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4. Radon-Nikodym Theorems
In this section we discuss a very important property which has many important applications.
Definition. Let X be a non-empty set, and let A be a σ-algebra on X. Given two measures μ and ν on A, we say that ν has the Radon-Nikodym property relative to μ, if there exists a measurable function f : X → [0, ∞], such that
(1) ν(A) =
A
f dμ, ∀ A ∈ A.
Here we use the convention which defines the integral in the right hand side by ∫
A
f dμ =
X f^ κA^ dμ^ if^ f^ κA^ ∈^ L
1 +(X,^ A, μ) ∞ if f κA 6 ∈ L^1 +(X, A, μ)
In this case, we say that f is a density for ν relative to μ.
The Radon-Nikodym property has an equivalent useful formulation. Proposition 4.1 (Change of Variables). Let X be a non-empty set, and let A be a σ-algebra on X, let μ and ν be measures on A, and let f : X → [0, ∞] be a measurable function. A. The following are equivalent (i) ν has the Radon-Nikodym property relative to μ, and f is a density for ν relative to μ; (ii) for every measurable function h : X → [0, ∞], one has the equality^1
(2)
X
h dν =
X
hf dμ.
B. If ν and f are as above, and K is either R or C, then the equality (2) also holds for those measurable functions h : X → K with h ∈ L^1 K(X, A, ν) and hf ∈ L^1 K(X, A, μ).
Proof. A. (i) ⇒ (ii). Assume property (i) holds, which means that we have (1). Fix a measurable function h : X → [0, ∞], and use Theorem III.3.2, to find a sequence (hn)∞ n=1 ⊂ A-ElemR(X), with
(a) 0 ≤ h 1 ≤ h 2 ≤ · · · ≤ h; (b) limn→∞ hn(x) = h(x), ∀ x ∈ X.
Of course, we also have
(a′) 0 ≤ h 1 f ≤ h 2 f ≤ · · · ≤ hf ; (b′) limn→∞ hn(x)f (x) = h(x)f (x), ∀ x ∈ X.
Using the Monotone Convergence Theorem, we then get the equalities
(3)
X
h dν = lim n→∞
X
hn dν and
X
hf dμ = lim n→∞
X
hnf dν
(^1) For the product hf we use the conventions 0 · ∞ = ∞ · 0 = 0, and t · ∞ = ∞ · t = ∞, ∀ t ∈ (0, ∞].
300
§4. Radon-Nikodym Theorems 301
Notice that, if we fix n and we write hn =
∑p k=1 αkκAk , for some^ A^1 ,... , Ap^ ∈^ A, and α 1 > · · · > αp > 0, then ∫
X
hn dν =
∑^ p
k=
αkν(Ak) =
∑^ p
k=
X
αkκAk f dμ =
X
hnf dμ,
so using (3), we immediately get (2). The implication (ii) ⇒ (i) is trivial, using functions of the form h = κA, A ∈ A. B. Suppose ν has the Radon-Nikodym property relative to μ, and f is a density for ν relative to μ, and let h : X → K be a measurable function with h ∈ L^1 K(X, A, ν) and hf ∈ L^1 K(X, A, μ). In the complex case, using the inequalities |Re h| ≤ |h| and |Im h| ≤ |h|, it is clear that both functions Re h and Im h belong to L^1 (X, A, ν), and also the products (Re h)f and (Im h)f belong to L^1 (X, A, μ). This shows that it suffices to prove (2) under the additional hypothesis that h is real-valued. In this case we consider the functions h±, defined by
h+^ = max{h, 0 } and h−^ = max{−h, 0 }.
Since we have 0 ≤ h±^ ≤ |h|, it follows that h±^ ∈ L^1 +(X, A, ν), as well as h±f ∈ L^1 +(X, A, μ). In particular, we get the equalities
(4)
X
h dν =
X
h+^ dν −
X
h−^ dν and
X
hf dν =
X
h+f dμ −
X
h−f dμ.
Since h±^ ≥ 0, we can use property A.(ii) above, and we have ∫
X
h±^ dν =
X
h±f dμ,
and then the desired equality (2) immediately follows from (4).
One important issue is the uniqueness of the density. For this purpose, it will be helpful to introduce the following.
Definition. Let T be one of the spaces [−∞, ∞] or C, and let r be some relation on T (in our case r will be either “=,” or “≥,” or “≤,” on [−∞, ∞]). Given a measurable space (X, A, μ), and two measurable functions f 1 , f 2 : X → T ,
f 1 r f 2 , μ-l.a.e.
if the set
A =
x ∈ X : f 1 (x) r f 2 (x)
belongs to A, and it has locally μ-null complement in X, i.e. μ
[X r A] ∩ F ) = 0, for every set F ∈ A with μ(F ) < ∞. (If r is one of the relations listed above, the set A automatically belongs to A, so all intersections [X r A] ∩ F , F ∈ A, also belong to A.) The abreviation “μ-l.a.e.” stands for “μ-locally-almost everywhere.” Remark that one has the implication
f 1 r f 2 , μ-a.e. ⇒ f 1 r f 2 , μ-l.a.e.
