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 [40]
titled Vector measures.
Definition 14.2.1Let (V,||⋅||) be a normed linear space and let (Ω,S) be a measure space. A functionμ : S→ V is a vector measure if μ is countably additive. That is, if {E_{i}}_{i=1}^{∞}is a sequence of disjoint setsof 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 14.2.2Let
(Ω, S)
be a measure space and let μ be a vector measure defined on S. A subset,π(E), of Sis called a partition of E if π(E) consists of finitely many disjoint 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 14.2.3If |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 1.13.3 on Page 66
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 |(E )− ε < ∑ ||μ(Ai)||i = 1,2.
i 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
(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 14.2.4Suppose μ is a complex measure on
(Ω,S )
where S is a σ algebra of subsets of Ω. Thatis, whenever,
{E }
i
is a sequence of disjoint sets of S,
∞ ∑∞
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 )|). (14.6)
F ∈π(E)
(14.6)
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
√ -
-22
|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},
∑
| iRe (zi)|
= ∑_{i}
|Re(zi)|
. 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
∑
| iIm zi|
= ∑_{i}
|Im zi|
is z_{i} are all in R_{2} or else all in
R_{4}. Then by 14.6, it follows that for some i,
∑
|μ (F)| > 5(1+ |μ(E)|). (14.7)
F∈πi
(14.7)
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 14.2.5Let
(Ω,S)
be a measure space and let λ : S→ ℂ be a complex vector measure. Thus|λ|(Ω) < ∞. Let μ : S→ [0,μ(Ω)] be a finite measure such that λ ≪ μ. Then there exists a uniquef ∈ 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, and |Im-λ|−-Im-(λ)
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 ∫
-(|Re λ|+ Re λ)(E ) = f1dμ,
2 ∫E
1(|Re λ|− Re λ)(E ) = f2dμ,
2 ∫E
1(|Im λ|+ Im λ)(E ) = gdμ,
2 E 1
1 ∫
2(|Im λ|− Im λ)(E ) = E g2dμ.
Now let f = f_{1}− f_{2} + i(g_{1}− g_{2}). ■
Next is the notion of representing the vector measure in terms of
|μ |
. It is like representing a complex
number in the form re^{iθ}. The proof requires the following lemma.
Lemma 14.2.6Suppose (Ω,S,μ) is a measure space and f is a function in L^{1}(Ω,μ) with the propertythat
||∫ ||
|| f dμ|| ≤ 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) = ∅.
PICT
Let E = f^{−1}(B(p,r)). In fact μ
(E)
= 0. If μ(E)≠0 then
|| 1 ∫ || || 1 ∫ ||
||----- fdμ− p|| = ||----- (f − p)dμ||
μ(E ) E μ(E)∫ E
≤ --1-- |f − p|dμ < r
μ(E ) E
because on E,
|f (ω)− p|
< r. Hence
μ1(E)
∫_{E}fdμ is closer to p than r and so
|| 1 ∫ ||
||----- f dμ|| > 1
μ(E ) E
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,
Corollary 14.2.7Let λ be a complex vector measure with |λ|(Ω) < ∞.^{1}Then there exists a unique f ∈ L^{1}(Ω) such that λ(E) = ∫_{E}fd|λ|. Furthermore,
|f|
= 1 for
|λ |
a.e.This is called the polar decomposition of λ.
Proof: First note that λ ≪|λ| and so such an L^{1} function exists and is unique by Theorem 14.2.5. It is
required to show