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I am reading Baby Rudin's definition on Riemann Integral, where partition $P$ is restricted to a finite set of points. When Rudin says "where the inf and the sup are taken over all partitions $P$ of $[a,b]$", I'm assuming he is meant to say over all finite partitions.

This is somewhat different from what I was taught, where $P$ essentially consists of countably many "anchor" points. For a function which has countably many discontinuities (e.g., a monotone function), I know it is Riemann integrable because the measure is 0, but does it fail if only finite partitions are allowed?

I've seen a somewhat related discussion: Proof that a function with a countable set of discontinuities is Riemann integrable without the notion of measure

So the compactness of [a,b] actually implies that a finite partition is sufficient?

Community
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David Tan
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3 Answers3

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In the theory of the Riemann integral on an interval $[a,b]$, it is completely standard that "partitions" of $[a,b]$ are necessarily finite. This is what Riemann did for his Riemann sums. What Rudin gives is not Riemann's approach but a slightly simpler one of G. Darboux, which uses upper and lower sums and upper and lower integrals: Darboux used finite partitions too. (Perhaps following Rudin, many people do not distinguish between the integrals of Riemann and Darboux: although they are defined differently, they can be shown to yield the same linear functional, in particular with the same domain of integrable functions: those which are bounded and with a zero measure set of discontinuities.)

Thus your question "Are finite partitions sufficient?" is a little strange from the standard perspective: sufficient for what? All the usual theorems and proofs use finite partitions.

In fact I own or have flipped through at least a dozen texts treating the Riemann integral, and to the best of my recollection I have never seen a "proper" Riemann integral using countably infinite partitions. Could you include in your question the precise definition you learned? Is there any textbook which uses this definition? (Are there any advantages to doing so?)

Pete L. Clark
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The last chapter of Baby Rudin, third edition, proves that a function is Riemann-integrable if and only if the measure of the set of its points of discontinuity is 0. He does that using the definition given in his book. So it's not be necessary to consider countably infinite partitions for that.

Michael Hardy
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First, the notion of "countable partitions" is actually somewhat tricky to define.

Once you've defined it, show that for any such countable partition, you can find a sequence of finite partitions that approaches this one in the limit in terms of upper and lower Riemann sums.

dfeuer
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Thomas Andrews
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