Remark that, when μ is σ-finite, then the other implication also holds:
f 1 r f 2 , μ-l.a.e. ⇒ f 1 r f 2 , μ-a.e.
With this terminology, one has the following uniqueness result.
§4. Radon-Nikodym Theorems 303
Remark 4.1. The local almost uniqueness result, given in Proposition 4.2, holds under slightly weaker assumptions. Namely, if (X, A, μ) is a measure space, and if f, g : X → [0, ∞] are measurable functions for which we have the equality ∫
A
f dμ =
A
g dμ,
for all A ∈ A with μ(A) < ∞, then we still have the equality f = g, μ-l.a.e. This follows actually from Proposition 4.2, applied to functions of the form f
A and^ g
A.
Let us introduce the following. Notations. For a measure space (X, A, μ) we define Aμ 0 = {N ∈ A : μ(N ) = 0}; Aμ fin = {F ∈ A : μ(F ) < ∞}; Aμ 0 ,loc = {A ∈ A : μ(A ∩ F ) = 0, ∀ F ∈ Aμ fin}.
With these notations, we have the inclusions
Aμ 0 = Aμ 0 ,loc ∩ Aμ fin ⊂ Aμ 0 ,loc ⊂ A,
and Aμ 0 and Aμ 0 ,loc are in fact σ-rings.
Comment. The “locally-almost everywhere” terminology is actually designed to “hide some pathologies under the rug.” For instance, if (X, A, μ) is a degenerate measure space , i.e. μ(A) ∈ { 0 , ∞}, ∀ A ∈ A, then “anything happens locally almost-everywhere,” which means that we have the equality Aμ 0 ,loc = A. At the other end, there is a particular type of measure spaces on which, even in the absence of σ-finiteness, the notions of “locally-almost everywhere” and ”almost everywhere” coincide, i.e. we have the equality Aμ 0 ,loc = Aμ 0. Such spaces are described by the following.
Definition. A measure space (X, A, μ) is said to be nowhere degenerate, or with finite subset property, if
(f) for every set A ∈ A with μ(A) > 0 , there exists some set F ∈ A, with F ⊂ A, and 0 < μ(F ) < ∞.
With this terminology, one has the following result.
Proposition 4.3. For a measure space (X, A, μ), the following are equivalent: (i) Aμ 0 ,loc = Aμ 0 ; (ii) (X, A, μ) has the finite subset property. Proof. (i) ⇒ (ii). Assume Aμ 0 ,loc = Aμ 0 , and let us prove that (X, A, μ) has the finite subset property. We argue by contradiction, so let us assume there exists some set A ∈ A, with μ(A) = ∞, such that μ(B) ∈ { 0 , ∞}, for every B ∈ A, with B ⊂ A. In particular, if we start with some arbitrary F ∈ Aμ fin, using the fact that μ(A ∩ F ) ≤ μ(F ) < ∞, we see that we must have μ(A ∩ F ) = 0. This proves precisely that A ∈ Aμ 0 ,loc. By assumption, it follows that A ∈ Aμ 0 , i.e. μ(A) = 0, which is impossible. (ii) ⇒ (i). Assume that (X, A, μ) has the finite subset property, and let us prove the equality (i). Since one inclusion is always true, all we need to prove is the inclusion Aμ 0 ,loc ⊂ Aμ 0 , which equivalent to the inclusion Aμ 0 ,loc ⊂ Aμ fin. Start with some set A ∈ Aμ 0 ,loc, but assume μ(A) = ∞. On the one hand, using the finite
304 CHAPTER IV: INTEGRATION THEORY
subset property, there exists some set F ∈ A with F ⊂ A and μ(F ) > 0. On the other hand, since A ∈ Aμ 0 ,loc, we have μ(F ) = 0, which is impossible.
Example 4.2. Take X be an uncountable set, let A = P(X), and let μ be the counting measure, i.e.
μ(A) =
Card A if A is finite ∞ if A is infinite
Then (X, P(X), μ) has the finite subset property, but is not σ-finite.
When we restrict to integrable functions, the two notions μ-l.a.e, and μ-a.e. coincide. More precisely, we have the following.
Proposition 4.4. Let (X, A, μ) be a measure space, let K be one of the fields R or C, and let p ∈ [1, ∞). For a function f ∈ Lp K(X, A, μ), the following are equivalent:
(i) f = 0, μ-l.a.e. (ii) f = 0, μ-a.e. Proof. Of course, we only need to prove the implication (i) ⇒ (ii). Assume f = 0, μ-l.a.e. Using the function g = |f |p, we can assume that p = 1 and f (x) ≥ 0, ∀ x ∈ X. Consider then the set N = {x ∈ X : f (x) > 0 }, and write it as a union N =
n=1 Nn, where Nn = {x ∈ X : f (x) ≥ (^1) n }, ∀ n ≥ 1.
