FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities". I like that characterization b/c it seems more precise than "almost everywhere", but I've heard both.

I mention that because when I read the first footnote, I thought this was a mistake:

> boundedness alone ensures the subinterval infima and suprema are finite.

But it wasn't. It does, in fact, insure that infima and suprema are finite. It just does NOT ensure that it is Riemann integrable (which, of course the last paragraph in the first section mentions).

Thanks for posting. This was a fun diversion down memory lane whilst having my morning coffee.

If anyone wants a rabbit hole to go down:

Think about why the Dirichlet function [1], which is bounded -- and therefore has upper and lower sums -- is not Riemann integrable (hint: its upper and lower sums don't converge. why?)

Then, if you want to keep going down the rabbit hole, learn how you _can_ integrate it (ie: how you _can_ assign a number to the area it bounds) [2]

[1] One of my favorite functions. It seems its purpose in life is to serve as a counter example. https://en.wikipedia.org/wiki/Dirichlet_function

[2] https://en.wikipedia.org/wiki/Lebesgue_integral

I feel similar about the trace of a matrix being equal to the sum of eigenvalues.

Probably this means I should sit with it more until it is obvious, but I also kind of like this feeling.