20.2. VECTOR MEASURES 609

because it is closer to p than r. (Refer to the picture.) However, this contradicts the as-sumption 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, {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 (x)| ≤ 1 a.e. as claimed. This proves the lemma.

Corollary 20.2.8 Let λ be a complex vector measure with |λ |(Ω) < ∞1 Then there existsa 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 .

Proof: First note that λ ≪ |λ | and so such an L1 function exists and is unique. It isrequired 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 20.2.7, | 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− 1n

)|λ |(En)

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

Corollary 20.2.9 Let λ be a complex vector measure such that λ ≪ µ where µ is σ finite.Then there exists a unique g ∈ L1(Ω,µ) such that λ (E) =

∫E gdµ .

1As proved above, the assumption that |λ |(Ω)< ∞ is redundant.