Download e-book for iPad: A Short Course on the Lebesgue Integral and Measure Theory by Steve Cheng

By Steve Cheng

ISBN-10: 0201510359

ISBN-13: 9780201510355

This text develops the fundamentals of the Lebesgue necessary and degree conception. by way of content material, it provides not anything new to any of the present textbooks at the topic. yet our strategy right here could be to prevent unduly abstractness and absolute generality, as an alternative concentrating on generating proofs of necessary effects as fast as attainable. a lot of the fabric right here comes from lecture notes from a brief actual research path I had taken, and the remainder are famous effects whose proofs I had labored out myself with tricks from quite a few assets. I typed this up often for my very own gain, yet i'm hoping it will likely be attention-grabbing for an individual thinking about the Lebesgue necessary (or greater arithmetic in general). i'll be delivering proofs of each theorem. when you are bored studying them, you're invited to do your personal proofs. The bibliography outlines the history you must comprehend this text.

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

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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. The finite intersection belongs to A by the previous step. So Ac ∈ A. 4. Finally, given Ai = j Ri,j , for a finite number of i, let D0 = ∅, and Di = Di−1 (Ai \ Di−1 ) ∈ A. Then i Ai = i Di = i (Di \ Di−1 ) ∈ A. By the way, the sigma algebra generated by A will contain the Borel sigma algebra: every open set U ∈ Rn obviously can be written as a union of open rectangles, and in fact we can use a countable union of open rectangles.

The norm can be arbitrary, for in Rn , every norm is equivalent: if · 1 and · 2 are any two given norms, then there always exist constants α, β > 0 such that α · 2 ≤ · 1 ≤ β · 2 . Thus being integrable in one norm implies integran bility in another norm. In particular, by using the norm x Σ = k=1 |xk |, we k see that f is integrable if and only if its components f are integrable. We want to show that f ≤ f ; this basic inequality will enable us to make estimates without having to separate components.

E. f = j=1 aj χEj for aj ∈ Rn . It is also trivial if f = ∞. To prove the inequality for the other f , we use the following easy lemma. 1. Let f : X → Rn be integrable. There exists a sequence of simple functions ϕj : X → Rn converging pointwise to f , with lim j→∞ ϕj − f = 0 . Proof. By equivalence of norms, it suffices to prove this only for the norm · Σ as defined above. For each f k+ , by the approximation theorem in R, there exists measurable k− simple ϕk+ increasing to f k+ . Similarly for f k− .

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


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