The next topic will use the Radon Nikodym theorem. It is the topic of vector and
complex measures. The main interest is in complex measures although a vector measure
can have values in any topological vector space. Whole books have been written on
this subject. See for example the book by Diestal and Uhl [?] titled Vector
measures.
Definition 18.2.1Let (V,||⋅||) be a normed linear space and let (Ω,S) be a measurespace. A function μ : S→ V is a vector measure if μ is countably additive. That is, if{E_{i}}_{i=1}^{∞}is a sequence of disjoint sets of S,
∞
μ(∪ ∞ Ei) = ∑ μ(Ei).
i=1 i=1
Note that it makes sense to take finite sums because it is given that μ has values
in a vector space in which vectors can be summed. In the above, μ
(Ei)
is a
vector. It might be a point in ℝ^{n} or in any other vector space. In many of the
most important applications, it is a vector in some sort of function space which
may be infinite dimensional. The infinite sum has the usual meaning. That
is
∑∞ ∑n
μ(Ei) = nl→im∞ μ(Ei)
i=1 i=1
where the limit takes place relative to the norm on V .
Definition 18.2.2Let
(Ω, S)
be a measure space and let μ be a vector measure definedon S. A subset, π(E), of Sis called a partition of E if π(E) consists of finitely manydisjoint sets of S and ∪π(E) = E. Let
|μ|(E) = sup{ ∑ ||μ(F)|| : π(E) is a partition of E }.
F∈π(E)
|μ| is called the total variation of μ.
The next theorem may seem a little surprising. It states that, if finite, the total
variation is a nonnegative measure.
Theorem 18.2.3If |μ|(Ω) < ∞, then |μ| is a measure on S. Even if
|μ|
(Ω)
= ∞,
|μ|
∞
(∪i=1Ei)
≤∑_{i=1}^{∞}
|μ|
(Ei)
. That is
|μ|
is subadditive and
|μ|
(A)
≤
|μ|
(B)
whenever A,B ∈S with A ⊆ B.
Proof: Consider the last claim. Let a <
|μ|
(A)
and let π
(A )
be a partition of A such
that
∑
a < ||μ (F )||.
F ∈π(A)
Then π
(A )
∪
{B ∖A }
is a partition of B and
∑
|μ |(B ) ≥ ||μ(F)||+ ||μ(B ∖A )|| > a.
F∈π(A)
Since this is true for all such a, it follows
|μ|
(B )
≥
|μ|
(A)
as claimed.
Let
{Ej}
_{j=1}^{∞} be a sequence of disjoint sets of S and let E_{∞} = ∪_{j=1}^{∞}E_{j}. Then
letting a <
|μ|
(E∞ )
, it follows from the definition of total variation there exists a
partition of E_{∞}, π(E_{∞}) = {A_{1},
⋅⋅⋅
,A_{n}} such that
∑n
a < ||μ(Ai)||.
i=1
Also,
∞
Ai = ∪ j=1Ai ∩Ej
and so by the triangle inequality, ||μ(A_{i})||≤∑_{j=1}^{∞}||μ(A_{i}∩ E_{j})||. Therefore, by the
above, and either Fubini’s theorem or Lemma 9.3.3 on Page 625
If the sets, E_{j} are not disjoint, let F_{1} = E_{1} and if F_{n} has been chosen, let
F_{n+1}≡ E_{n+1}∖∪_{i=1}^{n}E_{i}. Thus the sets, F_{i} are disjoint and ∪_{i=1}^{∞}F_{i} = ∪_{i=1}^{∞}E_{i}.
Therefore,
is always subadditive as claimed regardless of whether
|μ|
(Ω)
< ∞.
Now suppose
|μ |
(Ω )
< ∞ and let E_{1} and E_{2} be sets of S such that E_{1}∩E_{2} = ∅ and
let {A_{1}^{i}
⋅⋅⋅
A_{ni}^{i}} = π(E_{i}), a partition of E_{i} which is chosen such that
ni
|μ|(Ei) − ε < ∑ ||μ(Ai)||i = 1,2.
j=1 j
Such a partition exists because of the definition of the total variation. Consider the sets
which are contained in either of π
(E )
1
or π
(E )
2
, it follows this collection of sets is a
partition of E_{1}∪ E_{2} denoted by π(E_{1}∪ E_{2}). Then by the above inequality and the
definition of total variation,
In the case that μ is a complex measure, it is always the case that
|μ|
(Ω)
< ∞.
Theorem 18.2.5Suppose μ is a complex measure on
(Ω,S)
where S is a σalgebra of subsets of Ω. That is, whenever,
{Ei}
is a sequence of disjoint sets ofS,
∑∞
μ (∪ ∞i=1Ei) = μ(Ei).
i=1
Then
|μ|
(Ω)
< ∞.
Proof: First here is a claim.
Claim: Suppose
|μ |
(E )
= ∞. Then there are disjoint subsets of E, A and B such
that E = A ∪ B,
|μ(A)|
,
|μ (B)|
> 1 and
|μ|
(B )
= ∞.
