How do I prove this statement:
If $ f :[a,b]\to\mathbb{R} $ is a periodic and bounded on $ [a,b] $, then both $ f $ and $f^2$ are integrable on $ [a,b] $.
I have not found ideas or literature to prove the above statement. Thank you for your kindness.
Good ol' rational indicator function
$$\mathbf{1}_{\mathbb{Q}} = \begin{cases} 1 & x\in\mathbb{Q} \\ 0 & x\in\mathbb{R}\text{\\}\mathbb{Q} \\ \end{cases}$$
comes to the rescue as a counterexample. The problem with that statement is that we need some condition on local continuity.
One might grumble, "Well this function does not have a fundamental period, perhaps imposing that condition might be enough." But consider
$$f(x) = \begin{cases} \sin\left(\frac{\pi}{x-2k}\right) & x\in[2k-1,2k+1)\text{\\}\{2k\} \\ 0 & x=2k \\ \end{cases} $$
for $k\in\mathbb{Z}$. This function is bounded and has a fundamental period of $2$, yet is not Riemann integrable on any interval that contains a whole period.
I suspect that the correct additional constraint needed is that $f$ must be of bounded variation, but that then that has nothing to do with $f$ having any sort of periodicity. It is quite a mystery what your adviser might have meant by asking you to prove this.