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 =∫
Ω
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,∫
Ω
XF (h(x))J (x)dmn =∫
Ω
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.