Of course, all we need is the fact that μ(Nn) = 0, ∀ n ≥ 1. Fix n ≥ 1. On the one hand, the assumption on f , it follows that Nn ∈ Aμ 0 ,loc. On the other hand, the inequality (^1) n κNn ≤ f , forces the elementary function (^) n^1 κNn to be μ-integrable, i.e. μ(Nn) < ∞. Consequently we have
N ∈ Aμ 0 ,loc ∩ Aμ fin = Aμ 0. Comment. In what follows we will discuss several results, which all have as conclusion the fact that one measure has the Radon-Nikodym property with respect to another one. All such results will be called “Radon-Nikodym Theorems.”
The first result is in fact quite general, in the sense that it works for finite signed or complex measures.
Theorem 4.1 (“Easy” Radon-Nikodym Theorem). Let (X, A, μ) be a finite measure space, let K denote one of the fields R or C, and let C > 0 be some constant. Suppose ν is a K-valued measure on A, such that
|ν(A)| ≤ Cμ(A), ∀ A ∈ A.
Then there exists some function f ∈ L^1 K(X, A, μ), such that
(5) ν(A) =
A
f dμ, ∀ A ∈ A.
Moreover:
(i) Any function f ∈ L^1 K(X, A, μ), satisfying (5) has the property |f | ≤ C, μ- a.e. If ν is an “honest” measure, then one also has the inequality |f | ≥ 0 , μ-a.e. (ii) A function satisfying (5) is essentially unique, in the sense that, whenever f 1 , f 2 ∈ L^1 K(X, A, μ) satisfy (5), it follows that f 1 = f 2 , μ-a.e.
306 CHAPTER IV: INTEGRATION THEORY
(Here we use the abusive notation that identifies an element in L^1 with a function in L^1 , which is defined almost uniquely.) Moreover, one has the inequality ∫
X
|h| dνk ≤ C
X
|h| dμ, ∀ h ∈ L^1 K(X, A, μ), k = 1, 2 , 3 , 4 ,
in other words, the linear maps (8) are all continuous. For every k = 1, 2 , 3 , 4, let φk denote the integration map
φk : L^1 K(X, A, νk) 3 h 7 −→
X
h dνk ∈ K.
We know (see Remark 3.5) that the φk’s are continuous. In particular, the compo- sitions ψk = φk ◦ Φk : L^1 K(X, A, μ) → K, which are defined by
ψk : L^1 K(X, A, μ) 3 h 7 −→
X
h dνk, k = 1, 2 , 3 , 4 ,
are linear and continuous. We now use Proposition 3.3 which states that one has an inclusion
(9) Θ : L^2 K(X, A, μ) ↪→ L^1 K(X, A, μ),
which is in fact a linear continuous map. So if we consider the compositions θk = ψk ◦ Θ, which are defined by
θk : L^1 K(X, A, μ) 3 h 7 −→
X
h dνk, k = 1, 2 , 3 , 4 ,
then these compositions are linear and continuous. Apply then Riesz Theorem (in the form given in Remark 3.4), to find functions f 1 , f 2 , f 3 , f 4 ∈ L^2 K(X, A, μ), such that θk(h) = 〈fk, h〉, ∀ h ∈ L^2 K(X, A, μ), k = 1, 2 , 3 , 4.
In particular, using functions of the form h = κA, A ∈ A (which all belong to L^2 K(X, A, μ), due to the finiteness of μ), we get
νk(A) =
X
κA dνk =
X
fkκA dμ, ∀ A ∈ A, k = 1, 2 , 3 , 4.
Finally, if we define the function f = α 1 f 1 + α 2 f 2 + α 3 f 3 + α 4 f 4 ∈ L^2 K(X, A, μ), then the above equalities immediately give the equality (5). At this point we only know that f belongs to L^2 K(X, A, μ). Using the inclusion (9), it turns out that f indeed belongs to L^1 K(X, A, μ). Let us prove now the additional properties (i) and (ii). To prove the first assertion in (i), we start off by fixing some function f ∈ L^1 K(X, A, μ), which satisfies (5), and we define the set
A = {x ∈ X : |f (x)| > C
for which we must prove that μ(A) = 0. Since f is measurable, it follows that A belongs to A. Consider the “rational unit sphere” S^1 Q in K, defined as
(10) S Q^1 =
{− 1 , 1 } if K = R {e^2 πit^ : t ∈ Q} if K = C
The point is that S^1 Q is dense in the unit sphere S^1 in K:
S^1 = {α ∈ K : |α| = 1},
§4. Radon-Nikodym Theorems 307
so we immediately have the equality A =
α∈S^1 Q^ Aα, where Aα = {x ∈ X : Re[αf (x)] > C
Since S Q^1 is countable, in order to prove that μ(A) = 0, it then suffices to show that μ(Aα) = 0, ∀ α ∈ S Q^1. Fix then α ∈ S^1 Q, and consider the K-valued measure η = αν. It is clear that we still have
(11) |η(A)| = |ν(A)| ≤ Cμ(A), ∀ A ∈ A,
as well as the equality
(12) η(A) =
A
αf dμ, ∀ A ∈ A.
