35.6. CHANGE OF VARIABLES FORMULA LIPSCHITZ MAPS 1251

What if F is only Lebesgue measurable? Note there are no measurability problems with theabove expression because x→XF (h(x)) is Borel measurable due to the assumption that his continuous while J is given to be Lebesgue measurable. However, if F is Lebesgue mea-surable, not necessarily Borel measurable, then it is no longer clear that x→XF (h(x)) ismeasurable. In fact this is not always even true. However, x→XF (h(x))J (x) is measur-able and 35.6.24 holds.

Let F be Lebesgue measurable. Then by inner regularity, F = H ∪N where N hasmeasure zero, H is the countable union of compact sets so it is a Borel set, and H ∩N = /0.Therefore, letting N′ denote a Borel set of measure zero which contains N,

b(x)≡XH (h(x))J (x)≤XF (h(x))J (x)

= XH (h(x))J (x)+XN (h(x))J (x)≤ XH (h(x))J (x)+XN′ (h(x))J (x)≡ u(x)

Now since N′ is Borel,∫Ω

(u(x)−b(x))dmn =∫

XN′ (h(x))J (x)dmn

= mn(h(h−1 (N′)∩Ω

))= mn

(N′∩h(Ω)

)= 0

and this shows XH (h(x))J (x) = XF (h(x))J (x) except on a set of measure zero. Bycompleteness of Lebesgue measure, it follows x→XF (h(x))J (x) is Lebesgue measurableand also since h maps sets of measure zero to sets of measure zero,∫

XF (h(x))J (x)dmn =∫

XH (h(x))J (x)dmn

=∫

h(Ω)XH (y)dmn

=∫

h(Ω)XF (y)dmn.

It follows that if s is any nonnegative Lebesgue measurable simple function,∫Ω

s(h(x))J (x)dmn =∫

h(Ω)s(y)dmn (35.6.25)

and now, if f ≥ 0 is Lebesgue measurable, let sk be an increasing sequence of Lebesguemeasurable simple functions converging pointwise to f . Then since 35.6.25 holds for sk,the monotone convergence theorem applies and yields 35.6.23. This proves the theorem.

It turns out that a Lipschitz function defined on some subset of Rn always has a Lips-chitz extension to all of Rn. The next theorem gives a proof of this. For more on this sortof theorem see Federer [50]. He gives a better but harder theorem than what follows.

Theorem 35.6.13 If h : Ω→Rm is Lipschitz, then there exists h :Rn→Rm which extendsh and is also Lipschitz.

35.6. CHANGE OF VARIABLES FORMULA LIPSCHITZ MAPS 1251What if F is only Lebesgue measurable? Note there are no measurability problems with theabove expression because x > 2% (h(x)) is Borel measurable due to the assumption that his continuous while J is given to be Lebesgue measurable. However, if F is Lebesgue mea-surable, not necessarily Borel measurable, then it is no longer clear that x + 27 (h(x)) ismeasurable. In fact this is not always even true. However, x +2 (h(x))J (x) is measur-able and 35.6.24 holds.Let F be Lebesgue measurable. Then by inner regularity, F = H UN where N hasmeasure zero, H is the countable union of compact sets so it is a Borel set, and HNN =9.Therefore, letting N’ denote a Borel set of measure zero which contains N,b(x) = 2y(h(x))J (x) < % (h(x)) J (x)= 2Xy(h(x))J(x) + 2y (a(x)) J (x)< 2y(h(x))J (x) + Zw (h(x) J (x) = u(x)Now since WN’ is Borel,I (u(x) —b(x)) drm = I Zw (n(x) J (x) dm,=m, (h(h7! (N’) NQ)) =m, (N’Nb(Q)) =0and this shows 2% (h(x))J (x) = 2 (h(x))J (x) except on a set of measure zero. Bycompleteness of Lebesgue measure, it follows x > 2 (h(x)) J (x) is Lebesgue measurableand also since h maps sets of measure zero to sets of measure zero,| Xe (h(x))J(x)dm, = | Xy (h(x) J (x) dinQ Q| Xu (y)dmh(Q)| Xp (y)dmy.h(Q)It follows that if s is any nonnegative Lebesgue measurable simple function,| s(n(x)) J (x) dm = / s(y) dim, (35.6.25)Q h(Q)and now, if f > 0 is Lebesgue measurable, let s; be an increasing sequence of Lebesguemeasurable simple functions converging pointwise to f. Then since 35.6.25 holds for sx,the monotone convergence theorem applies and yields 35.6.23. This proves the theorem.It turns out that a Lipschitz function defined on some subset of IR” always has a Lips-chitz extension to all of R”. The next theorem gives a proof of this. For more on this sortof theorem see Federer [50]. He gives a better but harder theorem than what follows.Theorem 35.6.13 [fh : Q — R” is Lipschitz, then there exists h : R" + R™ which extendsh and is also Lipschitz.