78 1. Measure theory (i) Show that E is also B-measurable. (ii) If there exists a B-measurable set F of finite measure (i.e. μ(F ) ∞) that contains all of the En, show that limn→∞ μ(En) = μ(E). (Hint: Apply downward monotonicity to the sets nN (EnΔE).) (iii) Show that the previous part of this exercise can fail if the hypothesis that all the En are contained in a set of finite measure is omitted. Exercise 1.4.25. Let X be an at most countable set with the discrete σ-algebra. Show that every measure μ on this measurable space can be uniquely represented in the form μ = x∈X cxδx for some cx [0, +∞], thus μ(E) = x∈E cx for all E X. (This claim fails in the uncountable case, although showing this is slightly tricky.) A useful technical property, enjoyed by some measure spaces, is that of completeness: Definition 1.4.31 (Completeness). A null set of a measure space (X, B,μ) is defined to be a B-measurable set of measure zero. A sub-null set is any subset of a null set. A measure space is said to be complete if every sub-null set is a null set. Thus, for instance, the Lebesgue measure space (Rd, L[Rd],m) is com- plete, but the Borel measure space (Rd, B[Rd],m) is not (as can be seen from the solution to Exercise 1.4.16). Completion is a convenient property to have in some cases, particularly when dealing with properties that hold almost everywhere. Fortunately, it is fairly easy to modify any measure space to be complete: Exercise 1.4.26 (Completion). Let (X, B,μ) be a measure space. Show that there exists a unique refinement (X, B, μ), known as the completion of (X, B,μ), which is the coarsest refinement of (X, B,μ) that is complete. Furthermore, show that B consists precisely of those sets that differ from a B-measurable set by a B-subnull set. Exercise 1.4.27. Show that the Lebesgue measure space (Rd, L[Rd],m) is the completion of the Borel measure space (Rd, B[Rd],m). Exercise 1.4.28 (Approximation by an algebra). Let A be a Boolean al- gebra on X, and let μ be a measure on A.
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