1.1. Prologue: The problem of measure 7 whenever E1,...,Ek are disjoint elementary sets. We also have the obvious degenerate case m(∅) = 0. Finally, elementary measure clearly extends the notion of volume, in the sense that m(B) = |B| for all boxes B. From non-negativity and finite additivity (and Exercise 1.1.1) we con- clude the monotonicity property m(E) ≤ m(F ) whenever E ⊂ F are nested elementary sets. From this and finite additivity (and Exercise 1.1.1) we easily obtain the finite subadditivity property m(E ∪ F ) ≤ m(E) + m(F ) whenever E, F are elementary sets (not necessarily disjoint) by induction one then has m(E1 ∪ . . . ∪ Ek) ≤ m(E1) + . . . + m(Ek) whenever E1,...,Ek are elementary sets (not necessarily disjoint). It is also clear from the definition that we have the translation invariance m(E + x) = m(E) for all elementary sets E and x ∈ Rd. These properties in fact define elementary measure up to normalisation: Exercise 1.1.3 (Uniqueness of elementary measure). Let d ≥ 1. Let m : E(Rd) → R+ be a map from the collection E(Rd) of elementary 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 elementary sets E. In particular, if we impose the additional normalisation m ([0, 1)d) = 1, then m ≡ m. (Hint: Set c := m ([0, 1)d), and then compute m ([0, 1 n )d) for any positive integer n.) Exercise 1.1.4. Let d1,d2 ≥ 1, and let E1 ⊂ Rd1 , E2 ⊂ Rd2 be elementary sets. Show that E1 × E2 ⊂ Rd1+d2 is elementary, and md1+d2 (E1 × E2) = md1 (E1) × md2 (E2).

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