Real and Complex Analysis (Rudin)

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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» IIII 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}. IIII (2)
138 REAL AND COMPLEX ANALYSIS 7.S Weak I! Iff E I!(Rk) and A > 0, then (1) because, putting E = {I f I > A}, we have Am(E) ~ [If I dm ~ [ If I dm = IIfIIl' JE JRk Accordingly, any measurable functionffor which (2) A . m{ I f I > A} (3) is a bounded function of A on (0, 00) is said to belong to weak I!. Thus weak I! contains I!. That it is actually larger is shown most simply by the function l/x on (0, 1). We associate to each fE I!(Rk) its maximal function Mf: Rk~ [0,00], by setting (Mf)(x) = sup _(1) [ I f I dm. O
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CONTENTS Preface xiii Prologue: The
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BIBLIOGRAPHY 1. L. V. Ahlfors: "Com
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LIST OF SPECIAL SYMBOLS AND ABBREVI
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INDEX Absolute continuity, 120 of f
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INDEX 411 Double integral, 165 Dual
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INDEX 415 Separable space, 92, 247
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