For each integer n ≥ 1, let us define the set
Anα = {x ∈ X : Re[αf (x)] ≥ C + (^) n^1
so that we obviously have the equality Aα =
n=1 A
n α. In particular, in order to prove μ(Aα) = 0, it suffices to prove that μ(Anα) = 0, ∀ n ≥ 1. Fix for the moment n ≥ 1. Using (12), it follows that
Re η(Anα) = Re
[ ∫
Anα
αf dμ
]
Anα
Re[αf ] dμ =
X
Re[αf ]κAnα dμ.
Since we have Re[αf ]κAnα ≥ (C + (^) n^1 )κAnα , the above inequality can be continued with
Re η(Anα) ≥
X
(C + (^1) n )κAnα dμ = (C + (^1) n )μ(Anα).
Of course, this will give
|η(Anα)| ≥ Re η(Anα) ≥ (C + (^1) n )μ(Anα).
Note now that, using (11), this will finally give
Cμ(Anα) ≥ (C + (^1) n )μ(Anα),
which clearly forces μ(Anα) = 0. Having proven that |f | ≤ C, μ-a.e., let us turn our attention now to the unique- ness property (ii). Suppose f 1 , f 2 ∈ L^1 K(X, A, μ) are such that
ν(A) =
A
f 1 dμ =
X
f 2 dμ, ∀ A ∈ A.
Consider then the difference f = f 1 − f 2 and the trivial measure ν 0 = 0. Obviously we have |ν 0 (A)| ≤ (^) n^1 μ(A), ∀ A ∈ A,
for every integer n ≥ 1, as well as
ν 0 (A) =
A
f dμ, ∀ A ∈ A.
By the first assertion in (i), it follows that
|f 1 − f 2 | = |f | ≤
n
, μ-a.e.,
for every n ≥ 1. So if we take the sets (Nn)∞ n=1 ⊂ A defined by
Nn = {x ∈ X : |f 1 (x) − f 2 (x)| > (^) n^1 },
§4. Radon-Nikodym Theorems 309
uniqueness property (ii) is automatic, and we also have |f | ≤ 1, |ν|-a.e. To prove the fact that we have in fact the equality |f | = 1, |ν|-a.e., we define the set
A = {x ∈ X : |f (x)| < 1 },
which belongs to A, and we prove that |ν|(A) = 0. If we define the sequence of sets (An)∞ n=1 ⊂ A, by
An = {x ∈ X : |f (x)| ≤ 1 − (^1) n }, ∀ n ≥ 1 ,
then we clearly have A =
n=1 An, so all we have to show is the fact that^ |ν|(An) = 0, ∀ n ≥ 1. Fix n ≥ 1. For every B ∈ A, with B ⊂ An, we have
|f (x)| ≤ 1 − (^1) n , ∀ x ∈ B,
so using (13) we get
|ν(B)| =
B
f d|ν|
B
|f | d|ν| ≤
B
(1 − (^1) n ) d|ν| = (1 − (^) n^1 )|ν|(B).
Now if we take an arbitrary pairwise disjoint sequence (Bk)∞ k=1 ⊂ A, with
k=1 Bk^ = An, then the above estimate will give
∑^ ∞
k=
|ν(Bk)| ≤ (1 − (^1) n )
∑^ ∞
k=
|ν|(Bk) = (1 − (^1) n )|ν|(An).
Taking supremum in the left hand side, and using the definition of the variation measure, the above estimate will finally give
|ν|(An) ≤ (1 − (^) n^1 )|ν|(An),
which clearly forces |ν|(An) = 0.
Remark 4.2. The case K = R can be slighly generalized, to include the case of infinite signed measures. If ν is a signed measure on A and if we consider the Hahn-Jordan set decomposition (X+, X−), then the density f is simply the function
f (x) =
1 if x ∈ X+ − 1 if x ∈ X−
The equality (13) will then hold only for those sets A ∈ A with |ν|(A) < ∞. Since |ν| is allowed to be infinite, as explained in Example 4.1, the only version of uniqueness property (ii) will hold with “|ν|-l.a.e” in place of “|ν|-a.e” Likewise, the absolute value property (i) will have to be replaced with ”|f | = 1, |ν|-l.a.e”
Comment. Up to this point, it seems that the hypotheses from Theorem 4. are essential, particularly the dominance condition |ν| ≤ Cμ. It is worth discussing this property in a bit more detail, especially having in mind that we plan to weaken it as much as possible.
Notation. Suppose A is a σ-algebra on some non-empty set X, and suppose μ and ν are “honest” (not necessarily finite) measures on A. We shall write
ν b μ,
if there exists some constant C > 0, such that
ν(A) ≤ Cμ(A), ∀ A ∈ A. A few steps in the proof of Theorem 4.1 hold even without the finiteness as- sumption, as indicated by the follwing.
