A Primer of Lebesgue Integration by H. S. Bear

By H. S. Bear

The Lebesgue indispensable is now usual for either purposes and complex arithmetic. This books starts off with a overview of the normal calculus imperative after which constructs the Lebesgue necessary from the floor up utilizing a similar rules. A Primer of Lebesgue Integration has been used effectively either within the lecture room and for person study.

Bear offers a transparent and straightforward creation for these reason on extra learn in larger arithmetic. also, this e-book serves as a refresher supplying new perception for these within the box. the writer writes with a fascinating, common-sense sort that appeals to readers in any respect degrees.

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A Lebesgue integrable function is always absolutely integrable in the sense that if f is measurable, then f is integrable if and only \i\f\ is integrable. We use the same upper sum-lower sum approach to define /5 f for f unbounded or S of infinite measure, only now we allow countable partitions. We saw earlier (Problem 2, Chapter 5) that countable partitions make no changes for bounded functions and sets of finite measure, so none of our earlier results change. As before, integrability will be equivalent to measurability, provided both the positive and negative areas are finite.

Proof Let Fn = {x e S : \ fk(x) - f(x)\ > £ for some k > n}. The sets P„ are measurable, and decreasing, and f]Fn = 0 because fn{x) —> f{x) for all x. Since /x(Fi) < oc, lim/x(f J = 0. Let /x(F]v) < 8. For x e S - FN? I fki^) - f{x)\ < ^ foi" all k>N. Illll Problem 5. e. the same result holds. ""HI The next proposition gives the form in which it is easiest to remember and apply the above result. Proposition 7. (Egoroffs Theorem) If {fn} is a sequence of measurable functions on a finite measure set Sy and fn —> f pointwise on S, then for every 8 > 0 there is a measurable set E c S of measure less than 8 so that fn —> f uniformly on S — E.

6 PROPERTIES OF THE INTEGRAL 59 Proof. Let s > 0. Let £ be a measurable set of measure less than s such that fn —> f uniformly off E. Let \fn(x)\ < M for all n and all x e S. Then \l(fn-n\ f uniformly on S—£, there is Nso that \fn— f\ < £ on S — E a n > N. Hence if « > N, /. s (fn- f) < siJi(S - E) + IMs = s[fi(S-E)+2M] f pointwise with all fn remaining in some fixed finite area.

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