1062 CHAPTER 30. INTEGRATION OF DIFFERENTIAL FORMS

Proof: It is clear that Ω = ∂Ω∪ int(Ω). First consider the claim that ∂Ω∩ int(Ω) = /0.Suppose this does not happen. Then there would exist x ∈ ∂Ω∩ int(Ω). Therefore, therewould exist two mappings Ri and R j such that R jx ∈ Rn

0 and Rix ∈ Rn< with x ∈Ui ∩U j.

Now consider the map, R j ◦R−1i , a continuous one to one map from Rn

≤ to Rn≤ having a

continuous inverse. By continuity, there exists r > 0 small enough that,

R−1i B(Rix,r)⊆Ui∩U j.

Therefore, R j ◦R−1i (B(Rix,r)) ⊆ Rn

≤ and contains a point on Rn0,R jx. However, this

cannot occur because it contradicts the theorem on invariance of domain, Theorem 23.4.3,which requires that R j ◦R−1

i (B(Rix,r)) must be an open subset of Rn and this one isn’tbecause of the point on Rn

0. Therefore, ∂Ω∩ int(Ω) = /0 as claimed. This same argumentshows that the property of being in int(Ω) or ∂Ω does not depend on the choice of the atlas.

To verify that ∂ (∂Ω) = /0, let Si be the restriction of Ri to ∂Ω∩Ui. Thus

Si (x) = (0,(Rix)2 , · · · ,(Rix)n)

and the collection of such points for x ∈ ∂Ω∩Ui is an open bounded subset of

{u ∈ Rn : u1 = 0} ,

identified with Rn−1. Si (∂Ω∩Ui) is bounded because Si is the restriction of a continuousfunction defined on Rm and ∂Ω∩Ui ≡ Vi is contained in the compact set Ω. Thus if Si ismodified slightly, to be of the form

S′i (x) = ((Rix)2− ki, · · · ,(Rix)n)

where ki is chosen sufficiently large enough that (Ri (Vi))2−ki < 0, it follows that {(Vi,S′i)}is an atlas for ∂Ω as an n−1 dimensional manifold such that every point of ∂Ω is sent toto Rn−1

< and none gets sent to Rn−10 . It follows ∂Ω is an n− 1 dimensional manifold with

empty boundary. In case n = 1, the result follows by definition of the boundary of a 0dimensional manifold.

Next consider the claim that int(Ω) is open in Ω. If x ∈ int(Ω) , are all points ofΩ which are sufficiently close to x also in int(Ω)? If this were not true, there wouldexist {xn} such that xn ∈ ∂Ω and xn → x. Since there are only finitely many charts ofinterest, this would imply the existence of a subsequence, still denoted by xn and a singlemap, Ri such that Ri (xn) ∈ Rn

0. But then Ri (xn)→ Ri (x) and so Ri (x) ∈ Rn0 showing

x∈ ∂Ω, a contradiction to int(Ω)∩∂Ω = /0. Now it follows that ∂Ω is closed in Ω because∂Ω = Ω\ int(Ω). This proves the Theorem.

Definition 30.1.5 An n dimensional manifold with boundary, Ω is a Ck manifold withboundary for some k ≥ 1 if

R j ◦R−1i ∈Ck

(Ri (Ui∩U j);Rn

)and R−1

i ∈ Ck(RiUi;Rm

). It is called a continuous or Lipschitz manifold with boundary

if the mappings, R j ◦R−1i , R−1

i ,Ri are respectively continuous or Lipschitz continuous. In