1.2. Lebesgue measure 25 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). As a consequence of these facts, we also obtain the important dyadic nesting property: given any two closed dyadic cubes (possibly of different sidelength), either they are almost disjoint, or one of them is contained in the other. If E is open, and x ∈ E, then by definition there is an open ball centered at x that is contained in E, and it is easy to conclude that there is also a closed dyadic cube containing x that is contained in E. Thus, if we let Q be the collection of all the dyadic cubes Q that are contained in E, we see that the union Q∈Q Q of all these cubes is exactly equal to E. As there are only countably many dyadic cubes, Q is at most countable. But we are not done yet, because these cubes are not almost disjoint (for instance, any cube Q in Q will of course overlap with its child cubes). But we can deal with this by exploiting the dyadic nesting property. Let Q∗ denote those cubes in Q which are maximal with respect to set inclusion, which means that they are not contained in any other cube in Q. From the nesting property (and the fact that we have capped the maximum size of our cubes) we see that every cube in Q is contained in exactly one maximal cube in Q∗, and that any two such maximal cubes in Q∗ are almost disjoint. Thus, we see that E is the union E = Q∈Q∗ Q of almost disjoint cubes. As Q∗ is at most countable, the claim follows (adding empty boxes if necessary to pad out the cardinality). We now have a formula for the Lebesgue outer measure of any open set: it is exactly equal to the Jordan inner measure of that set, or of the total volume of any partitioning of that set into almost disjoint boxes. Finally, we have a formula for the Lebesgue outer measure of an arbitrary set: Lemma 1.2.12 (Outer regularity). Let E ⊂ Rd be an arbitrary set. Then one has m∗(E) = inf E⊂U,U open m∗(U). Proof. From monotonicity one trivially has m∗(E) ≤ inf E⊂U,U open m∗(U)

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