What does it means by "integrable on the circle"

Question

As far that i have known, i understand the notion "a function on the circle" by each one of the followings (both equivalent):

1. A function is defined on $\mathbb{R}$ that is $2\pi-$periodic.
2. A function that is defined on $[a,b]$ with $b-a=2\pi$ and $f(a)=f(b).$ So we can extend this function to get a $2\pi-$periodic function.

And my problem is: i am confused every time the author (of the book i have been reading) uses the notion "integrable on the circle". So, by "a function that is integrable on the circle", do i have to get the meaning in which way:

1. A function that is integrable on all the interval of length $2\pi.$ Just like this http://math.uchicago.edu/~may/REU2017/REUPapers/Xue.pdf (page 1)
2. A function that is needed firstly be a function on the circle and then, it is integrable on some interval of length $2\pi$ (because the integral gets the same value over any interval of length $2\pi$) like this

I ask the question because of the passage 1 and the passage 2. So do i need a function getting the same value at the end-points of every interval of lenth $2\pi?$ By the way, the author approachs the integral in the Riemann sense, if it helps.

Answer 1

Actually the two conditions are equivalent (unless you require the functions to be continuous).

In the second case, you could equally well say that $f$ should be defined on $[a,b)$ to begin with, and forget about the requirement $f(a)=f(b)$, because it will be satisfied by definition when you extend the function to a $2\pi$-periodic function on $\mathbb{R}$. And then it should be clear that this is the same thing as the first case.

But to require a function to be “continuous on the circle” means that the $2\pi$-periodic extension must be continuous on $\mathbb{R}$. This is of course not necessarily true if you extend a function which is continuous on $[a,b)$. But if $f$ is continuous on $[a,b]$ and satisfies $f(a)=f(b)$, then the extension will be continuous (and conversely).

Anyway, the value at a single point doesn't affect integrals, and a function need not be continuous in order to be integrable or to have a Fourier series, so perhaps one shouldn't worry too much about continuity when only talking about what “a function on the circle” means. Of course, when studying convergence of the Fourier series, it's interesting to talk about continuity, but that's a later question.

Answer 2

The whole theory of Fourier series is built to study periodic functions. The correct interpretation of "integrable function on the circle" is 2.