By Leadbetter R., Cambanis S., Pipiras V.

ISBN-10: 1107020409

ISBN-13: 9781107020405

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**Sample text**

Finally, we obtain a result of general use, which will be applied ﬁrst in the coming sections, giving conditions on which a measure on a generated σ-ring S(E) is determined by its values on the generating class E. 7 Let E be a class (containing ∅) which is closed under intersections, and write S = S(E). Let μ be a measure on S which is σﬁnite on E. Then μ is σ-ﬁnite on S. If μ1 is another measure on S with μ1 (E) = μ(E) for all E ∈ E, then μ1 (E) = μ(E) for all E ∈ S. Proof Let A be any ﬁxed set in E such that μ(A) < ∞.

23 Let E, F be two subsets of X and E = {E, F}. Write down D(E) and show that D(E) = S(E) if and only if either (i) E ∩ F = ∅ or (ii) E ⊃ F or (iii) F ⊃ E. 1 Set functions, measure A set function is a function deﬁned on a class of sets; that is, for every set in a given class, a (ﬁnite or inﬁnite) function value is deﬁned. e. values in R = (–∞, ∞). The sets of the class are mapped into R by the function. For example, the class might consist of all bounded intervals and the set function might be their lengths.

N→∞ Proof If μ(Em ) < ∞ then μ(En ) < ∞ for n ≥ m and μ(lim En ) < ∞ since lim En ⊂ Em . Now (Em – En ) is monotone increasing in n, and lim (Em – En ) = ∪n (Em – En ) = Em – ∩n En = Em – lim En ∈ R. 4, μ(Em ) – μ(lim En ) = μ{lim(Em – En )} = lim μ(Em – En ) n n→∞ = lim {μ(Em ) – μ(En )} (μ(En ) < ∞, En ⊂ Em for n ≥ m) n→∞ = μ(Em ) – lim μ(En ). n→∞ Since μ(Em ) is ﬁnite, subtracting it from each side yields the desired result. The two preceding theorems may be expressed in terms of notions of set function continuity.

### A Basic Course in Measure and Probability: Theory for Applications by Leadbetter R., Cambanis S., Pipiras V.

by David

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