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

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**Extra info for A Basic Course in Measure and Probability: Theory for Applications**

**Sample text**

A measure μ may be unambiguously deﬁned on S by the equation μ(E ∪ N) = μ(E), E ∈ S, N ⊂ A ∈ S, μ(A) = 0. μ is then a complete measure on S, extending μ on S. The σ-ring S is thus “slightly” enlarged by adjoining subsets of zero measure, to sets of S. Proof We show ﬁrst that μ is well deﬁned. That is, if E1 ∪ N1 = E2 ∪ N2 , where E1 , E2 ∈ S, N1 ⊂ A1 ∈ S, N2 ⊂ A2 ∈ S and μ(A1 ) = μ(A2 ) = 0, then we must show that μ(E1 ) = μ(E2 ). To see this, note that E1 – E2 is clearly a subset of N2 , hence of A2 , and thus μ(E1 – E2 ) = 0.

Compactness), the bounded closed interval on the left is contained in a ﬁnite number of the open intervals on the right, and hence for some n, [a0 + , b0 ] ⊂ ∪ni=1 (ai , bi + /2i ). 2, b0 – a0 – ≤ n i=1 (bi Since is arbitrary, b0 – a0 ≤ – ai + ∞ i=1 (bi 2i ) ≤ ∞ i=1 (bi – ai ) + . – ai ), as required. 4 There is a unique measure μ on the σ-ﬁeld B of Borel sets, such that μ{(a, b]} = b – a for all real a < b. μ is σ-ﬁnite and is called Lebesgue measure on B. 38 Measures: general properties and extension Proof Deﬁne μ on P by μ{(a, b]} = b – a.

May be written as a sequence covering 2 E = ∪∞ 1 En . Hence μ* (E) ≤ = ∞ ∞ n=1 m=1 μ(Enm ) ∞ * n=1 μ (En ) + . ≤ ∞ * n=1 (μ (En ) + /2n ) * Since > 0 is arbitrary, μ* (E) ≤ ∞ n=1 μ (En ). On the other hand this is * trivially true if μ (En ) = ∞ for one or more values of n. Thus μ* is an outer measure, as required. 1). However, we are primarily interested in obtaining a measure on S(R). This may be done by restricting μ* further to S(R) (a subclass of S* by the next lemma). Then the set function μ on S(R), deﬁned by μ(E) = μ* (E), will be a measure on S(R), again extending μ on R.