < ∞.

Deﬁnition(Signed Measure) (Ω, F) is a measurable space, a set map ν : F → (−∞, +∞] ∞ P∞ 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 ≤ +∞ while R − R f dµ < +∞. Then for ∀E ∈ 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 relations 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), ∀E ∈ 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 ∃A, B ∈ F,A ∩ B = ∅ such that µ(E) = µ(E ∩ A) and ν(E) = ν(E ∩ B) for all E ∈ 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 decomposition 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 |l(f)| ≤ |f(x)|dν(x) ≤ |f(x)|dρ(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 ∈ L2(Ω, ρ) such that Z Z l(f) = f(x)dν(x) = f(x)g(x)dρ(x), ∀f ∈ L2(Ω, ρ) Ω Ω R For E ∈ 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, ∀E ∈ 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 = {x ∈ Ω, 0 ≤ g(x) < 1},B = {x ∈ Ω, g(x) = 1} 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 ∈ A, then (1 − gn+1)(x) → 1; if x ∈ B, then (1 − gn+1)(x) → 0. When x ∈ A, Pn n g g i=0 g → 1−g . Let n → ∞, 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→∞ n→∞ 1 − g E E i=0 E That is Z g νa(E) = fdµ, where f = E 1 − g 1 At this time, f ∈ L (Ω, µ) since νa(Ω) ≤ ν(Ω) < ∞.

4 Now consider the condition that µ and ν are σ-finite positive measures. Obviously, we can find Ej ∈ F, ∪jEj = Ω and µ(Ej) < ∞, ν(Ej) < ∞, ∀j 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. Then it suﬃces P P P to set νa = j νj,a, νs = j νj,s, f = j fj such that ν = νa + νs where νs ⊥ µ, νa µ and R νa = fdµ.

Finally, if ν is a σ-ﬁnite signed measure, use the Jordan decomposition theorem, we sepa- rately consider ν+ and ν− as above.

0 0 It remains to prove that the decomposition is unique. Suppose ν = νa + νs = νa + νs, we 0 0 0 0 0 ﬁnd νa − νa = νs − νs. However, νa µ, νa µ, νs ⊥ µ and νs ⊥ µ, so νa − νa µ and 0 0 0 νs − νs ⊥ µ. Thus, νa − νa = νs − νs = 0.

As we have seen in the proof, when ν is ﬁnite, it’s easy to see the Radon-Nikodym derivative is µ-integrable. This case is extremely useful in probability theory because the probability measure is always positive and ﬁnite.

Halmos pointed out that the theorem remains true if ν is a signed measure without the condition that ν is σ-finite in [7]. The proof of this case can be found in [3]. In addition, the result is also true for any complex measure ν. The proof of this condition can be found in [4]. But µ is σ-finite is necessary. To see this, consider Ω = [0, 1] and F is the set of all Lebesgue measurable sets in [0, 1]. Let ν be the Lebesgue measure, then it’s finite. Define µ by µ(∅) = 0 and µ(A) = ∞ for all ∅ 6= A ∈ F. Suppose the Radon-Nikodym theorem still holds, we can find f ∈ L1(Ω, µ) such that dν = fdµ, where f is µ-integrable because ν is finite. However, L1(Ω, µ) = {0}, so we must have ν = 0, a contradiction!

3 Some applications

3.1 Conditional expectation One of the most important applications of Radon-Nikodym theorem is the existence of conditional expectation. In probability theory, everything happens on (Ω, F,P ), where Ω is called the sample space, F is called the events ﬁeld, and P is the probability measure. A random variable X : (Ω, F,P ) → (R, B) is a Borel-measurable function. If X is also integrable, we can deﬁne the expectation by Z E[X] = X(ω)dP (ω) Ω

5 In advanced probability theory, we deﬁne conditional expectation in a strange way.

Deﬁnition(Conditional Expectation) Suppose X ∈ L1(Ω, F,P ), and G ⊂ F is a σ- 1 algebra. If Y ∈ L (Ω, G,P ) such that E[X1G] = E[Y 1G], ∀G ∈ G, then we say Y is the conditional expectation of X given G. We denote this by Y = E[X|G].

Note that here Y must be G-integrable. If we only require Y to be F-integrable, then just take Y = X, then we are done. Thus, the existence of Y is non-trivial. However, if X is G-integrable, it’s trivial that E[XG] = X. And we claim if X is G-integrable, Y is F-integrable, then E[XY |G] = XE[Y |G]. Besides, form the deﬁnition we immediately know E[E[X|G]] = E[X] by taking G = Ω. Now, let’s go back to the theorem.

Theorem Suppose X ∈ L1(Ω, F,P ), and G ⊂ F is a σ-algebra, then exists Y ∈ L1(Ω, G,P ) ¯ 1 such that E[X1G] = E[Y 1G], ∀G ∈ G. In addition, if there is another Y ∈ L (Ω, G,P ) satisfying the conditions, we must have Y = Y¯ almost everywhere.

