# A question about the definition of partition in Riemann Integral

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?

• Yes, he means finite partitions. "Finitely many partitions" is different than "all finite partitions." There are infinitely many "finite partitions." – Thomas Andrews Jul 14 '13 at 23:14
• Yes, you are right. I meant to say "finite partitions". – David Tan Jul 14 '13 at 23:31

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.)