482 CHAPTER 18. DIFFERENTIAL FORMS
18.4 Lipschitz MapsThis will be based on the approximation with C1 maps. Let r : [a,b]⊆Rk→Rp where r isLipschitz and k ≤ p as before. The idea is to extend the above Stokes theorem to this casewhere r is Lipschitz rather than C1. First extend r as follows. For t ≥ bl and ul ∈ [a,b]l ,let
r (u1, ...,ul−1, t,ul+1, ...,uk)≡ r (u1, ...,ul−1,bl ,ul+1, ...,uk)
and also if t ≤ al and ul ∈ [a,b]l , let
r (u1, ...,ul−1, t,ul+1, ...,uk)≡ r (u1, ...,ul−1,al ,ul+1, ...,uk)
This is done for each l. Then define for h > 0
rh (u)≡(
12h
)k ∫ u1+2h
(b1−a1)(u1−a1)
−2h+u1+2h
(b1−a1)(u1−a1)
· · ·∫ uk+
2h(bk−ak)
(uk−ak)
−2h+uk+2h
(bk−ak)(uk−ak)
r (t)dtk · · ·dt1 (18.7)
Consider those integrals. When u1 = a1 you are integrating over [a1−2h,a1] and whenu1 = b1 you are integrating over [b1,b1 +2h] . Of course it is similar for the other [al ,bl ] .In general, each iterated integral is taken over an interval of length 2h. For example,
u1 +2h
(b1−a1)(u1−a1)−
(−2h+u1 +
2h(b1−a1)
(u1−a1)
)= 2h
Now consider the right end of the lth interval in [a,b] where ul = bl . Then this is describ-ing one of the two faces corresponding to the lth interval. By Fubini’s theorem and theconstruction of the extention of r, 18.7 implies that this equation simplifies to
rh (ul (bl)) =
(1
2h
)k−1 ∫ u1+2h
(b1−a1)(u1−a1)
−2h+u1+2h
(b1−a1)(u1−a1)
· · ·∫ uk+
2h(bk−ak)
(uk−ak)
−2h+uk+2h
(bk−ak)(uk−ak)
·
r (t1, ...,bl , ...tk)dtk · · · d̂tl · · ·dt1
Now by the fundamental theorem of calculus, for i ̸= l,
rhui(ul (bl)) =
(1
2h
)k−2 ∫ u1+2h
(b1−a1)(u1−a1)
−2h+u1+2h
(b1−a1)(u1−a1)
· · ·∫ uk+
2h(bk−ak)
(uk−ak)
−2h+uk+2h
(bk−ak)(uk−ak)
·
r
(t1, ...,ui +
2h(bi−ai)
(ui−a1) , ...bl , ...tk
)1
2h
(1+
2hbi−ai
)−
r
(t1, ...,−2h+ui +
2h(bi−ai)
(ui−ai) , ...bl , ...tk
)12h
(1+
2hbi−ai
)·
dtk · · · d̂ti, · · · , d̂tl · · ·dt1
By the material on Rademacher’s theorem, that integrand is(1+
2hbi−ai
)1
2h
∫ ui+2h
(bi−ai)(ui−ai)
−2h+ui+2h
(bi−ai)(ui−ai)
rui (t1, ..., ti, ...bl , ...tk)dti