74 1. Measure theory impossible) to exhibit a specific set that is not Borel measurable. Indeed, a large majority of the explicitly constructible sets that one actually encoun- ters in practice tend to be Borel measurable, and one can view the property of Borel measurability intuitively as a kind of “constructibility” property. (Indeed, as a very crude first approximation, one can view the Borel mea- surable sets as those sets of “countable descriptive complexity” in contrast, sets of finite descriptive complexity tend to be Jordan measurable (assuming they are bounded, of course). Exercise 1.4.17. Let E, F be Borel measurable subsets of Rd1 , Rd2 , re- spectively. Show that E × F is a Borel measurable subset of Rd1+d2 . (Hint: First establish this in the case when F is a box, by using Remark 1.4.15. To obtain the general case, apply Remark 1.4.15 yet again.) The above exercise has a partial converse: Exercise 1.4.18. Let E be a Borel measurable subset of Rd1+d2 . (i) Show that for any x1 ∈ Rd1 , the slice {x2 ∈ Rd2 : (x1,x2) ∈ E} is a Borel measurable subset of Rd2 . Similarly, show that for every x2 ∈ Rd2 , the slice {x1 ∈ Rd1 : (x1,x2) ∈ E} is a Borel measurable subset of Rd1 . (ii) Give a counterexample to show that this claim is not true if “Borel” is replaced with “Lebesgue” throughout. (Hint: The Cartesian product of any set with a point is a null set, even if the first set was not measurable.) Exercise 1.4.19. Show that the Lebesgue σ-algebra on Rd is generated by the union of the Borel σ-algebra and the null σ-algebra. 1.4.3. Countably additive measures and measure spaces. Having set out the concept of a σ-algebra a measurable space, we now endow these structures with a measure. We begin with the finitely additive theory, although this theory is too weak for our purposes and will soon be supplanted by the countably additive theory. Definition 1.4.19 (Finitely additive measure). Let B be a Boolean algebra on a space X. An (unsigned) finitely additive measure μ on B is a map μ: B → [0, +∞] that obeys the following axioms: (i) (Empty set) μ(∅) = 0. (ii) (Finite additivity) Whenever E, F ∈ B are disjoint, then μ(E∪F ) = μ(E) + μ(F ).

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