Proof of the claim: From the definition of
|μ |
, there exists a partition of E,π
(E)
such that
∑
|μ(F )| > 20 (1 + |μ (E )|). (18.2.6)
F∈π(E )
(18.2.6)
Here 20 is just a nice sized number. No effort is made to be delicate in this argument.
Also note that μ
(E )
∈ ℂ because it is given that μ is a complex measure. Consider the
following picture consisting of two lines in the complex plane having slopes 1 and -1
which intersect at the origin, dividing the complex plane into four closed sets, R_{1},R_{2},R_{3},
and R_{4} as shown.
PICT
Let π_{i} consist of those sets, A of π
(E)
for which μ
(A )
∈ R_{i}. Thus, some sets, A of
π
(E )
could be in two of the π_{i} if μ
(A)
is on one of the intersecting lines. This is
not important. The thing which is important is that if μ
(A)
∈ R_{1} or R_{3}, then
√2-
2
|μ(A)|
≤
|Re (μ (A ))|
and if μ
(A )
∈ R_{2} or R_{4} then
√2-
2
|μ(A)|
≤
|Im (μ(A))|
and
Re
(z)
has the same sign for z in R_{1} and R_{3} while Im
(z)
has the same sign for z in R_{2}
or R_{4}. Then by 18.2.6, it follows that for some i,
∑
|μ(F)| > 5(1+ |μ(E )|). (18.2.7)
F∈πi
(18.2.7)
Suppose i equals 1 or 3. A similar argument using the imaginary part applies if i equals 2
or 4. Then,
= ∞, let B = C and A = D.
Otherwise, let B = D and A = C. This proves the claim.
Now suppose
|μ |
(Ω )
= ∞. Then from the claim, there exist A_{1} and B_{1} such that
|μ|
(B1)
= ∞,
|μ (B1)|
,
|μ (A1 )|
> 1, and A_{1}∪ B_{1} = Ω. Let B_{1}≡ Ω ∖ A play the same
role as Ω and obtain A_{2},B_{2}⊆ B_{1} such that
|μ|
(B2 )
= ∞,
|μ (B2)|
,
|μ (A2 )|
> 1, and
A_{2}∪B_{2} = B_{1}. Continue in this way to obtain a sequence of disjoint sets,
{Ai}
such that
|μ (Ai)|
> 1. Then since μ is a measure,
∞ ∞∑
μ (∪i=1Ai) = μ(Ai)
i=1
but this is impossible because lim_{i→∞}μ
(Ai)
≠0. This proves the theorem.
Theorem 18.2.6Let
(Ω, S)
be a measure space and let λ : S→ ℂ be a complex vectormeasure. Thus |λ|(Ω) < ∞. Let μ : S→ [0,μ(Ω)] be a finite measure such that λ ≪ μ.Then there exists a unique f ∈ L^{1}(Ω) such that for all E ∈S,
∫
fdμ = λ(E ).
E
Proof: It is clear that Reλ and Imλ are real-valued vector measures on S. Since
|λ|(Ω) < ∞, it follows easily that |Reλ|(Ω) and |Imλ|(Ω) < ∞. This is clear
because
|λ (E)| ≥ |Re λ (E )|,|Im λ(E)|.
Therefore, each of
|Re λ|+ Re λ |Re λ|− Re(λ) |Im λ|+ Im λ |Im λ|− Im (λ)
-----------,-------------,------------, and--------------
2 2 2 2
are finite measures on S. It is also clear that each of these finite measures are absolutely
continuous with respect to μ and so there exist unique nonnegative functions in
L^{1}(Ω),f_{1,}f_{2},g_{1},g_{2} such that for all E ∈S,
∫
1
2(|Reλ|+ Re λ)(E ) = E f1dμ,
1 ∫
2(|Reλ|− Re λ)(E ) = f2dμ,
∫E
1(|Im λ|+ Im λ)(E ) = g1dμ,
2 ∫E
1(|Im λ|− Im λ)(E ) = g2dμ.
2 E
Now let f = f_{1}− f_{2} + i(g_{1}− g_{2}).
The following corollary is about representing a vector measure in terms of its total
variation. It is like representing a complex number in the form re^{iθ}. The proof requires
the following lemma.
Lemma 18.2.7Suppose (Ω,S,μ) is a measure space and f is a function in L^{1}(Ω,μ)
with the property that
∫
| fdμ| ≤ μ(E )
E
for all E ∈S. Then |f|≤ 1a.e.
Proof of the lemma: Consider the following picture.
PICT
where B(p,r) ∩ B(0,1) = ∅. Let E = f^{−1}(B(p,r)). In fact μ
(E )
= 0. If μ(E)≠0 then
| ∫ | | ∫ |
||-1-- fdμ− p|| = ||--1-- (f − p)dμ||
|μ (E ) E | |μ(E) E |
1 ∫
≤ μ-(E-) E |f − p|dμ < r
because on E,
|f (x)− p|
< r. Hence
1 ∫
|μ(E-) fdμ| > 1
E
because it is closer to p than r. (Refer to the picture.) However, this contradicts the
assumption of the lemma. It follows μ(E) = 0. Since the set of complex numbers, z such
that
|z|
> 1 is an open set, it equals the union of countably many balls,