614 CHAPTER 20. REPRESENTATION THEOREMS

Next let c be a number. Then |c f (x)| = |c| | f (x)| ≤ |c| || f ||∞

and so ||c f ||∞≤ |c| || f ||

∞.

Therefore since c is arbitrary, || f ||∞= ||c(1/c) f ||

∞≤∣∣ 1

c

∣∣ ||c f ||∞

which implies |c| || f ||∞≤

||c f ||∞

. Thus || ||∞ is a norm as claimed.To verify completeness, let { fn} be a Cauchy sequence in L∞(Ω) and use the above

claim to get the existence of a set of measure zero, Enm such that for all x /∈ Enm,

| fn(x)− fm(x)| ≤ || fn− fm||∞

Let E = ∪n,mEnm. Thus µ(E) = 0 and for each x /∈ E, { fn(x)}∞n=1 is a Cauchy sequence in

C. Let

f (x) ={

0 if x ∈ Elimn→∞ fn(x) if x /∈ E = lim

n→∞XEC(x) fn(x).

Then f is clearly measurable because it is the limit of measurable functions. If

Fn = {x : | fn(x)|> || fn||∞}

and F = ∪∞n=1Fn, it follows µ(F) = 0 and that for x /∈ F ∪E,

| f (x)| ≤ lim infn→∞| fn(x)| ≤ lim inf

n→∞|| fn||∞ < ∞

because {|| fn||∞} is a Cauchy sequence. (||| fn||∞−|| fm||∞| ≤ || fn− fm||∞ by the triangleinequality.) Thus f ∈ L∞(Ω). Let n be large enough that whenever m > n,

|| fm− fn||∞ < ε .

Then, if x /∈ E,

| f (x)− fn(x)| = limm→∞| fm(x)− fn(x)|

≤ limm→∞

inf || fm− fn||∞ < ε .

Hence || f − fn||∞ < ε for all n large enough. This proves the theorem.The next theorem is the Riesz representation theorem for

(L1 (Ω)

)′.Theorem 20.3.4 (Riesz representation theorem) Let (Ω,S ,µ) be a finite measure space.If Λ ∈ (L1(Ω))′, then there exists a unique h ∈ L∞(Ω) such that

Λ( f ) =∫

Ω

h f dµ

for all f ∈ L1(Ω). If h is the function in L∞(Ω) representing Λ ∈ (L1(Ω))′, then ||h||∞ =||Λ||.

Proof: Just as in the proof of Theorem 20.3.1, there exists a unique h ∈ L1(Ω) suchthat for all simple functions, s,

Λ(s) =∫

hs dµ . (20.3.11)