29.12. THE CASE OF W 1,p 1057

measure. Thus for some 1/l > 0 there must be infinitely many of these disjoint sets havingmeasure larger than 1/l, l ∈ N a contradiction to the set having finite measure. This provesthe claim.

Now from the claim, consider bk,k∈Z given by bk = k2−(m+1) and for each i= 1, · · · ,n,let aik denote a value of the claim, |aik−bk| ≤ 4−m. Thus

m(h([pix = aik]∩Q)) = 0, i = 1, · · · ,n

Thus also ∣∣aik−ai(k+1)∣∣≤ |aik−bk|+

∣∣bk+1−ai(k+1)∣∣+ |bk+1−bk|

≤ 4−m +4−m +2−(m+1) < 2−m

Consider boxes of the form ∏ni=1[aik,ai(k+1)

]. Denote these boxes as Bm. Thus they

are non overlapping boxes the sides of which are of length less than 2−m. Let Ω1 denote theunion of the finitely many boxes of B1 which are contained in Ω. Next let Ω2 denote theunion of the boxes of B2∪B1 which are contained in Ω and so forth. Then ∪∞

k=1Ωk ⊆Ω.Suppose now that p ∈ Ω. Then it is at positive distance from ∂Ω. Let k be the first suchthat p is contained in a box of Bk which is contained in Ω. Then p ∈ Ωk. Therefore, thishas shown that Ω is a countable union of non overlapping closed boxes B which have theproperty that ∂B,h(∂B) have measure zero. Denote these boxes as {Bk}.

First assume f is nonnegative and Borel measurable. Then from Corollary 29.12.5,∫h(Bk)

f (y)dy =∫

Bk

|det(Dh(x))| f (h(x))dx

Since h(∂Bk) has measure zero,∫h(∪m

k=1Bk)f (y)dy =

m

∑k=1

∫h(Bk)

f (y)dy

=m

∑k=1

∫Bk

|det(Dh(x))| f (h(x))dx

=∫∪m

k=1Bk

|det(Dh(x))| f (h(x))dx

and now letting m→ ∞ and using the monotone convergence theorem,∫h(Ω)

f (y)dy =∫

Ω

|det(Dh(x))| f (h(x))dx (29.12.75)

Next assume in addition that f is also in L1 (h(Ω)). Recall that from properties of thedegree, d (y,U,h) is constant on h(U) for U a component of Ω. Since h is one to one,Proposition 23.6.4 implies this constant is either −1 or 1. Let the components of Ω be{Ui}∞

i=1 . Also from Theorem 23.2.2 and the assumption that h is one to one, if y ∈ h(Ui) ,then y /∈ h

(∪ j ̸=iU j

)and

d (y,Ui,h)+d(y,∪ j ̸=iU j,h

)= d (y,Ω,h)