324 CHAPTER 13. REPRESENTATION THEOREMS

so |λ |(E) =∫

E |h|dµ .2.) Now let |

∫E f dµ| ≤ µ(E) for all E. Consider the following picture where

B(p,r)∩B(0,1) = /0

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. If |∫

E f dµ|= µ(E) for all E then from Part 1.), µ (E) =|µ|(E) = |

∫E f dµ|=

∫E | f |dµ and so | f |= 1 a.e.

3.) Clearly λ ≪ |λ | so there exists a unique g in L1 (Ω,λ ) such that λ (E) =∫

E gd |λ | .From Part 1.), |λ |(E) =

∫E |g|d |λ | for all E and so |g| = 1 a.e. Now if also λ (E) =∫

E gd |λ |=∫

E hdµ, let sn be simple functions converging pointwise to g. Then∫

E gsnd |λ |=∫E snhdµ. From the dominated convergence theorem,

∫E d |λ | =

∫E ghdµ. Thus gh ≥ 0

µ a.e. and |g| = 1. Therefore, |h| = |gh| = gh. More formally, it is assumed gd |λ | =hdµ, |g|= 1 so d |λ |= ghdµ and so we must have gh≥ 0 hence equal to |h|. ■

13.3 The Dual Space of Lp (Ω)

This is on representation of the dual space of Lp (Ω).

Theorem 13.3.1 (Riesz representation theorem) Let ∞ > p > 1 and let (Ω,S ,µ)be a finite measure space. If Λ∈ (Lp(Ω))′, then there exists a unique h∈ Lq(Ω) ( 1

p +1q = 1)

such that

Λ f =∫

Ω

h f dµ .

This function satisfies ∥h∥q = ∥Λ∥ where ∥Λ∥ is the operator norm of Λ.