310 CHAPTER IV: INTEGRATION THEORY
Exercise 1*. Suppose A is a σ algebra on some non-empty set X, and suppose μ and ν are “honest” measures on A. Prove the following.
(i) If ν b μ, then one has the inclusions NK(X, A, μ) ⊂ NK(X, A, ν) and Lp K(X, A, μ) ⊂ Lp K(X, A, ν), ∀ p ∈ [1, ∞). Consequently (see the proof of Theorem 4.1) one has linear maps Lp K(X, A, μ) 3 h 7 −→ h ∈ Lp K(X, A, ν), ∀ p ∈ [1, ∞). Show that these linear maps are continuous. (ii) Conversely, assuming one has the inclusion Lp K^0 (X, A, μ) ⊂ Lp K^0 (X, A, ν), for some p 0 ∈ [1, ∞), prove that ν b μ.
Hint: To prove (ii) show first one has the inclusion L^1 +(X, A, μ) ⊂ L^1 +(X, A, μ). Then show that the quantity
C = sup
{ ∫ X
h dν : h ∈ L^1 +(X, A, μ),
∫ X
h dμ ≤ 1
}
is finite. If C = ∞, there exists some sequence (hn)∞ n=1 ⊂ L^1 +(X, A, μ), with ∫
X
hn dμ ≤ 1 and
∫
X
h dν ≥ 4 n, ∀ n ≥ 1.
Consider then the series ∑∞ n= 1 2 n^ hn, and get a contradiction. Finally prove that^ ν(A)^ ≤^ Cμ(A), ∀ A ∈ A.
It is the moment now to introduce the following relation, which is a highly non-trivial weakening of the relation b.
Definition. Let A is a σ-algebra on some non-empty set X, and suppose μ and ν are “honest” (not necessarily finite) measures on A. We say that ν is absolutely continuous with respect to μ, if for every A ∈ A, one has the implication
(14) μ(A) = 0 =⇒ ν(A) = 0.
In this case we are going to use the notation
ν μ.
It is obvious that one always has the implication
ν b μ ⇒ ν μ. Remarks 4.3. Let (X, A, μ) be a measure space. A. If ν is an “honest” measure on A, which has the Radon-Nikodym property relative to μ, then ν μ. This is pretty obvious, since if we pick f : X → [0, ∞] to be a density for ν realtive to μ, then for every A ∈ A with μ(A) = 0, we have f κA = 0, μ-a.e., so we get
ν(A) =
A
f dμ =
X
f κA dμ = 0.
B. For an “honest” measure ν on A, the relation ν μ is equivalent to the inclusion NK(X, A, μ) ⊂ NK(X, A, ν).
By Exercise 1, this already suggests that the relation is much weaker than b (see Exercise 2 below). C. If ν is either a signed or a complex measure on A, then the following are equivalent:
312 CHAPTER IV: INTEGRATION THEORY
On the other hand, since λn+1 = μ + λn, ∀ n ≥ 1, using Lemma III.8.2, we get the relations X 1 + ⊂ μ
X 2 + ⊂
μ
(Recall that the notation D ⊂ μ
E stands for μ(D r E) = 0.) Since ν μ, we also
have the relations A ∩ X 1 + ⊂ ν
A ∩ X 2 + ⊂
ν
so using Proposition III.4.3, one gets the equality
ν(A ∩ X ∞+) = lim n→∞ ν(A ∩ X n+ ).
Combining this with the inequalities (15) and (17) then yields the inequality
(18) ν(A) ≥ lim sup n→∞
νn(A) ≥ lim inf n→∞ νn(A) ≥ ν(A ∩ X ∞+) + lim n→∞
[
nμ(A ∩ X∞−)
]
There are two posibilities here.
Case I: μ(A ∩ X−∞) > 0.
In this case, the estimate (18) forces
ν(A) = lim sup n→∞
νn(A) = lim inf n→∞ νn(A) = ∞.
Case II: μ(A ∩ X−∞) = 0.
In this case, using absolute continuity, we get ν(A ∩ X∞−) = 0, and the equality A = (A ∩ X ∞+) ∪ (A ∩ X∞−) yields
ν(A) = ν(A ∩ X+ ∞).
Then (18) forces lim sup n→∞
νn(A) = lim inf n→∞ νn(A) = ν(A).
In either case, the concluison is the same: limn→∞ νn(A) = ν(A).
After the above preparation, we are now in position to prove the following. Theorem 4.2 (Radon-Nikodym Theorem: the finite case). Let (X, A, μ) be a finite measure space. A. If ν is an “honest” measure on A, with ν μ, then there exists a measurable function f : X → [0, ∞], such that
(19) ν(A) =
A
f dμ, ∀ A ∈ A.
Moreover, such a function is essentially unique, in the sense that, whenever f 1 , f 2 : X → [0, ∞] are measurable functions, that satisfy (19), it follows that f 1 = f 2 , μ-a.e. B. Let K be either R or C. If λ is a K-valued measure on A, with λ μ, then there exists a function f ∈ L^1 K(X, A, μ), such that
(20) λ(A) =
A
f dμ, ∀ A ∈ A.
