19.7. MEASURES AND POSITIVE LINEAR FUNCTIONALS 517

Lemma 19.7.4 µ is a well-defined outer measure.

Proof: First it is necessary to verify that µ is well defined because there are two de-scriptions of it on open sets. Suppose then that µ1 (V )≡ inf{µ(U) : U ⊇V and U is open}.It is required 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. 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 assumed µ(Ei) < ∞ for each i since if

not so, there is nothing to prove. Let Vi ⊇ Ei with µ(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). It is clear from the definition that if A⊆ B, then

µ (A)≤ µ (B). ■

Lemma 19.7.5 Let K be compact, g≥ 0, g ∈Cc(Ω), and g = 1 on K. Then µ(K)≤ Lg.Also µ(K)< ∞ whenever K is compact.

Proof: Let α ∈ (0,1) and Vα = {x : g(x)> α} so Vα ⊇ K and let h≺Vα .

g > α

VαK

Then h≤ 1 on Vα while gα−1 ≥ 1 on Vα and so gα−1 ≥ h which implies L(gα−1)≥ Lhand that therefore, since L is linear, Lg ≥ αLh. Since h ≺ Vα is arbitrary, and K ⊆ Vα ,Lg≥ αµ (Vα)≥ αµ (K) . Letting α ↑ 1 yields Lg≥ µ(K). This proves the first part of thelemma. The second assertion follows from this and Theorem 19.1.23. If K is given, letK ≺ g≺Ω and so from what was just shown, µ (K)≤ Lg < ∞. ■

Lemma 19.7.6 If A and B are disjoint compact subsets of Ω, then µ(A∪B) = µ(A)+µ(B).

Proof: By Theorem 19.1.23, there exists h ∈Cc (Ω) such that A ≺ h ≺ BC. Let U1 =h−1(( 1

2 ,1]), V1 = h−1([0, 12 )). Then A⊆U1,B⊆V1 and U1∩V1 = /0.

B V1A U1