Real and Complex Analysis (Rudin)

DIFFERENTIATION, 153
Since r" = m(B(O, r»/m(B(O, 1», (9) implies that
lim m(T(B(O, r») = O.
,-0 m(B(O, r»
This proves (1), since L\(T'(O» = L\(A) = O.
(10)
IIII
7.25 Lemma Suppose E c Rk, m(E) = 0, T maps E into R\ and
. I T(y) - T(x) I
1m sup I I < 00
for every x E E, as y tends to x within E.
Then m(T(E» = O.
1
y-x
PROOF Fix positive integers nand p, let F = F n, p be the set of all x E E such
that
I T(y) - T(x) I :S; nly - xl
for all y E B(x, lip) () E, and choose € > O. Since m(F) = 0, F can be covered
by balls Bi = B(Xi' ri), where Xi E F, ri < lip, in such a way that L m(BJ < €.
(To do this, cover F by an open set W of small measure, decompose W into
disjoint boxes of small diameter, as in Sec. 2.19, and cover each of those that
intersect F by a ball whose center lies in the box and in F.)
If x E F () Bi then I Xi - x I < ri < lip and Xi E F. Hence
I T(xi) -
so that T(F () Bi) c B(T(x;), nr;). Therefore
The measure of this union is at most
T(x) I :S; nlxi - xl < nri
T(F) c U B(T(xi), nr;).
i
L m(B(T(x;), nri) = nk L m(B;) < nk€.
i
i
Since Lebesgue measure is complete and € was arbitrary, it follows that T(F)
is measurable and m(T(F)) = O.
To complete the proof, note that E is the union of the countable collection
{Fn,p}'
IIII
Here is a special case of the lemma:
If V is open in Rk and T: V -4 Rk is differentiable at every point of V, then T
maps sets of measure 0 to sets of measure O.
We now come to the change-of-variables theorem.
7.26 Theorem Suppose that
(i) X eVe Rk, V is open, T: V -4 Rk is continuous;