Chapter 10

Change of VariablesLemma 10.0.1 Every open set in Rp is the countable disjoint union of half open boxesof the form

p

∏i=1

(ai,ai +2−k]

where ai = l2−k for some integers, l,k where k ≥ m. If Bm denotes this collection of halfopen boxes, then every box of Bm+1 is contained in a box of Bm or equals a box of Bm.

Proof: Let m ∈ N be given and let k ≥ m. Let Ck denote all half open boxes of theform ∏

pi=1(ai,ai+2−k] where ai = l2−k for some integer l. Thus Ck consists of a countable

disjoint collection of boxes whose union is Rp. This is sometimes called a tiling of Rp.Think of tiles on the floor of a bathroom and you will get the idea. Note that each box hasEuclidean diameter no larger than 2−k√p. This is because if we have two points, x,y ∈

∏pi=1(ai,ai + 2−k], then |xi− yi| ≤ 2−k. Therefore, |x−y| ≤

(∑

pi=1

(2−k)2)1/2

= 2−k√p.Also, a box of Ck+1 is either contained in a box of Ck or it has empty intersection with thisbox of Ck.

Let U be open and let B1 ≡ all sets of C1 which are contained in U . If B1, · · · ,Bkhave been chosen, Bk+1 ≡ all sets of Ck+1 contained in

U \∪(∪k

i=1Bi

).

Let B∞ = ∪∞i=1Bi. I claim ∪B∞ =U . Clearly ∪B∞ ⊆U because every box of every Bi is

contained in U . If p∈U , let k be the smallest integer such that p is contained in a box fromCk which is also a subset of U . Thus p ∈ ∪Bk ⊆∪B∞. Hence B∞ is the desired countabledisjoint collection of half open boxes whose union is U . The last claim follows from theconstruction. ■

10.1 Linear TransformationsLemma 10.1.1 Let A : Rp → Rp be linear and invertible. Then A maps open sets toopen sets.

Proof: This follows from the observation that if B is any linear transformation, thenB is continuous. Indeed, it is realized by matrix multiplication and so it is clear that ifxn → x, then Bxn → Bx. Then for U open, A(U) =

(A−1

)−1(U) which is open because

A−1 is continuous. ■First is a general result.

Proposition 10.1.2 Let h : U → Rp is continuous where U is an open subset of Rp.Also suppose h is differentiable on H ⊆U where H is Lebesgue measurable. Then if E is aLebesgue measurable set contained in H, then h(E) is also Lebesgue measurable. Also ifN ⊆ H is a set of measure zero, then h(N) is a set of measure zero. In particular, a linearfunction A maps measurable sets to measurable sets.

Proof: Consider the second claim first. Let N be a set of measure zero contained in Hand let

Nk ≡ {x ∈ N : ∥Dh(x)∥ ≤ k}

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