Kenneth Kuttler

1058 CHAPTER 29. THE AREA FORMULA

Since y /∈ h(∪ j ̸=iU j

)the second term on the left is 0 and so d (y,Ui,h) = d (y,Ω,h) . There-

fore, by Corollary 29.12.5,∫h(Ω)

f (y)d (y,Ω,h)dy =∞

∑i=1

∫h(Ui)

f (y)d (y,Ω,h)dy

=∞

∑i=1

∫h(Ui)

f (y)d (y,Ui,h)dy =∞

∑i=1

∫XUi (x) f (h(x))det(Dh(x))dx (29.12.76)

From 29.12.75

∞

∑i=1

∫XUi (x) f (h(x)) |det(Dh(x))|dx

=∫ ∞

∑i=1

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

=∫

XΩ f (h(x)) |det(Dh(x))|dx < ∞

and so by Fubini’s theorem, the sum and the integral may be interchanged in 29.12.76 toobtain from the dominated convergence theorem,∫ ∞

∑i=1

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

=∫

Ω

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

which shows ∫h(Ω)

f (y)d (y,Ω,h)dy =∫

Ω

f (h(x))det(Dh(x))dx (29.12.77)

Now if f is Borel measurable and in L1 (Ω) , the above may be applied to the positiveparts of the real and imaginary parts of f to obtain 29.12.77 for such f . This proves thetheorem.

Not surprisingly, it is not necessary to assume f is Borel measurable.

Corollary 29.12.7 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 Lebesgue measurable. Thenx→|det(Dh(x))| f (h(x)) is Lebesgue measurable and∫

h(Ω)f (y)dy =

∫Ω

|det(Dh(x))| f (h(x))dx (29.12.78)

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

∫Ω

det(Dh(x)) f (h(x))dx (29.12.79)

1058 CHAPTER 29. THE AREA FORMULASince y ¢ h(Uj,;U;) the second term on the left is 0 and so d (y,U;,h) =d (y,Q,h) . There-fore, by Corollary 29.12.5,[gf eoma=¥ | ro)a.2m\a= X | uy f (y)d(y,Uj,h) dy = L / Xy; (x) f (h(x))det(Dh(x))dx (29.12.76)From 29.12.75y/ Xu, (x) f (n(x) [det (Dh (x))| dxi=]~ Tiny (x)) |det (Dh (x))| dx_ | FoF tht )) |det (Dh (x)) dx < 00and so by Fubini’s theorem, the sum and the integral may be interchanged in 29.12.76 toobtain from the dominated convergence theorem,[Yeuvor (x)) det (Dh (x)) dx= [raw )) det (Dh (x)) dxwhich shows/ f (y)d(y,Q,h) dy = [rence )) det (Dh (x)) dx (29.12.77)h(Q)Now if f is Borel measurable and in L' (Q) , the above may be applied to the positiveparts of the real and imaginary parts of f to obtain 29.12.77 for such f. This proves thetheorem.Not surprisingly, it is not necessary to assume f is Borel measurable.Corollary 29.12.7 Let h be one to one on Q and in W'? (IR";R"),p >n. Let Q be abounded, open set in IR". Let f € L'(h(Q)) where f is Lebesgue measurable. Thenx — |det (Dh (x))| f (h(x)) is Lebesgue measurable and| fly)dy = | |det (Dh (x))| f (hh (x)) dx (29.12.78)n(Q) QIt can also be written in the form/ f (yd (y,Q,h)dy = I det (Dh (x)) f (h(x)) dx (29.12.79)