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?