Moreover:
(i) A function f ∈ L^1 K(X, A, μ) satisfying (20) is essentially unique, in the sense that, whenever f 1 , f 2 ∈ L^1 K(X, A, μ) satisfy (20), it follows that f 1 = f 2 , μ-a.e.
§4. Radon-Nikodym Theorems 313
(ii) If f ∈ L^1 K(X, A, μ) is any function satisfying (20), then the variation measure |λ| of λ is given by
|λ|(A) =
A
|f | dμ, ∀ A ∈ A.
Proof. A. Use Lemma 4.1 to find a sequence (νn)∞ n=1 of “honest” measures on A, such that
- νn b μ, ∀ n ≥ 1; in particular the measures νn, n ≥ 1 are all finite;
- ν 1 ≤ ν 2 ≤... ;
- limn→∞ νn(A) = ν(A), ∀ A ∈ A.
For each n ≥ 1, we apply the “Easy” Radon-Nikodym Theorem 4.1, to find some measurable function fn : X → R, such that
νn(A) =
A
fn dμ, ∀ A ∈ A.
Claim: The sequence (fn)∞ n=1 satisfies 0 ≤ fn ≤ fn+1, μ-a.e., ∀ n ≥ 1.
Fix n ≥ 1. On the one hand, since the νn’s are “honest” finite measures, and νn b μ, by part (i) of Theorem 4.1, it follows that fn ≥ 0, μ-a.e. On other hand, since νn+1 − νn is also an “honest” finite measure with νn+1 − νn b μ, and with density fn+1 − fn, again by part (i) of Theorem 4.1, it follows that fn+1 − fn ≥ 0, μ-a.e. Having proven the above Claim, let us define the function f : X → [0, ∞], by f (x) = lim inf n→∞
[
max{fn(x), 0 }
]
∀ x ∈ X.
It is obvious that f is measurable. By the Claim, we have in fact the equality
f = μ-a.e.- lim n→∞ fn.
Since we also have f κA = μ-a.e.- lim n→∞ fnκA, ∀ A ∈ A,
using the Claim and the Monotone Convergence Theorem, we get ∫
A
f dμ =
X
f κA dμ = lim n→∞
X
fnκA dμ = lim n→∞
A
fn dμ =
= lim n→∞ νn(A) = ν(A), ∀ A ∈ A.
Having shown that f satisfies (19), let us observe that the uniqueness property stated in part A is a consequence of Proposition 4.2. B. Let λ be a K-valued. In particular, the variation measure |λ| is finite, so by the Polar Decomposition (Proposition 4.3) there exists some measurable function h : X → K, such that
(21) λ(A) =
A
h d|λ|, ∀ A ∈ A,
and such that |h| = 1, |λ|-a.e. Replacing h with the measurable function h′^ : X → K, defined by
h′(x) =
h(x) if |h(x)| = 1 1 if |h(x)| 6 = 1
§4. Radon-Nikodym Theorems 315
Corollary 4.1 (Radon-Nikodym Theorem: the σ-finite case). Let (X, A, μ) be a σ-finite measure space. A. If ν is an “honest” measure on A, with ν μ, then there exists a measurable function f : X → [0, ∞], such that
(23) ν(A) =
A
f dμ, ∀ A ∈ A.
Moreover, such a function is essentially unique, in the sense that, whenever f 1 , f 2 : X → [0, ∞] are measurable functions, that satisfy (19), it follows that f 1 = f 2 , μ-a.e. B. Let K be either R or C. If λ is a K-valued measure on A, with λ μ, then there exists a function f ∈ L^1 K(X, A, μ), such that
(24) λ(A) =
A
f dμ, ∀ A ∈ A.
Moreover:
(i) A function f ∈ L^1 K(X, A, μ) satisfying (20) is essentially unique, in the sense that, whenever f 1 , f 2 ∈ L^1 K(X, A, μ) satisfy (24), it follows that f 1 = f 2 , μ-a.e. (ii) If f ∈ L^1 K(X, A, μ) is any function satisfying (24), then the variation measure |λ| of λ is given by
|λ|(A) =
A
|f | dμ, ∀ A ∈ A.
