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 [7] titled Vector
measures.
Definition 25.2.1Let (V,||⋅||) be a normed linear space and let (Ω,S) be ameasure space. 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(∪i=1Ei) = μ (Ei).
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) = nli→m∞ μ (Ei)
i=1 i=1
where the limit takes place relative to the norm on V .
Definition 25.2.2Let
(Ω,S )
be a measure space and let μ be a vector measuredefined on 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 25.2.3If |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 1.5.4 on Page 34
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
n∑i
|μ|(Ei)− ε < ||μ(Aij)||i = 1,2.
j=1
Such a partition exists because of the definition of the total variation. Consider the sets which
are contained in either of π
(E1)
or π
(E2)
, 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 25.2.5Suppose μ is a complex measure on
(Ω,S)
where S is a σalgebra of subsets of Ω. That is, whenever,
{E }
i
is a sequence of disjoint sets of S,
∞
μ(∪∞ E ) = ∑ μ (E ).
i=1 i i=1 i
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 )|). (25.6)
F∈ π(E)
(25.6)
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. Let
π_{i} consist of those sets, A of
PICT
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
√-
22-
|μ(A)|
≤
|Re (μ(A))|
and if μ
(A)
∈ R_{2} or R_{4} then
√2-
2
|μ(A)|
≤
|Im (μ(A))|
and both R_{1} and R_{3} have complex numbers z contained in these
sets all have the same sign for Re
(z)
. Thus, for z_{i}∈ R_{1},
|∑ Re(z )|
i i
= ∑_{i}
|Re (z)|
i
. A
similar statement holds for z_{i}∈ R_{3}. In the case of R_{2},R_{4}, similar considerations hold for the
imaginary parts. Thus
|∑ Im z |
i i
= ∑_{i}
|Im z|
i
is z_{i} are all in R_{2} or else all in R_{4}. Then by
25.6, it follows that for some i,
∑ |μ (F )| > 5(1+ |μ(E)|). (25.7)
F∈π
i
(25.7)
Suppose i equals 1 or 3. A similar argument using the imaginary part applies if i equals 2 or
4. Then, since Re
= ∞, 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. ■
Theorem 25.2.6Let
(Ω,S)
be a measure space and let λ : S → ℂ be acomplex vector measure. Thus |λ|(Ω) < ∞. Let μ : S→ [0,μ(Ω)] be a finite measuresuch 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
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(|Re λ|+ Re λ)(E ) = f1dμ,
2 ∫E
1(|Re λ|− Re λ)(E ) = f dμ,
2 ∫E 2
1
2(|Im λ|+ Im λ)(E ) = Eg1dμ,
1 ∫
2(|Im λ|− Im λ)(E ) = g2dμ.
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 25.2.7Suppose (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 where B(p,r) ∩ B(0,1) = ∅.