Real and Complex Analysis (Rudin)

DIFFERENTIATION 137
7.3 Lemma If W is the union of a finite collection of balls B(Xi' ri), 1 ~ i ~ N,
then there is a set S c {I, ... , N} so that
(a) the balls B(Xi' ri) with i E S are disjoint,
(b) We U B(Xi' 3ri), and
i eS
(c) m(W) ~ 3 k L m(B(xi' ri»·
ieS
PROOF Order the balls Bi = B(x;, ri) so that r l ~ r2 ~ ... ~ r N • Put i l = 1.
Discard all B j that intersect Bit. Let Bi2 be the first of the remaining B i , if
there are any. Discard all B j with j > i2 that intersect Bi2 , let Bi3 be the first
of the remaining ones, and so on, as long as possible. This process stops after
a finite number of steps and gives S = {ib i2, ... }.
It is clear that (a) holds. Every discarded B j is a subset of B(x;, 3ri) for
some i E S, for if r' ~ rand B(x', r') intersects B(x, r), then B(x', r') c B(x, 3r).
This proves (b), and (c) follows from (b) because
m(B(x, 3r» = 3 k m(B(x, r»
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The following theorem says, roughly speaking, that the maximal function of
a measure cannot be large on a large set.
7.4 Theorem If Jl. is a complex Borel measure on Rk and A is a positive number,
then
(1)
Here 1IJl.11 = I Jl.1 (Rk), and the left side of (1) is an abbreviation for the
more cumbersome expression
We shall often simplify notation in this way.
PROOF Fix Jl. and A. Let K be a compact subset of the open set {MJl. > A}.
Each x E K is the center of an open ball B for which
I Jl.1 (B) > Am(B).
Some finite collection of these B's covers K, and Lemma 7.3 gives us a disjoint
subcollection, say {B b ... , B.}, that satisfies
• •
m(K) ~ 3 k L m(Bi) ~ 3 k ;,-l L 1Jl.I(BJ ~ 3 k ;'-lllJl.lI.
1 1
The disjointness of {B b ... , B.} was used in the last inequality.
Now (1) follows by taking the supremum over all compact
K c {MJl. > A}.
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(2)