Lebesgue PointsTheorem 6.10 (Lebesgue’s Differentiation Theorem). Let µ be a Radon measureon Rn. For f ∈ L1loc(µ),limr→01µ(Br(x)) ZBr(x)f dµ = f(x),for µ-a.e. x.This result is due to Lebesgue for µ = L1 and the general case is due to Besicovitch.It is also called Lebesgue-Besicovitch differentiation theorem.Proof. Clearly it is sufficient to prove the theorem by assuming f ≥ 0 and f ∈ L1(Rn).Define a positive linear functional Λ beΛϕ =Zϕf dµ, ϕ ∈ Cc(Rn) .By Riesz representation theorem, there is a Radon measure ν satisfyingZϕdν =Zϕf dµ .By a routine argument,ν(E) = ZEf dµ, E µ-measurable .By Theorem 6.6,limr→01µ(Br(x)) ZBr(x)f(y) dµ(y) = Dµν(x),for µ-a.e. x. As ν << µ, by Theorem 6.8,ν(E) = ZEf dµ =ZEDµν dµ, E µ-measurable ,hence f = Dµν µ-a.e.. Combining these we get the theorem.The corollary gives a stronger result.14 CHAPTER 6. DIFFERENTIATION OF MEASURES AND FUNCTIONSCorollary 6.11. Let f ∈ L1loc(µ) where µ is a Radon measure. There exists a set L offull measure such that for every x ∈ L, f(x) is finite andlimr→01µ(Br(x)) ZBr(x)|f − f(x)| dµ = 0.Proof. Order all rational numbers in a sequence {qj} and apply Theorem 6.10 to eachgj = |f − qj| to get1µ(Br(x)) ZBr(x)|f(y) − qj| dµ(y) → |f(x) − qj| , ∀x /∈ Nj, (6.1)as r → 0 where µ(Nj ) = 0. Letting N =[∞j=1Nj, µ(N) = 0 and (6.1) holds for all x /∈ N.Given ε > 0 and x /∈ N such that f(x) is finite, pick qj such that |f(x) − qj| < ε. Wehave1µ(Br(x)) ZBr(x)|f(y) − f(x)| dµ(y) ≤1µ(Br(x)) ZBr(x)|f(y) − qj| dµ(y)+1µ(Br(x)) ZBr(x)|qj − f(x)| dµ(y)=1µ(Br(x)) ZBr(x)|f(y) − qj| dµ(y) + |qj − f(x)| .By (6.1), for ε > 0, we can find some r0 such that for all r ∈ (0, r0),1µ(Br(x)) ZBr(x)|f(y) − qj| dµ(y) < ε.Therefore, we have1µ(Br(x)) ZBr(x)|f − f(x)| dµ(y) ≤ 2ε.A point x at which this corollary holds is called a Lebesgue point for the function fand the set of all Lebesgue points forms the Lebesgue set of f. It is implicitly assumedthat the function is finite at a Lebesgue point. Moreover, the Lebesgue point depends onthe pointwise definition of f and therefore is not a concept attached to f as an equivalenceclass. Every point of continuity of f is a Lebesgue point, but the converse may be nottrue. It is not hard to construct Lebesgue points at which the function is discontinuous.So far we have been considering taking the average of a function over balls. Now weconsider taking average over other sets. Let {Ej} be a sequence of µ-measurable sets. Wecall it shrinks regularly to x if there is Brj(x), rj → 0, such thatαµ(Ej ) ≥ µ(Brj(x)), Ej ⊂ Brj(x),

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Measures and Distribution Functions
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