Calculus of Real and Complex Variables

14.2 Vector Measures

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.

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

is a vector. It might be a point in ℝⁿ 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 .

The next theorem may seem a little surprising. It states that, if finite, the total variation is a nonnegative measure.

Proof: Consider the last claim. Let a <

and let π be a partition of A such that

∑ a < ||μ (F )||. F∈π(A)

Then π

∪ 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

≥ as claimed.

Let

_j=1^∞ be a sequence of disjoint sets of S and let E_∞ = ∪_j=1^∞E_j. Then letting a < , it follows from the definition of total variation there exists a partition of E_∞, π(E_∞) = {A₁,,A_n} such that

∑n a < ||μ(Ai)||. i=1

Also,

Ai = ∪∞j=1Ai ∩Ej

and so by the triangle inequality,

≤∑ _j=1^∞||μ(A_i ∩ E_j)||. Therefore, by the above, and either Fubini’s theorem or Lemma 1.13.3 on Page 66

◜--≥||μ(◞A◟i)||--◝ n∑ ∞∑ ∑∞ ∑n ∑∞ a < ||μ(Ai ∩ Ej)|| = ||μ(Ai ∩ Ej)|| ≤ |μ|(Ej) i=1 j=1 j=1 i=1 j=1

because

_i=1ⁿ is a partition of E_j.

Since a is arbitrary, this shows

∞ ∞∑ |μ|(∪j=1Ej) ≤ |μ|(Ej). j=1

If the sets, E_j are not disjoint, let F₁ = E₁ and if F_n has been chosen, let F_n+1 ≡ E_n+1 ∖∪_i=1ⁿE_i. Thus the sets, F_i are disjoint and ∪_i=1^∞F_i = ∪_i=1^∞E_i. Therefore,

( ) ( ) ∑∞ ∑∞ |μ| ∪∞j=1Ej = |μ| ∪∞j=1Fj ≤ |μ|(Fj) ≤ |μ|(Ej ) j=1 j=1

and proves

is always subadditive as claimed regardless of whether < ∞.

Now suppose

< ∞ and let E₁ and E₂ be sets of S such that E₁ ∩ E₂ = ∅ and let {A₁ⁱA_niⁱ} = 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

or π, it follows this collection of sets is a partition of E₁ ∪ E₂ denoted by π(E₁ ∪ E₂). Then by the above inequality and the definition of total variation,

∑ |μ|(E1 ∪ E2) ≥ ||μ(F)|| > |μ|(E1) +|μ|(E2)− 2ε, F∈π(E1∪E2)

which shows that since ε > 0 was arbitrary,

|μ|(E1 ∪ E2) ≥ |μ|(E1) +|μ|(E2 ). (14.5)

(14.5)

Then 14.5 implies that whenever the E_i are disjoint, |μ|(∪_j=1ⁿE_j) ≥∑ _j=1ⁿ|μ|(E_j). Therefore,

∞ n ∑ |μ|(Ej) ≥ |μ|(∪∞ Ej) ≥ |μ|(∪n Ej) ≥ ∑ |μ|(Ej). j=1 j=1 j=1 j=1

Since n is arbitrary,

∞ ∞∑ |μ|(∪ j=1Ej ) = |μ |(Ej) j=1

which shows that

is a measure as claimed. ■

In the case that μ is a complex measure, it is always the case that

< ∞.

Proof: First here is a claim.

Claim: Suppose

= ∞. Then there are disjoint subsets of E, A and B such that E = A ∪ B, , > 1 and = ∞.

Proof of the claim: From the definition of

, there exists a partition of E,π 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

∈ 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, R₁,R₂,R₃, and R₄ as shown. Let π_i consist of those sets, A of

PICT

π for which μ ∈ R_i. Thus, some sets, A of π could be in two of the π_i if μ is on one of the intersecting lines. This is not important. The thing which is important is that if μ ∈ R₁ or R₃, then ≤ and if μ ∈ R₂ or R₄ then ≤ and both R₁ and R₃ have complex numbers z contained in these sets all have the same sign for Re. Thus, for z_i ∈ R₁, = ∑ _i. A similar statement holds for z_i ∈ R₃. In the case of R₂,R₄, similar considerations hold for the imaginary parts. Thus = ∑ _i is z_i are all in R₂ or else all in R₄. 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

always has the same sign,

Now letting C be the union of the sets in π_i,

| | |μ (C)| = ||∑ μ(F)||> 5 (1+ |μ (E)|) > 1. (14.8) ||F∈π || 2 i

(14.8)

Define D such that C ∪ D = E. Then, since μ is a measure,

PICT

and so, subtracting

from both sides,

1 < 5 + 3|μ(E )| < |μ(D )|. 2 2

Now since

= ∞, it follows from Theorem 14.2.4 that ∞ = ≤ + and so either = ∞ or = ∞. If = ∞, let B = C and A = D. Otherwise, let B = D and A = C. This proves the claim.

Now suppose

= ∞. Then from the claim, there exist A₁ and B₁ such that = ∞,, > 1, and A₁ ∪B₁ = Ω. Let B₁ ≡ Ω ∖A play the same role as Ω and obtain A₂,B₂ ⊆ B₁ such that = ∞,, > 1, and A₂ ∪ B₂ = B₁. Continue in this way to obtain a sequence of disjoint sets, such that > 1. Then since μ is a measure,

∞∑ μ (∪ ∞i=1Ai) = μ(Ai) i=1

but this is impossible because lim_i→∞μ

≠0. ■

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 L¹(Ω),f_1,f₂,g₁,g₂ such that for all E ∈S,

Now let f = f₁ − f₂ + i(g₁ − g₂). ■

Next is the notion of representing the vector measure in terms of

. It is like representing a complex number in the form re^iθ. The proof requires the following lemma.

Proof of the lemma: Consider the following picture where B(p,r) ∩ B(0,1) = ∅.

Let E = f⁻¹(B(p,r)). In fact μ = 0. If μ(E)≠0 then because on E, < r. Hence ∫ _Efdμ 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

> 1 is an open set, it equals the union of countably many balls, _i=1^∞. Therefore,

Thus ≤ 1a.e. as claimed. ■

Proof: First note that λ ≪|λ| and so such an L¹ function exists and is unique by Theorem 14.2.5. It is required to show

= 1a.e. If |λ|(E)≠0,

|| λ(E) || || 1 ∫ || |||λ|(E-)|| = |||λ|(E-) fd|λ||| ≤ 1. E