Real and Complex Analysis (Rudin)

DIFFERENTIATION 139
PROOF Define
for x E R\ r > 0, and put
('1; fXx) = m(~)
r If - f(x) I dm
r J8(%. r)
(Tf)(x) = lim sup (T,. f)(x). (2)
r-O
We have to prove that Tf = 0 a.e. Em].
Pick y > O. Let n be a positive integer. By Theorem 3.14, there exists
g E C(Rk) so that IIf - gill < lin: Put h =f - g.
Since g is continuous, Tg = O. Since
we have
(T,. h)(x) ::; m(1 ) r I h I dm + I h(x) I
Br J8(%.r)
Since T,. f::; T,. g + T,. h, it follows that
Therefore
(1)
(3)
Th::; Mh + Ihl. (4)
Tf::; Mh + Ihl. (5)
{Tf> 2y} c {Mh > y} u {Ihl > y}. (6)
Denote the union on the right of (6) by E(y, n). Since Ilhlll < lin,
Theorem 7.4 and the inequality 7.5(1) show that
The left side of (6) is independent of n. Hence
m(E(y, n» ::; W + l)/(yn). (7)
ao
{Tf> 2y} c n E(y, n). (8)
n=l
This intersection has measure 0, by (7), so that {Tf> 2y} is a subset of a set
of measure O. Since Lebesgue measure is complete, {Tf> 2y} is Lebesgue
measurable, and has measure O. This holds for every positive y. Hence Tf = 0
a.e. Em].
IIII
Theorem 7.7 yields interesting information, with very little effort, about
topics such as
(a) differentiation of absolutely continuous measures,
(b) differentiation using sets other than balls,
(c) differentiation of indefinite integrals in RI,
(d) metric density of measurable sets.