1038 CHAPTER 29. THE AREA FORMULA
= lim infj→∞
∞
∑i=1
β (s)(
r(
B ji
))sH m
(f(
B ji
)).
Recall the equality of H m and Lebesgue measure on Rm, (Recall this was how β (m) waschosen. Theorem 28.2.4) Then the above is
≤ lim infj→∞
∞
∑i=1
β (s)(
r(
B ji
))smm
(f(
B ji
))
≤ lim infj→∞
∞
∑i=1
β (s)α (m)Lip(f)m r(
B ji
)m(r(
B ji
))s
= Lip(f)mβ (s)α (m) lim inf
j→∞
∞
∑i=1
r(
B ji
)m(r(
B ji
))s
= Lip(f)m β (s)α (m)
β (m+ s)lim inf
j→∞
∞
∑i=1
β (m+ s)r(
B ji
)m+s
≤ Lip(f)m β (s)α (m)
β (m+ s)H s+m (A)
from 29.9.46. However, it was shown earlier that α (m) = β (m).This last identification that α (m) = β (m) depended on the technical material involving
isodiametric inequality. It isn’t all that important to know the exact value of this constantβ (s)α(m)β (m+s) so one could simply write C (s,m) in its place and suffer no lack of utility.
Can one change∫ ∗ to
∫? The next lemma will enable the change in notation.
Lemma 29.9.5 Let A⊆ Rn be Lebesgue measurable and
f : Rn→ Rm
be Lipschitz, m < n. Theny→H n−m (A∩ f−1 (y)
)is Lebesgue measurable. If A is compact, this function is Borel measurable.
Proof: Suppose first that A is compact. Then A∩ f−1 (y) is also and so it is H n−m
measurable. Suppose H n−m(A∩ f−1 (y)
)< t. Then for all δ > 0,
H n−mδ
(A∩ f−1 (y)
)< t
and so there exist sets Si, satisfying
r (Si)< δ , A∩ f−1 (y)⊆ ∪∞i=1Si,
∞
∑i=1
β (n−m)(r (Si))n−m < t
Replacing Si with the open set Ŝi ≡ Si +B(0,η i) where the Ŝi satisfy the above inequality,it can be assumed each Si is open.
Claim: If z is close enough to y, then A∩ f−1 (z)⊆ ∪∞i=1Si.