664 CHAPTER 33. THE LEBESGUE MEASURE AND INTEGRAL IN Rp
Lemma 33.6.1 Let U and V be bounded open sets in Rp and let h,h−1 be C1 definedrespectively on Û ⊇ Ū and V̂ ⊇ V̄ such that h(U) = V and let f be a bounded uniformlycontinuous function defined on U. Then∫
Vf (y)dmp =
∫U
f (h(x)) |det(Dh(x))|dmp
Proof: Let x ∈U. By the assumption that h and h−1 are C1,
h(x+v)−h(x) = Dh(x)v+o(v)
= Dh(x)(v+Dh−1 (h(x))o(v)
)Let an upper bound for
∥∥Dh−1 (h(x))∥∥ be C. It exists because V is compact and h−1 is
C1 on an open set containing this compact set. Therefore, since all the boxes in Bm are indiameter less than δ ,
h(B(x,r))−h(x) =
h(x+B(0,r))−h(x)⊆ Dh(x)(B(0,(1+Cε)r)) . (33.8)
Then choose m still larger if necessary so that f (y) is uniformly approximated by
∑i
f (h(x(mi)))Xh(Rmi )
(y) , x(mi) ∈ Rmi ,
to within ε. Let rmi be half the diameter of Rm
i . Thus ∑i mp (B(0,rmi )) = mp (U). This is
by the formula for the measure of a box. It is just the product of the lengths of the sides.Recall the norm is ∥·∥
∞so the balls are boxes.
mp (Rmi ) = mp (B(x(mi) ,rm
i )) = mp (B(0,rmi ))
Then ∫V f (y)dmp = ∑
∞i=1∫h(Rm
i )f (y)dmp
≤ εmp (V )+∑∞i=1∫h(Rm
i )f (h(x(mi)))dmp
≤ εmp (V )+∑∞i=1 f (h(x(mi)))mp (h(Rm
i ))
≤ εmp (V )+∑∞i=1 f (h(x(mi)))mp (Dh(x(mi))(B(0,(1+Cε)ri)))
= εmp (V )+(1+Cε)p∑
∞i=1∫
Rmi
f (h(x(mi))) |det(Dh(x(mi)))|dmp
≤ εmp (V )+(1+Cε)p∑
∞i=1
(∫Rm
if (h(x)) |det(Dh(x))|dmp +2εmp (Rm
i ))
≤ εmp (V )+(1+Cε)p∑
∞i=1∫
Rmi
f (h(x)) |det(Dh(x))|dmp
+(1+Cε)p 2εmp (U)
Since ε > 0 is arbitrary, this shows∫V
f (y)dmp ≤∫
Uf (h(x)) |det(Dh(x))|dmp (33.9)
whenever f is uniformly continuous and bounded on V. Now
x→ f (h(x)) |det(Dh(x))|