I am really confused about discontinuous points and how it affects the integrability. We proved in the lecture that if f is bounded and contunuous except for finite numbers of discontinuous points at [a,b] then f is integrable at [a,b]. The lecturer proved this theorem with induction on the discontinuous points which troubles me a lot. I either don't get it because I am looking at it at the wrong way. When he does induction he says n=0, in other words when we dont have discontinuous points. then he assumes for n and proves for n+1. However when you use induction it's for every n that is natural number so this proof implies that if the function had infinite number of discontinuous points it would still be true since it's for every n? What am I not understanding here?
Or does he mean, for any given n, he proves it and he says that had the function had n+1 number of discontinuous points it would still be true?
And what is the reason that if the function has infinite amount of number discontinuous points it's not integrable? As I understand , it would be a problem with the partition. In other words if the function has infinite amount of number discontinuous points then no matter what partition we choose there at least exists two consecutive points in the partition that has discontinuous points. Is my logic correct?