23.2. VECTOR MEASURES 625

1(0,0)

•p

B(p,r)

Let E = f−1(B(p,r)). In fact µ (E) = 0. If µ(E) ̸= 0 then∣∣∣∣ 1µ(E)

∫E

f dµ− p∣∣∣∣= ∣∣∣∣ 1

µ(E)

∫E( f − p)dµ

∣∣∣∣≤ 1µ(E)

∫E| f − p|dµ < r

because on E, | f (ω)− p|< r. Hence 1µ(E)

∫E f dµ is closer to p than r and so∣∣∣∣ 1

µ(E)

∫E

f dµ

∣∣∣∣> 1.

Refer to the picture. However, this contradicts the assumption of the lemma. It followsµ(E) = 0. Since the set of complex numbers z such that |z|> 1 is an open set, it equals theunion of countably many balls, {Bi}∞

i=1 . Therefore,

µ(

f−1({z ∈ C : |z|> 1})= µ

(∪∞

k=1 f−1 (Bk))≤

∞

∑k=1

µ(

f−1 (Bk))= 0.

Thus | f (ω)| ≤ 1 a.e. as claimed. ■Note that the above argument would work with essentially no change ifCwere replaced

with V a separable normed linear space.

Corollary 23.2.8 Let λ be a complex vector measure with |λ |(Ω) < ∞.1 Then thereexists a unique f ∈ L1(Ω) such that λ (E) =

∫E f d|λ |. Furthermore, | f | = 1 for |λ | a.e.

This is called the polar decomposition of λ . We write dλ = f d |λ | sometimes.

Proof: Letting µ = |λ | in Theorem 23.1.3, the first claim follows because λ ≪ |λ | andso such an L1 function exists and is unique. It is required to show | f |= 1 a.e. If |λ |(E) ̸= 0,∣∣∣ λ (E)|λ |(E)

∣∣∣= ∣∣∣ 1|λ |(E)

∫E f d|λ |

∣∣∣≤ 1. Therefore by Lemma 23.2.7, | f | ≤ 1, |λ | a.e. Now let

En =

[| f | ≤ 1− 1

n

].

Let {F1, · · · ,Fm} be a partition of En. Then

m

∑i=1|λ (Fi)|=

m

∑i=1

∣∣∣∣∫Fi

f d|λ |∣∣∣∣≤ m

∑i=1

∫Fi

| f |d|λ |

≤m

∑i=1

∫Fi

(1− 1

n

)d |λ |=

m

∑i=1

(1− 1

n

)|λ |(Fi) = |λ |(En)

(1− 1

n

).

Then taking the supremum over all partitions, |λ |(En) ≤(1− 1

n

)|λ |(En) which shows

|λ |(En) = 0. Hence |λ |([| f |< 1]) = 0 because [| f |< 1] = ∪∞n=1En. ■

Next is a specific case which leads to complex measures.1As proved above, the assumption that |λ |(Ω)< ∞ is redundant.