36 1. Measure theory (i) If d ≥ 2, show that every continuously differentiable curve has Lebesgue measure zero. (Why is the condition d ≥ 2 necessary?) (ii) Conclude that if d ≥ 2, then the unit cube [0, 1]d cannot be covered by countably many continuously differentiable curves. We remark that if the curve is only assumed to be continuous, rather than continuously differentiable, then these claims fail, thanks to the existence of space-filling curves. 1.2.3. Non-measurable sets. In the previous section we have set out a rich theory of Lebesgue measure, which enjoys many nice properties when applied to Lebesgue measurable sets. Thus far, we have not ruled out the possibility that every single set is Lebesgue measurable. There is good reason for this: A famous theorem of Solovay [So1970] asserts that, if one is willing to drop the axiom of choice, there exist models of set theory in which all subsets of Rd are measurable. So any demonstration of the existence of non-measurable sets must use the axiom of choice in some essential way. That said, we can give an informal (and highly non-rigorous) motivation as to why non-measurable sets should exist, using intuition from probability theory rather than from set theory. The starting point is the observation that Lebesgue sets of finite measure (and, in particular, bounded Lebesgue sets) have to be “almost elementary”, in the sense of Exercise 1.2.16. So all we need to do to build a non-measurable set is to exhibit a bounded set which is not almost elementary. Intuitively, we want to build a set which has oscillatory structure even at arbitrarily fine scales. We will non-rigorously do this as follows. We will work inside the unit interval [0, 1]. For each x ∈ [0, 1], we imagine that we flip a coin to give either heads or tails (with an independent coin flip for each x), and let E ⊂ [0, 1] be the set of all the x ∈ [0, 1] for which the coin flip came up heads. We suppose for contradiction that E is Lebesgue measurable. Intuitively, since each x had a 50% chance of being heads, E should occupy about “half” of [0, 1] applying the law of large numbers (see e.g. [Ta2009, §1.4]) in an extremely non-rigorous fashion, we thus expect m(E) to equal 1/2. Moreover, given any subinterval [a, b] of [0, 1], the same reasoning leads us to expect that E ∩ [a, b] should occupy about half of [a, b], so that m(E ∩ [a, b]) should be |[a, b]|/2. More generally, given any elementary set F in [0, 1], we should have m(E ∩ F ) = m(F )/2. This makes it very hard for E to be approximated by an elementary set indeed, a little algebra then shows that m(EΔF ) = 1/2 for any elementary F ⊂ [0, 1]. Thus E is not Lebesgue measurable.

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.