20.6. THE DUAL SPACE OF C0 (X) 631

Now since Cc (X) is dense in C0 (X) , there exists f ∈Cc (X) such that || f || ≤ 1 and

|Λ f |+ ε > ||Λ||= ||L||

Then also f ≺ X and so||L||− ε < |Λ f |= Λ f ≤ µ (X)

Since ε is arbitrary, this shows ||L||= µ (X). This proves the lemma.What follows is the Riesz representation theorem for C0(X)′.

Theorem 20.6.5 Let L ∈ (C0(X))′ for X a locally compact Hausdorf space. Then thereexists a finite Radon measure µ and a function σ ∈ L∞(X ,µ) such that for all f ∈C0 (X) ,

L( f ) =∫

Xf σdµ.

Furthermore,µ (X) = ||L|| , |σ |= 1 a.e.

and if

ν (E)≡∫

Eσdµ

then µ = |ν |

Proof: From the above there exists a unique Radon measure µ such that for all f ∈Cc (X) ,

Λ f =∫

Xf dµ

Then for f ∈Cc (X) ,

|L f | ≤ Λ(| f |) =∫

X| f |dµ = || f ||L1(µ).

Since µ is both inner and outer regular thanks to it being finite, Cc(X) is dense in L1(X ,µ).(See Theorem 15.2.4 for more than is needed.) Therefore L extends uniquely to an elementof (L1(X ,µ))′, L̃. By the Riesz representation theorem for L1 for finite measure spaces,there exists a unique σ ∈ L∞(X ,µ) such that for all f ∈ L1 (X ,µ) ,

L̃ f =∫

Xf σdµ

In particular, for all f ∈C0 (X) ,

L f =∫

Xf σdµ

and it follows from Lemma 20.6.4, µ (X) = ||L||.It remains to verify |σ |= 1 a.e. For any f ≥ 0,

Λ f ≡∫

Xf dµ ≥ |L f |=

∣∣∣∣∫Xf σdµ

∣∣∣∣