1.1. Prologue: The problem of measure 11 Exercise 1.1.15 (Uniqueness of Jordan measure). Let d≥1. Let m : J (Rd) R+ be a map from the collection J (Rd) of Jordan-measurable subsets of Rd to the non-negative reals that obeys the non-negativity, finite additiv- ity, and translation invariance properties. Show that there exists a constant c R+ such that m (E) = cm(E) for all Jordan measurable sets E. In particular, if we impose the additional normalisation m ([0, 1)d) = 1, then m m. Exercise 1.1.16. Let d1,d2 1, and let E1 Rd1 , E2 Rd2 be Jordan measurable sets. Show that E1 × E2 Rd1+d2 is Jordan measurable, and md1+d2 (E1 × E2) = md1 (E1) × md2 (E2). Exercise 1.1.17. Let P, Q be two polytopes in Rd. Suppose that P can be partitioned into finitely many sub-polytopes P1,...,Pn which, after being rotated and translated, form new polytopes Q1,...,Qn which are an almost disjoint cover of Q, which means that Q = Q1 . . . Qn, and for any 1 i j n, Qi and Qj only intersect at the boundary (i.e. the interior of Qi is disjoint from the interior of Qj). Conclude that P and Q have the same Jordan measure. The converse statement is true in one and two dimensions d = 1, 2 (this is the Bolyai-Gerwien theorem), but false in higher dimensions (this was Dehn’s negative answer [De1901] to Hilbert’s third problem). The above exercises give a fairly large class of Jordan measurable sets. However, not every subset of Rd is Jordan measurable. First of all, the unbounded sets are not Jordan measurable, by construction. But there are also bounded sets that are not Jordan measurable: Exercise 1.1.18. Let E Rd be a bounded set. (1) Show that E and the closure E of E have the same Jordan outer measure. (2) Show that E and the interior E◦ of E have the same Jordan inner measure. (3) Show that E is Jordan measurable if and only if the topological boundary ∂E of E has Jordan outer measure zero. (4) Show that the bullet-riddled square [0, 1]2\Q2, and set of bullets [0, 1]2 ∩Q2, both have Jordan inner measure zero and Jordan outer measure one. In particular, both sets are not Jordan measurable. Informally, any set with a lot of “holes”, or a very “fractal” boundary, is unlikely to be Jordan measurable. In order to measure such sets we will need to develop Lebesgue measure, which is done in the next set of notes. Exercise 1.1.19 (Carath´ eodory type property). Let E Rd be a bounded set, and let F Rd be an elementary set. Show that m∗,(J)(E) = m∗,(J)(E F ) + m∗,(J)(E\F ).
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