is measurable and has m_{n}
measure zero. See Proposition 19.9.6. Now let k →∞ to conclude that m_{n}
(f (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
(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 ℝ^{n}.
Lemma 21.5.2Let h be continuous and map B
(0,r)
⊆ ℝ^{n}to ℝ^{n}. Suppose that for all x ∈B
(0,r)
,
|h (x )− x| < εr
Then it follows that
(------)
h B (0,r) ⊇ B (0,(1 − ε)r)
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.
Lemma 21.5.3Let h : U → ℝ^{n}where U is an open bounded set and suppose that Dh
(x)
exists and isinvertible and continuous on U and that h is one to one on U. Let A ⊆ U be a Lebesgue measurable set.Then
∫
m (h(A )) = |detDh (x)|dm
n A n
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}
(V ∖ A)
< 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}
(G ∖A)
= 0. Say G = ∩_{i}V_{i}. Then since h is one to one,
mn (h (G))− mn (h(A)) = mn (h(G) ∖h(A))
= m (h(G ∖A )) = 0
n
Then
m (h(A )) = m (h (G )) = lim m (h (V ))
n n m→ ∞ n m
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
(x)
≠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
||x||
≡ max
{|xi|,i = 1,⋅⋅⋅,n}
.
Lemma 21.5.6(Sard) Let U be an open set in ℝ^{n}and let h : U → ℝ^{n}be differentiable.Let
Z ≡ {x ∈ U : detDh (x) = 0}.
Then m_{n}
(h (Z ))
= 0.
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
(x,r)
⊆ V
and also,
h (B(x,r)) ⊆ h(x)+ Dh (x)(B (0,r))+ B (0,rη), η < 1.
Regard Dh
(x)
as an m × m matrix, the matrix of the linear transformation Dh
(x)
with respect to the
usual coordinates. Since x ∈ Z, it follows that there exists an invertible matrix A such that ADh
(x)
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)