A Short Course on the Lebesgue Integral and Measure Theory by Steve Cheng PDF

By Steve Cheng

ISBN-10: 0201510359

ISBN-13: 9780201510355

This text develops the fundamentals of the Lebesgue necessary and degree thought. when it comes to content material, it provides not anything new to any of the present textbooks at the topic. yet our method right here could be to prevent unduly abstractness and absolute generality, in its place targeting generating proofs of worthy effects as speedy as attainable. a lot of the fabric the following comes from lecture notes from a brief actual research direction I had taken, and the remaining are recognized effects whose proofs I had labored out myself with tricks from a number of resources. I typed this up as a rule for my very own profit, yet i'm hoping will probably be fascinating for a person considering the Lebesgue critical (or larger arithmetic in general). i'll be supplying proofs of each theorem. when you are bored analyzing them, you're invited to do your personal proofs. The bibliography outlines the heritage you must comprehend this text.

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Extra info for A Short Course on the Lebesgue Integral and Measure Theory

Sample text

The following theorem gives a classic example (and should also serve to convince you why our strenuous efforts are necessary). 3 (Vitali). There exists a non-measurable set in [0, 1] using Lebesgue measure. In other words, Lebesgue measure cannot be defined consistently for all subsets of [0, 1]. Proof. The key fact in this proof is translation-invariance. In particular, given any measurable H ⊆ [0, 1], define its “shift with wrap-around”: H ⊕ x = {h + x : h ∈ H, h + x ≤ 1} ∪ {h + x − 1 : h ∈ H, h + x > 1} .

It is easy to see, although tiresome to write down formally, that the collection of all rectangles is indeed a semi-algebra. 1. The set A of all finite disjoint unions of elements of a semialgebra R is an algebra on X. Proof. We check the properties for an algebra: 1. The empty set is trivially in A. 2. If A = i Ri , and B = j Sj , where Ri and Si denote a finite number of sets chosen from R, then A ∩ B = i Ri ∩ j Sj = i,j Ri ∩ Sj ∈ A. 3. If A = i Ri , then Ac = i Ric = i j Si,j for some Si,j ∈ R.

Similarly, we have µ∗ (B c ) ≥ ν(B c ), so that µ∗ (X) − µ∗ (B) ≥ ν(X) − ν(B) = µ∗ (X) − ν(B), or µ∗ (B) ≤ ν(B). ) But the hypothesis that µ(X) < ∞ is clearly too restrictive. 1. A measure space (X, B, µ) is sigma-finite, if there are measurable sets X1 , X2 , . . ⊆ X, such that n Xn = X and µ(Xn ) < ∞ for all n. 6. 5 holds also in the case that (X, B, µ) is sigma-finite. Proof. Let {Xn } X, µ(Xn ) < ∞ as in the definition of sigma-finiteness. 5 says that µ∗ (B ∩ Xn ) = ν(B ∩ Xn ). Taking limits as n → ∞ gives µ∗ (B) = ν(B).

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A Short Course on the Lebesgue Integral and Measure Theory by Steve Cheng

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