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Warpvader Posted:
o its very simple
it means “for all rings R, for all A within M(R ), there exists a sequence of sets A_n which are elements of R such that their union is A.”
to see wat im talking about, here is an example in which it is implicitly stated: http://www-maths.swan.ac.uk/staff/vl/c98.pdf go to page 18, theorem 4.2
Theorem 4.2 Let A be an algebra of sets, μ and ν two measures defined on the σ-algebra A_σ generated by A. Then μ ↾ A = ν ↾ A implies μ = ν. Proof. Choose S ∈ A_σ, then S=lim (n⇒∞) S_n, S_n ∈A, for A_σ=M(A ). Using continuity of measure, one has μ(S) = lim (n⇒∞) μ(S_n) = lim (n⇒∞) ν(S_n) = ν(S). a dallop a daisy edited this message on 03/07/2008 9:01PM |
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Posted On: 03/07/2008 8:59PM | View a dallop a daisy...'s Profile | # |