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Example text

Define a closed dyadic cube to be a cube Q of the form Q= id id + 1 i1 i1 + 1 , × ... × n, 2n 2n 2 2n for some integers n, i1 , . . , id . To avoid some technical issues we shall restrict attention here to “small” cubes of sidelength at most 1, thus we restrict n to the non-negative integers, and we will completely ignore “large” cubes of sidelength greater than one. Observe that the closed dyadic cubes of a fixed sidelength 2−n are almost disjoint, and cover all of Rd . Also observe that each dyadic cube of sidelength 2−n is contained in exactly one “parent” cube of sidelength 2−n+1 (which, conversely, has 2d “children” of sidelength 2−n ), giving the dyadic cubes a structure analogous to that of a binary tree (or more precisely, an infinite forest of 2d -ary trees).

Now we consider the case when the elementary E is not closed. Then we can write E as the finite union Q1 ∪ . . ∪ Qk of disjoint boxes, which need not be closed. But, similarly to before, we can use the epsilon of room strategy: for every ε > 0 and every 1 ≤ j ≤ k, one can find a closed sub-box Qj of Qj such that |Qj | ≥ |Qj | − ε/k (say); then E contains the finite union of Q1 ∪ . . ∪ Qk disjoint closed boxes, which is a closed elementary set. By the previous discussion and the finite additivity of elementary measure, we have m∗ (Q1 ∪ .

An ∈{0,2} i=1 n 1 ai ai , + n ]. i i 3 i=1 3 3 ∞ n=1 In Let C := be the intersection of all the elementary sets In . Show that C is compact, uncountable, and a null set. 10. ) Show that the half-open interval [0, 1) cannot be expressed as the countable union of disjoint closed intervals. (Hint: It is easy to prevent [0, 1) from being expressed as the finite union of disjoint closed intervals. Next, assume for sake of contradiction that [0, 1) is the union of infinitely many closed intervals, and conclude that [0, 1) is homeomorphic to the middle thirds Cantor set, which is absurd.