8.2. POSITIVE LINEAR FUNCTIONALS AND MEASURES 189

inner and outer regularity. A measure µ satisfying the above conditions is called a Radonmeasure.

The plan is to define an outer measure and then to show that it, together with the σ

algebra of sets measurable in the sense of Caratheodory, satisfies the conclusions of thetheorem. Always, K will be a compact set and V will be an open set.

Definition 8.2.2 µ(V )≡ sup{L f : f ≺V} for V open, µ( /0) = 0. µ(E)≡ inf{µ(V ) :V ⊇ E} for arbitrary sets E.

Lemma 8.2.3 µ is a well-defined outer measure.

Proof: First it is necessary to verify µ is well defined because there are two descriptionsof it on open sets. Suppose then that µ1 (V ) ≡ inf{µ(U) : U ⊇ V and U is open}. It isrequired to verify that µ1 (V ) = µ (V ) where µ is given as sup{L f : f ≺ V}. If U ⊇ V,then µ (U)≥ µ (V ) directly from the definition. Hence from the definition of µ1, it followsµ1 (V ) ≥ µ (V ) . On the other hand, V ⊇ V and so µ1 (V ) ≤ µ (V ) . This verifies µ is welldefined.

It remains to show that µ is an outer measure. First I show that it acts like and outermeasure on open sets. Let V =∪∞

i=1Vi and let f ≺V . Then spt( f )⊆∪ni=1Vi for some n. Let

ψ i ≺Vi, ∑ni=1 ψ i = 1 on spt( f ).

L f =n

∑i=1

L( f ψ i)≤n

∑i=1

µ(Vi)≤∞

∑i=1

µ(Vi).

Hence µ(V )≤ ∑∞i=1 µ(Vi) since f ≺V is arbitrary.

Now let E = ∪∞i=1Ei. Is µ(E) ≤ ∑

∞i=1 µ(Ei)? Without loss of generality, it can be as-

sumed µ(Ei) < ∞ for each i since if not so, there is nothing to prove. Let Vi ⊇ Ei withµ(Ei)+ ε2−i > µ(Vi).

µ(E)≤ µ(∪∞i=1Vi)≤

∞

∑i=1

µ(Vi)≤ ε +∞

∑i=1

µ(Ei).

Since ε was arbitrary, µ(E)≤ ∑∞i=1 µ(Ei) which proves the lemma. ■

Lemma 8.2.4 Let K be compact, g≥ 0, g ∈Cc(X), and g = 1 on K. Then µ(K)≤ Lg.Also µ(K)< ∞ whenever K is compact.

Proof: Let Vα ≡ {x : g(x) > 1−α} where α is small. I want to compare µ (Vα) withµ (K). Thus let h≺Vα .

g > 1−α

VαK

Then h ≤ 1 on Vα while g(1−α)−1 ≥ 1 on Vα and so g(1−α)−1 ≥ h which impliesL(g(1−α)−1)≥ Lh and that therefore, since L is linear,

Lg≥ (1−α)Lh.