1056 CHAPTER 29. THE AREA FORMULA

Corollary 29.12.5 Let h be one to one and in W 1,p (Rn;Rn) , p > n. Let Ω be a bounded,open, connected set in Rn and suppose ∂Ω,h(∂Ω) have measure zero. Let f ∈ L1 (h(Ω))where f is also Borel measurable. Then∫

h(Ω)f (y)dy =

∫Ω

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

It can also be written in the form∫f (y)d (y,Ω,h)dy =

∫Ω

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

Note this is a general area formula under somewhat more restrictive hypotheses than theusual area formula because it involves an assumption that Ω is connected and a troublesomecondition on the measure of h(∂Ω) ,∂Ω being zero, but it does not require h to be Lipschitz.It looks like a strange result because |detDh(x)| is not in L∞ and so it is not clear why theintegral on the right should even be finite just because f is in L1. If the result is correct, itis surprising.

The condition on the measure of ∂Ω and h(∂Ω) is not necessary. Neither is it necessaryto assume Ω is connected. This is shown next.

Theorem 29.12.6 Let h be one to one on Ω and in W 1,p (Rn;Rn) , p > n. Let Ω be abounded, open set in Rn. Let f ∈ L1 (h(Ω)) where f is also Borel measurable. Then∫

h(Ω)f (y)dy =

∫Ω

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

It can also be written in the form∫f (y)d (y,Ω,h)dy =

∫Ω

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

Proof: Let Ω ⊆ [−R,R]n ≡ Q. Let pi (x) ≡ xi where x =(x1, · · · ,xn)T . Then here is a

claim.Claim: Let b be given. There exists a, |a−b| ≤ 4−k such that

m(h([pix = a]∩Q)) = 0.

Here [pix = a] is short for {x : pix = a}.Proof of claim: If this is not so, then for every a in an interval centered at b,

m(h([pix = a]∩Q))> 0

However,

m(∪a∈[b−4−k,b+4−k]h([pix = a]∩Q)

)= m

(h

(∏

jA j

))where A j = [−R,R] if j ̸= i and Ai =

[b−4−k,b+4−k

]. This is finite because h

(∏ j A j

)is

a compact set, being the continuous image of such a set. Since h is one to one, this com-pact set would then be the union of uncountably many disjoint sets, each having positive