1.7. Outer measures, pre-measures, product measures 155 Combining all of these estimates, we obtain μ∗(A ∩ E) + μ∗(A\E) ≤ μ∗(A) + ε since ε 0 was arbitrary, the claim follows. Finally, we have to show that μ∗(E) = μ0(E). Since E covers itself, we certainly have μ∗(E) ≤ μ0(E). To show the converse inequality, it suﬃces to show that ∞ n=1 μ0(En) ≥ μ0(E) whenever E1,E2,... ∈ B0 cover E. By replacing each En with the smaller set En\ n−1 m=1 Em (which still lies in B0, and still covers E), we may assume without loss of generality (thanks to the monotonicity of μ0) that the En are disjoint. Similarly, by replacing each En with the smaller set En ∩ E we may assume without loss of generality that the union of the En is exactly equal to E. But then the claim follows from the hypothesis that μ0 is a pre-measure (and not merely a finitely additive measure). Let us call the measure μ constructed in the above proof the Hahn- Kolmogorov extension of the pre-measure μ0. Thus, for instance, from Ex- ercise 1.7.2, the Hahn-Kolmogorov extension of elementary measure (with the convention that co-elementary sets have infinite elementary measure) is Lebesgue measure. This is not quite the unique extension of μ0 to a count- ably additive measure, though. For instance, one could restrict Lebesgue measure to the Borel σ-algebra, and this would still be a countably additive extension of elementary measure. However, the extension is unique within its own σ-algebra: Exercise 1.7.7. Let μ0 : B0 → [0, +∞] be a pre-measure, let μ: B → [0, +∞] be the Hahn-Kolmogorov extension of μ0, and let μ : B → [0, +∞] be another countably additive extension of μ0. Suppose also that μ0 is σ- finite, which means that one can express the whole space X as the countable union of sets E1,E2,... ∈ B0 for which μ0(En) ∞ for all n. Show that μ and μ agree on their common domain of definition. In other words, show that μ(E) = μ (E) for all E ∈ B∩B . (Hint: First show that μ (E) ≤ μ∗(E) for all E ∈ B .) Exercise 1.7.8. The purpose of this exercise is to show that the σ-finite hypothesis in Exercise 1.7.7 cannot be removed. Let A be the collection of all subsets in R that can be expressed as finite unions of half-open intervals [a, b). Let μ0 : A → [0, +∞] be the function such that μ0(E) = +∞ for non-empty E and μ0(∅) = 0. (i) Show that μ0 is a pre-measure. (ii) Show that A is the Borel σ-algebra B[R].

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