Proof: For all G ∈ G, define ν(G) = E[X1G]. It’s not difficult to verify that ν is a finite signed measure on (Ω, G). And we define µ(G) = P (G), the restriction of probability measure P on G. Then µ is a finite positive measure and ν µ because Z ν(G) = E[X1G] = XdP , P = µ on G G By Radon-Nikodym theorem, there exists an µ-measurable function Y such that dν = Y dµ. That is to say Z Z E[X1G] = ν(G) = Y dµ = Y dP = E[Y 1G] G G Note that ν is finite here, so Y is also integrable, i.e. Y ∈ L1(Ω, G,P ). If there is another Y¯ , ¯ ¯ ¯ we have E[(Y − Y )1G] = 0 for all G ∈ G. Because both Y and Y are measurable, Y − Y is also measurable. Take G = {Y − Y¯ > 0} and G = {Y − Y¯ < 0}, we know Y = Y¯ almost everywhere.

3.2 The dual space of Lp(Ω, F, µ) In real analysis and functional analysis, we all know the dual space of Lp is Lq where 1 1 p + q = 1, 1 < p < ∞. Here, I give a proof of this result on an abstract measurable space by Radon-Nikodym theorem. The main reference for this part is [4]. First, we need a lemma.

Lemma Suppose µ is a σ-ﬁnite measure on (Ω, F), then there exists ω ∈ L1(Ω, F, µ) such that ∀x ∈ Ω, 0 < ω(x) < 1.

6 Proof: µ is σ-ﬁnite, so we can ﬁnd En ∈ F such that Ω = ∪nEn and µ(En) < +∞. When 1 Pn x ∈ Ω − E , let ω (x) = 0; when x ∈ E , let ω (x) = n . Then ω(x) = ω (x) is n n n n 2 (1+µ(En)) i=1 n what we want.

The key point of this lemma is that we can construct a finite measure dµ¯ = ωdµ to replace the σ-finite measure. It’s not difficult to verify thatµ ¯ is a finite measure by the definition of ω.

Now, we come to the main theorem of this part. The special case is well-known in elementary real analysis. Here is the abstract version.

Theorem µ is a σ-ﬁnite measure on (Ω, F), ϕ is a bounded linear functional on Lp(Ω, F, µ), q R 1 1 then there exists a unique g ∈ L (Ω, F, µ) such that ϕ(f) = Ω fgdµ where p + q = 1. In addition, ||ϕ|| = ||g||q.

Proof: The uniqueness of g is obvious. Otherwise, suppose ϕ(f) = R fgdµ = R fgdµ¯ , R R Ω Ω we must have Ω f(g − g¯)dµ = 0. Take f = χE, we know E(g − g¯)dµ = 0. Then take E = {g − g¯ > 0} and E = {g − g¯ < 0}, we immediately get g =g ¯ almost everywhere.

First, we consider the condition µ(Ω) < ∞. For any E ∈ F, deﬁne ν(E) = ϕ(χE). If A, B ∈ F are disjoint, then χA∪B = χA + χB, thus ν(A ∪ B) = ϕ(χA∪B) = ϕ(A) + ϕ(B) = ν(A) + ν(B). So ν is ﬁnite addictive. To see ν is countable addictive, suppose E = ∪nEn k where En ∈ F are disjoint. Let Ak = ∪n=1Ek, note that

Z 1 ||χ − χ || = ( χ ∞ dµ) p → 0 E Ak p ∪n=k+1En Ω

By the continuity of ϕ, ν(Ak) = ϕ(χAk ) → ϕ(χE) = ν(E). We now proved ν is countable additive, so ν is indeed a measure. By the deﬁnition of ν, we know ν µ. Then by Radon- 1 R Nikodym theorem, there exists g ∈ L (Ω, F, µ) such that for all E ∈ F, ν(E) = E gdµ, that is Z ϕ(χE) = χEgdµ Ω R Because the linearity and continuity of ϕ, we know ϕ(f) = Ω fgdµ holds for any f ∈ Lp(Ω, F, µ).

q q−1 It remains to show that g ∈ L (Ω, F, µ). Deﬁne En = {x : |g(x)| ≤ n} and f = χEn |g| . p q ∞ Then we have |f| = |g| one En and f ∈ L (Ω, F, µ). Z Z Z q q 1 |g| dµ = fgdµ = ϕ(f) ≤ ||ϕ|| · ||f||p = ||ϕ||( |g| ) p En Ω Ω Thus, we ﬁnd R |g|q ≤ ||ϕ||q. Apply monotone convergence theorem, we know ||g|| ≤ ||ϕ||. En q q So we have proved g ∈ L (Ω, F, µ). However, by Holder’s inequality, ||ϕ(f)|| ≤ ||f||p||g||q implies ||ϕ|| ≤ ||g||q. Thus we must have ||ϕ|| = ||g||q.

