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

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

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In particular this holds if μ is a measure on P. Proof If E ∈ P, F ∈ P and E ⊂ F, then F – E = ∪n1 Ei for disjoint sets Ei ∈ P. 1) since μ is nonnegative. Hence μ is monotone. If also F–E ∈ P and μ(E) is finite, then F = E∪(F–E) and μ(F) = μ(E)+ μ(F – E) so that μ(F) – μ(E) = μ(F – E), showing that μ is subtractive. 2 If μ is a measure on a ring R, if E ∈ R, and {Ei } is any ∞ sequence of sets of R such that E ⊂ ∪∞ 1 μ(Ei ). 3). Thus μ(E) = ∞ 1 μ(Gi ) ≤ ∞ 1 μ(Ei ) since μ is monotone and Gi ⊂ E ∩ Ei ⊂ Ei .

Choose 0 < b0 > a0 ). Then (a0 , b0 ] ⊂ ∪∞ i=1 (ai , bi ] so that clearly < b0 – a0 (assuming i [a0 + , b0 ] ⊂ ∪∞ i=1 (ai , bi + /2 ). e. 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.

4, as asserted. 5 shows that μ{a} = limn→∞ μ{(a – = 0. Consequently any countable set has Lebesgue measure zero. g. μ{[a, b]} = μ{(a, b]} + μ({a}) = b – a). Lebesgue measure on B provides a generalized notion of “length”, for sets of B which need not be intervals. 1 to obtain μ on a σ-field B ⊃ B. B consists of sets of the form B ∪ N where B ∈ B and N ⊂ A for some A ∈ B, μ(A) = 0. B is called the σ-field of Lebesgue measurable sets, and the completion μ on B is called Lebesgue measure on the class (B) of Lebesgue measurable sets.

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