Warpvader Posted:

I would love to help you with your math problem, but I don’t understand Greek/Latin/or any other dead languages.

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).