7 Now, we consider the condition µ(Ω) = ∞. But µ is σ-finite, so by the lemma, we can find a µ-integrable ω such that dµ¯ = ωdµ define a finite measure on (Ω, F). We map each f ∈ Lp(Ω, F, µ¯) to ω1/pf ∈ Lp(Ω, F, µ) and define a linear function ψ on Lp(Ω, F, µ¯) by ψ(f) = ϕ(ω1/pf). Obviously, ||ψ|| = ||ϕ||. Use the result of the first part, we know there q R 1/q existsg ¯ ∈ L (Ω, F, µ¯) such that ψ(f) = Ω fgd¯ µ¯. Then let g = ω g¯, Z Z Z ϕ(f) = ψ(ω−1/pf) = ω−1/pfgd¯ µ¯ = ω1/qfgdµ¯ = fgdµ Ω Ω Ω In addition, Z Z |g|qdµ = |g¯|qdµ¯ = ||ψ||q = ||ϕ||q Ω Ω

So the same as above, ||ϕ|| = ||g||q.

Above all, we have ﬁnished the proof.

q R p Note that for all g ∈ L (Ω, F, µ), ϕ(f) = Ω fgdµ is a linear functional on L (Ω, F, µ). From the theorem above we know every linear functional must be of this form. Thus we ﬁnd an isometric isomorphism between the dual space of Lp(Ω, F, µ) and Lq(Ω, F, µ), i.e. we can regard (Lp(Ω, F, µ))∗ = Lq(Ω, F, µ) Moreover, Lp(Ω, F, µ) is a reﬂexive space.

3.3 Change of measure Let’s see a probabilistic application of the Radon-Nikodym theorem, more speciﬁcally, the Radon-Nikodym derivative. The change of measure is the fundament of Girsanov’s theorem, which is one of the most important theorem in stochastic analysis and ﬁnancial engineering. First, we introduce the concept of equivalent measures.

Deﬁnition(Equivalent Measures) Suppose P and Q are two measures on (Ω, F), they are equivalent if Q P and P Q.

By the definition we immediately know that if P,Q are equivalent, then P (A) = 0 ⇔ Q(A) = 0. Thus, we can say almost everywhere without telling the measure. In this part, we just take P,Q as probability measures. When defining expectation (and conditional expec- R R tation), we use EP [X] to denote Ω XdP and EQ[X] to denote Ω XdQ. Before I state the theorem, we still need some basic definitions.

Deﬁnition(Filtration) Ft is called a ﬁltration if they are a increasing chain of σ-algebra of Ω. That is to say, Ft is a σ-algebra for anyt and t < s implies Ft ⊂ Fs.

8 Deﬁnition(Adapted Process) Xt is an Ft-adapted process if Xt is a random variable on (Ω, Ft) where Ft is a ﬁltration.

Theorem If P and Q are equivalent probability measures, and Xt is an Ft-adapted process, dQ dQ let dP be the Radon-Nikodym derivative and Ls = EP [ dP |Fs], then for s < t dQ E [X ] = E [ X ] Q t P dP t −1 EQ[Xt|Fs] = Ls EP [LtXt|Fs]

Proof: The ﬁrst equality is trivial by the deﬁnition of Radon-Nikodym derivative: Z Z dQ dQ EQ[Xt] = XtdQ = Xt dP = EP [ Xt] Ω Ω dP dP

The second equality is well-deﬁned because P ({Ls = 0}) = 0. Indeed, dQ Q({Ls = 0}) = Q({EP [ |Fs] = 0}) = EQ[1{E [ dQ |F ]=0}] dP P dP s By the ﬁrst equality, we have dQ EQ[1{E [ dQ |F ]=0}] = EP [ 1{E [ dQ |F ]=0}] P dP s dP P dP s

Conditioning on Fs, we know dQ dQ dQ EP [ 1{E [ dQ |F ]=0}] = Ep[EP [ 1{E [ dQ |F ]=0}|Fs]] = Ep[EP [ |Fs]1{E [ dQ |F ]=0}] = 0 dP P dP s dP P dP s dP P dP s

To prove the second equality, we need to recall the deﬁnition of conditional expectation. −1 For ∀A ∈ Fs, we need to prove EQ[Xt1A] = EQ[1ALs EP [LtXt|Fs]]. By the ﬁrst equality, dQ E [1 L−1E [L X |F ]] = E [ 1 L−1E [L X |F ]] Q A s P t t s P dP A s P t t s

Conditioning on Fs, dQ dQ E [ 1 L−1E [L X |F ]] = E [E [ 1 L−1E [L X |F ]|F ] P dP A s P t t s P P dP A s P t t s s

EP [EP [LtXt|Fs]1A] = EP [LtXt1A]

Expand Lt, we note both 1A and Xt are Ft-integrable, dQ E [L X 1 ] = E [E [ |F ]X 1 ] = E [X 1 ] P t t A P P dP t t A Q t A 4 Reference

[1] Elias M. Stein, Rami Shakarchi, Real Analysis

[2] Emmanuele DiBenedetto, Real Analysis

[3] Hu XiaoYu, Advanced Probability

[4] Walter Rudin, Real Analysis and Complex Analysis

[5] Gerald B. Folland, Real Analysis

[6] Peter Lax, Functional Analysis

[7] Paul Halmos, Measure Theory

[8] Charalambos D. Aliprantis, Owen Burkinshaw, Principles of Real Analysis

[9] Alan Bain, Stochastic Analysis