By Molk J. (ed.)

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Then µ0 has an extension to a complete locally determined measure µ, defined on every member of K, inner regular with respect to K, and such that whenever E ∈ dom µ and µE < ∞ there is an E0 ∈ Σ0 such that µ(E E0 ) = 0. proof (a) Set T0 = {E : E ∈ Σ0 , µ0 E < ∞}, ν0 = µ0 T0 . Then ν0 , T0 satisfy the conditions of 413N; take ν1 , T1 as in 413N. If K, L ∈ K and L ⊆ K, then ν1 L + sup{ν1 K : K ∈ K, K ⊆ K \ L} = ν1 L + ν1 (K \ L) = ν1 K. So ν1 K satisfies the conditions of 413M and there is a complete locally determined measure µ, extending ν1 K, and inner regular with respect to K.

Iii) If E ∈ T, then ν∗ (E \ M ) = sup{νF : F ∈ T, F ⊆ E \ M } = sup{νE − ν(E \ F ) : F ∈ T, F ⊆ E \ M } = sup{νE − νF : F ∈ T, E ∩ M ⊆ F ⊆ E} = νE − inf{νF : F ∈ T, E ∩ M ⊆ F ⊆ E} = νE − ν ∗ (E ∩ M ). So ν E = ν ∗ (E ∩ M ) + ν∗ (E \ M ) = νE. Thus ν extends ν. (iv) If H ∈ T and > 0, express H as (E ∩ M ) ∪ (F \ M ), where E, F ∈ T. Then we can find (α) a K ∈ K ∩ T such that K ⊆ E and ν(E \ K) ≤ (β) an F ∈ T such that F ⊆ F \ M and νF ≥ ν∗ (F \ M ) − (γ) a K ∈ K ∩ T such that K ⊆ F and νK ≥ νF − .

Define θ : PX → [0, ∞] by setting θA = inf{ ∞ n=0 νEn : En n∈N is a sequence in Σ covering A} for A ⊆ X, interpreting inf ∅ as ∞ if necessary. Show that θ is an outer measure. Let µθ be the measure defined from θ by Carath´eodory’s method. d. version of µθ . d. ) > (m) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative additive functional. Show that the following are equiveridical: (i) ν has an extension to a measure on X; (ii) limn→∞ νEn = 0 whenever ∞ En n∈N is a non-increasing sequence in Σ with empty intersection; (iii) ν( n∈N En ) = n=0 νEn whenever En n∈N is a disjoint sequence in Σ such that n∈N En ∈ Σ.