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1996 | OriginalPaper | Buchkapitel

Measures

verfasst von : Andrew Browder

Erschienen in: Mathematical Analysis

Verlag: Springer New York

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In Chapter 5 we defined the Riemann integral of a real function f over a bounded interval [a, b] by $$\int_a^b {f\left( x \right)} dx = \lim \sum\limits_{j = 1}^n {f\left( {\xi _j } \right)} \left( {x_j - x_{j - 1} } \right),$$ where xj-1 ≤ ξj ≤ xj for each j, and the limit is taken over increasingly fine partitions a = x0 < x1 < … < xn= b of the interval. We found that this limit existed whenever f was continuous on [a,b], in fact, whenever f was bounded, with a set of discontinuities D which was “small,” in the sense that for any ε > 0, there existed a finite collection of open intervals $$\left\{ {\left( {a_k ,b_k } \right):k = 1, \ldots r} \right\}$$ such that $$D \subset \mathop \cup \limits_{k = 1}^r \left( {a_k ,b_k } \right)and\sum\limits_{k = 1}^r {\left( {b_k - a_k } \right)} < \epsilon$$ This is a fairly rich class of functions, including as it does not only every continuous function, but also some functions which have infinitely many, even uncountably many, discontinuities (recall that the Cantor set is small in the above sense.) However, the class of Riemann integrable functions does have at least one glaring weakness: it is not stable under pointwise convergence. That is, if fn is Riemann integrable for each n, and if fn(x) → f(x) for every x, a ≤x ≤b, it is entirely possible that f is not Riemann integrable. (For instance, take a = 0 and b=1, and set fn(x) = 1 if x = m/n! for some integer m, and fn(x) = 0 otherwise. Then each fn is Riemann integrable, and fn converges pointwise to the function f, where f(x) = 1 if x is rational, and f(x) = 0 when x is irrational. We have seen that f is not Riemann integrable.

Metadaten
Titel
Measures
verfasst von
Andrew Browder
Copyright-Jahr
1996
Verlag
Springer New York
DOI
https://doi.org/10.1007/978-1-4612-0715-3_9