By Leadbetter R., Cambanis S., Pipiras V.
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Extra resources for A Basic Course in Measure and Probability: Theory for Applications
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Fn are disjoint sets of P (since E ∈ R). Since Fj = ∪∞ k=1 Ek ∩ Fj (a union of disjoint sets of R), the above result implies μ(Fj ) = ∞ k=1 ν(Fj ∩ Ek ). Hence ν(E) = n j=1 μ(Fj ) = ∞ k=1 n j=1 ν(Fj ∩ Ek ) = ∞ k=1 ν(Ek ) (Ek = ∪nj=1 Fj ∩Ek and ν is finitely additive on R) so that countable additivity follows. g. summing “by diagonals”. 4 Measures from outer measures 29 (iii) If μ is finite, ν clearly is also. If μ is σ-finite, and E ∈ R, then E = ∪n1 Fi for some Fi ∈ P. Each Fi may be covered by a countable sequence of sets of P (⊂ R) with finite μ-values.
E. μF may have an atom at a. In fact, μF {a} = limn→∞ {F(a) – F(a – 1n )} = F(a) – F(a – 0). Thus μF ({a}) is zero if F is continuous at a, and otherwise its value is the magnitude of the jump of F at a. We see also that for open and closed intervals, μF {(a, b)} = F(b – 0) – F(a), (writing (a, b) = (a, b] – {b}, μF {[a, b]} = F(b) – F(a – 0) [a, b] = (a, b] ∪ {a}). e. μF = μG , then F – G Exercises 41 is constant. The converse of this is clear – if F and G differ by a constant then certainly μF = μG on P and hence on B.
One may ask whether there are in fact (a) any Lebesgue measurable sets which are not Borel sets, and (b) any sets at all which are not Lebesgue measurable. The answer is, in fact affirmative in both cases (the former may be proved by a cardinality argument and the latter by using the “axiom of choice”), but we shall not pursue the matter here. 9. It is worth noting that both Borel and Lebesgue measurable sets of finite measure, may be approximated by finite unions of intervals. That is if E ∈ B or E ∈ L and m(E) < ∞ there are, given > 0, intervals I1 , I2 , .