⋃^ Proof.^ Since^ μ^ is^ σ-finite, there exists a sequence (An)∞^ n=1^ ⊂^ Aμ^ fin, with ∞ n=1 An^ =^ X.^ Put^ X^1 =^ A^1 and^ Xn^ =^ An^ r^ (A^1 ∪ · · · ∪^ An−^1 ),^ ∀^ n^ ≥^ 2.^ Then (Xn)∞ n=1 ⊂ Aμ fin is pairwise disjoint, and we still have
n=1 Xn^ =^ X. The Corol- lary follows then immediately from Theorem 4.2, applied to the measure spaces (Xn, A
Xn , μ
Xn ) and the measures^ ν
Xn and^ λ
Xn respectively. What is used here is the fact that, if K denotes one of the sets [0, ∞], R or C, then for a function f : X → K the fact that f is measurable, is equivalent to the fact that f
Xn^ is measurable for each n ≥ 1. Moreover, given two functions f 1 , f 2 : X → K, the con- dition f 1 = f 2 , μ-a.e. is equivalent to the fact that f 1
Xn =^ f^2
Xn ,^ μ-a.e.,^ ∀^ n^ ≥^ 1. Finally, for f : X → K(= R, C), the condition f ∈ L^1 K(X, A, μ), is equivalent to the fact that f
Xn^ ∈^ L
1 K(Xn,^ A
Xn^ , μ
Xn^ ),^ ∀^ n^ ≥^ 1, and ∑^ ∞
n=
Xn
f
Xn
d
μ
Xn
Comment. The σ-finite case of the Radon-Nikodym Theorem, given above, is in fact a particular case of a more general version (Theorem 4.3 below). In order to formulate this, we need a concept which has already appeared earlier in III.5. Recall that a measure space (X, A, μ) is said to be decomposable, if there exists a pairwise disjoint subcollection F ⊂ Aμ fin, such that
(i)
F ∈F F^ =^ X;
(ii) for a set A ⊂ X, the condition A ∈ A is equiavelnt to the condition
A ∩ F ∈ A, ∀ F ∈ F;
316 CHAPTER IV: INTEGRATION THEORY
(iii) one has the equality
μ(A) =
F ∈F
μ(A ∩ F ), ∀ Aμ fin.
Such a collection F is then called a decomposition of (X, A, μ). Condition (ii) is referred to as the patching property, because it characterizes measurability as follows.
(p) Given a measurable space (Y, B), a function f : (X, A) → (Y, B) is mea- surable, if and only if all restrictions F
F : (F,^ A
F )^ →^ (Y,^ B),^ F^ ∈^ F, are measurable. Theorem 4.3 (Radon-Nikodym Theorem: the decomposable case). Let (X, A, μ) be a decomposable measure space. Let Aμσ-fin be the collection of all μ-σ-finite sets in A, that is,
Aμσ-fin =
A ∈ A : there exists (An)∞ n=1 ⊂ Aμ fin, with A =
⋃^ ∞
n=
An
A. If ν is an “honest” measure on A, with ν μ, then there exists a measurable function f : X → [0, ∞], such that
(25) ν(A) =
A
f dμ, ∀ A ∈ Aμσ-fin.
Moreover, such a function is locally essentially unique, in the sense that, whenever f 1 , f 2 : X → [0, ∞] are measurable functions, that satisfy (25), it follows that f 1 = f 2 , μ-l.a.e. B. Let K be either R or C. If λ is a K-valued measure on A, with λ μ, then there exists a function f ∈ L^1 K(X, A, μ), such that
(26) λ(A) =
A
f dμ, ∀ A ∈ Aμσ-fin.
Moreover:
(i) A function f ∈ L^1 K(X, A, μ) satisfying (26) is essentially unique, in the sense that, whenever f 1 , f 2 ∈ L^1 K(X, A, μ) satisfy (26), it follows that f 1 = f 2 , μ-a.e. (ii) If f ∈ L^1 K(X, A, μ) is any function satisfying (26), then the variation measure |λ|, of λ, satisfies
|λ|(A) =
A
|f | dμ, ∀ A ∈ Aμσ-fin.
Proof. Fix F to be a decomposition for (X, A, μ). A. For every F ∈ F, we apply Theorem 4.2 to the measure space (F, A
f , μ
F )
and the measure ν
F , to find some measurable function^ fF^ :^ F^ →^ [0,^ ∞], such that
ν(A) =
A
fF dμ, ∀ A ∈ A
F.
Using the patching property, there exists a measurable function f : X → [0, ∞], such that f
F =^ fF^ ,^ ∀^ F^ ∈^ F. The key feature we ar going to prove is a particular case of (25).
Claim 1: ν(A) =
A f dμ,^ ∀^ A^ ∈^ A
μ fin.
318 CHAPTER IV: INTEGRATION THEORY
At this point let us remark that the local almost uniqueness of f already follows from Remark 4.1. Let us prove now the equality (25). Start with some set A ∈ Aμσ-fin, and choose a sequence (An)∞ n=1 ⊂ Aμ fin, such that A =
n=1 An. Define the sequence (Bn)
∞ n= by
Bn = A 1 ∪ · · · ∪ An, ∀ n ≥ 1 ,
so that we still have Bn ∈ Aμ fin, ∀ n ≥ 1, as well as A =
n=1 Bn, but moreover we have B 1 ⊂ B 2 ⊂.... For each n ≥ 1, using Claim 1, we have the equality
ν(Bn) =
Bn
f dμ.
Using these equalities, combined with
- 0 ≤ f κB 1 ≤ f κB 2 ≤... ,
- limn→∞(f κBn )(x) = (f κB )(x), ∀ x ∈ X,
the Monotone Convergence Theorem, combined with continuity yields ∫
B
f dμ =
X
f κB dμ = lim n→∞
X
f κBn dμ = lim n→∞
Bn
dμ = lim n→∞ ν(Bn) = ν(A).
