Linear Algebra and Analysis

21.5 Change Of Variables

Here is an interesting proposition.

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

Nk ≡ {x ∈ N : ∥Df (x)∥ ≤ k}

There is an open set V ⊇ N_k such that m_n

< ε. For each x ∈ N_k, there is a ball B_x centered at x with radius r_x < 1 such that B_x ⊆ V and for y ∈ B_x,

f (y) ∈ f (x )+ Df (x)B (0,rx)+ B (0,εrx)

Thus

By the Besicovitch covering theorem, there are balls of this sort such that

Mn Nk ⊆ ∪j=1 ∪ {B ∈ Gj}

where G_j is a countable disjoint collection of these balls. Thus,

Since ε is arbitrary, it follows that m_n = 0 and so in fact f is measurable and has m_n measure zero. See Proposition 19.9.6. Now let k →∞ to conclude that m_n = 0. Now the conclusion is shown the same as in Proposition 21.2.5. You exploit inner regularity, and what was just shown, to obtain the conclusion that f is measurable if F is. ■

Recall Lemma 18.11.1 which was based on the Brouwer fixed point theorem. The version of use here is stated below. In what follows,

is the usual Euclidean norm on ℝⁿ.

Now the Besicovitch covering theorem for a Vitali cover is used to give an easy treatment of the change of variables for multiple integrals. This will be based on the following lemma.

Proof: In what follows, ε will be a small positive number. Let A be a Lebesgue measurable subset of U and

A ⊆ {x ∈ U : |detDh (x)| < k} (**)

(**)

let V be an open set containing A such that m_n

< ε. We can also assume that

mn (h(V)∖ h(A)) < ε

The reason for this is as follows. There exists G ⊇ A such that G is the countable intersection of nested open sets which each contain A and m_n

= 0. Say G = ∩_iV _i. Then since h is one to one,

Then

m (h(A )) = m (h (G )) = lim m (h (V )) n n m→ ∞ n m

so eventually, for large enough m,

mn (h (A))+ ε > mn (h(Vm ))

Then for x ∈ A,

h(x + v) = h(x) +Dh (x)v + o(v) (21.18)

(21.18)

Thus, for all v small enough, ≤ r,

| | ||Dh (x)−1(h (x + v)− h (x))− v|| < εr

It follows from Lemma 21.5.2 that for each x ∈ A, there is a ball B

⊆ V

this holding for all r_x small enough. Here λ > 1 is arbitrary. Thus

n mn (h (B(x,λrx))) ≥ |det(Dh (x))|(1 − ε) mn (B (x,rx))

Letting λ ↓ 1, and using that h maps sets of measure zero to sets of measure zero,

n mn (h(B (x,rx))) ≥ |det(Dh (x ))|(1− ε) mn (B (x,rx))

As explained earlier, even if you have a general Radon measure, you could assume that r_x is such

has measure zero, although in the case of Lebesgue measure, this can be shown for all r_x. It follows from Proposition 21.1.4 for example.

Also from 21.18, we can assume that r_x is small enough that

and so

mn (h(B (x,rx))) ≤ |detDh (x )|(1+ ε)nmn (B (x,rx))

Thus, whenever r_x is small enough,

At this point, assume a little more on h. Assume it is actually C¹

. Then by making r_x smaller if necessary, it can be assumed that for y ∈ B,

||detDh (x)|− |detDh (y)|| < ε

This is a Vitali cover for A. Therefore, there is a countable disjoint sequence of these balls

such that m_n = 0. Then since h is one to one,

∑∞ mn (h(A)) ≤ mn (h (Bi )) ≤ mn (h (V )) < mn (h(A))+ ε i=1

Now the following chain of inequalities follow from the above.

∑∞ ≤ (1+ ε)n|detDh (xi)|mn (Bi)+ Cε i=1

∑∞ ∫ = (1+ ε)n |detDh (xi)|dmn + Cε i=1 Bi

∞ ∫ ≤ ∑ (1+ ε)n |detDh (x)|dm + Cε i=1 Bi n

∫ n n ≤ |det Dh (x)|dmn (1+ ε) + εk(1+ ε) + C ε A

Since ε is arbitrary, the ends of this string and m_n

+ ε + εⁿm_n in the middle show that

∫ mn (h (A)) = A |Dh (x)|dmn

Now we remove the assumption ∗∗. Letting A be arbitrary and measurable, let

Ak ≡ A∩ {x : |detDh (x)| ≤ k}

From what was just shown,

∫ m (h(A )) = |Dh (x)|dm n k Ak n

Let k →∞ and use monotone convergence theorem to prove the lemma. ■

You can remove the assumption that h is C¹ but it is a little more trouble to do so. It is easy to remove the assumption that U is bounded.

Proof: Let A ⊆ U and let A_k ≡ B

∩ A,U_k ≡ B ∩ U. Then the above lemma shows

∫ mn (h (Ak )) = |detDh (x )|dmn Ak

Now use the monotone convergence theorem to obtain the conclusion of the lemma. ■

Now it is easy to prove the change of variables formula.

Proof: Let f

= lim_k→∞s_k where s_k is a nonnegative simple function. Say

m∑k k sk (y ) = ciXFki (y) i=1

Then

m∑k m∑k sk(h(x)) = ckiXFki (h (x )) = ckiXh−1(Fki )(x) i=1 i=1

It follows from the above corollary that

∫ ∑mk ∫ sk(h(x))|det Dh (x)|dmn = cki |det Dh (x)|dmn U i=1 h−1(Fki )

m∑k ( ) ∫ m∑k ∫ = ckimn F ki = ckiXFk (y )dmn = sk (y )dmn i=1 h(U)i=1 i h(U )

Now apply monotone convergence theorem to obtain the desired result. ■

It is a good idea to remove the requirement that detDh

≠0. This is also fairly easy from the Besicovitch covering theorem.

The following is Sard’s lemma. In the proof, it does not matter which norm you use in defining balls but it may be easiest to consider the norm

≡ max.

Proof: For convenience, assume the balls in the following argument come from

_∞. First note that Z is a Borel set because h is continuous and so the component functions of the Jacobian matrix are each Borel measurable. Hence the determinant is also Borel measurable.

Suppose that U is a bounded open set. Let ε > 0 be given. Also let V ⊇ Z with V ⊆ U open, and

mn (Z)+ ε > mn (V ).

Now let x ∈ Z. Then since h is differentiable at x, there exists δ_x > 0 such that if r < δ_x, then B

⊆ V and also,

h (B(x,r)) ⊆ h(x)+ Dh (x)(B (0,r))+ B (0,rη), η < 1.

Regard Dh

as an m × m matrix, the matrix of the linear transformation Dh with respect to the usual coordinates. Since x ∈ Z, it follows that there exists an invertible matrix A such that ADh is in row reduced echelon form with a row of zeros on the bottom. Therefore,

mn (A (h(B (x,r)))) ≤ mn (ADh (x)(B(0,r)) + AB (0,rη)) (21.20)

(21.20)

The diameter of ADh

is no larger than 2r and it lies in ℝⁿ⁻¹ ×. The diameter of AB