1.2. Lebesgue measure 37 Unfortunately, the above argument is terribly non-rigorous for a number of reasons, not the least of which is that it uses an uncountable number of coin flips, and the rigorous probabilistic theory that one would have to use to model such a system of random variables is too weak12 to be able to assign meaningful probabilities to such events as “E is Lebesgue measurable”. So we now turn to more rigorous arguments that establish the existence of non-measurable sets. The arguments will be fairly simple, but the sets constructed are somewhat artificial in nature. Proposition 1.2.18. There exists a subset E ⊂ [0, 1] which is not Lebesgue measurable. Proof. We use the fact that the rationals Q are an additive subgroup of the reals R, and so partition the reals R into disjoint cosets x + Q. This creates a quotient group R/Q := {x + Q : x ∈ R}. Each coset C of R/Q is dense in R, and so has a non-empty intersection with [0, 1]. Applying the axiom of choice, we may thus find an element xC ∈ C ∩ [0, 1] for each C ∈ R/Q. We then let E := {xC : C ∈ R/Q} be the collection of all these coset representatives. By construction, E ⊂ [0, 1]. Let y be any element of [0, 1]. Then it must lie in some coset C of R/Q, and thus differs from xC by some rational number in [−1, 1]. In other words, we have (1.4) [0, 1] ⊂ q∈Q∩[−1,1] (E + q). On the other hand, we clearly have (1.5) q∈Q∩[−1,1] (E + q) ⊂ [−1, 2]. Also, the different translates E +q are disjoint, because E contains only one element from each coset of Q. We claim that E is not Lebesgue measurable. To see this, suppose for contradiction that E was Lebesgue measurable. Then the translates E + q would also be Lebesgue measurable. By countable additivity, we thus have m( q∈Q∩[−1,1] (E + q)) = q∈Q∩[−1,1] m(E + q), and thus by translation invariance and (1.4), (1.5) 1 ≤ q∈Q∩[−1,1] m(E) ≤ 3. 12For some further discussion of this point, see [Ta2009, §1.10].

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