B. We start off by choosing a measurable function h : X → K, with |h| = 1, such that
λ(A) =
A
h d|λ|, ∀ A ∈ A.
Using part A, there exists some measurable function g 0 : X → [0, ∞], such that
(28) |λ|(A) =
A
g 0 dμ, ∀ A ∈ Aμσ-fin.
At this point, g 0 may not be integrable, but we have the freedom to perturb it (μ- l.a.e.) to try to make it integrable. This is done as follows. Consider the collection
F 0 =
F ∈ F : |λ|(F ) > 0
Since |λ| is finite, it follows that F 0 is at most countable. Define then the set X 0 =
F ∈F 0 F^ ∈^ A
μ σ-fin. Since^ X^0 is^ μ-σ-finite, every set^ A^ ∈^ A^ with^ A^ ⊂^ X^0 , is μ-σ-finite, so we have
|λ|(A) =
A
g 0 dμ, ∀ A ∈ A
X 0.
Applying the σ-finite version of the Radon-Nikodym Theorem to the σ-finite mea- sure space (X 0 , A
X 0 , μ
X 0 ) and the finite measure^ λ
X 0 , it follows that the density g 0
X 0 belongs to^ L
1 +(X^0 ,^ A
X 0 , μ
X 0 ), which means that the function^ g^ =^ g^0 κX^0 belongs to L^1 +(X, A, μ). With this choice of g, let us prove now that the equality (28) still holds, with g in place of g 0. Exactly as in the proof of part A, it suffices to prove only the equality
(29) |λ|(A) =
A
g dμ, ∀ A ∈ Aμ fin.
Claim 2: |λ|(A) = |λ|(A ∩ X 0 ), ∀ A ∈ Aμσ-fin.
§4. Radon-Nikodym Theorems 319
Since (use the fact that |λ| is finite) the equality is equivalent to
|λ|(A r X 0 ) = 0, ∀ A ∈ Aμσ-fin,
it suffices to prove it only for A ∈ Aμ fin. If A ∈ Aμ fin, using the properties of the decomposition F, we have
|λ|(A) =
F ∈F
|λ|(A ∩ F ) =
F ∈F 0
|λ|(A ∩ F ) +
F ∈FrF 0
|λ|(A ∩ F ) =
= |λ|
F ∈F 0
[A ∩ F ]
F ∈FrF 0
|λ|(A ∩ F ) =
= |λ|(A ∩ X 0 ) +
F ∈F′(A)
|λ|(A ∩ F ).
Notice now that, for F ∈ F r F 0 , we have |λ|(F ) = 0, which gives |λ|(A ∩ F ) = 0, so the Claim follows immediately from the above computation. Having proven the above Claim, let us prove now (29). Fix A ∈ Aμ fin. The desired equality is now immediate from Claim 2, combined with (28):
|λ|(A) = |λ|(A ∩ X 0 ) =
A∩X 0
g 0 dμ =
X
g 0 κA∩X 0 dμ =
X
g 0 κX 0 κA dμ =
X
gκA dμ =
A
g μ.
Define now the function f 0 = hg. Since |f 0 | = g ∈ L^1 +(X, A, μ), it follows that f 0 ∈ L^1 K(X, A, μ). Let us prove that f 0 satisfies the equality (26). Start with some A ∈ Aμσ-fin. On the one hand, using Claim 2, we have
|λ(A r X 0 )| ≤ |λ|(A r X 0 ) = 0,
so we get λ(A) = λ(A ∩ X 0 ). Using the σ-finite version of the Radon-Nikodym Theorem for (X 0 , A
X 0 , μ
X 0 ) and^ λ
X 0 , we then have
λ(A) = λ(A ∩ X 0 ) =
A∩X 0
hg 0 dμ =
X
hg 0 κA∩X 0 dμ =
X
hg 0 κX 0 κA dμ =
X
hgκA dμ =
A
hg dμ =
A
f 0 dμ.
We now prove the uniqueness property (i) of f (μ-a.e.!). Assume f ∈ L^1 K(X, A, μ) is another function, such that
λ(A) =
A
f dμ, ∀ A ∈ Aμσ-fin.
Claim 3: f = f 0 , μ-l.a.e.
What we need to show here is the fact that
f κB = f 0 κB , μ-a.e., ∀ B ∈ Aμ fin.
But this follows immediately from the uniqueness from part B of Theorem 4.2, applied to the finite measure space (B, A
B , μ
B ) and the measure^ λ
B , which has both f
B and^ f^0
B as densities. Using Claim 3, we now have f − f 0 ∈ L^1 K(X, A, μ), with f − f 0 = 0, μ-l.a.e., so we can apply Proposition 4.4, which forces f − f 0 = 0, μ-a.e., so we indeed get f = f 0 , μ-a.e.
