I WTS that if $f$ is periodic and Riemann integrable on $[-\pi, \pi]$, then $f$ is integrable on any closed interval and for every real number $x,$ we have $\int_{x-\pi}^{x+\pi}f = \int_{-\pi}^{\pi}f$.
I know that the quickest proof of this comes from the change-of-variable formula.
But since I haven't seen the proof for change-of-variables without the assumption that $f$ is continuous, I want to prove the above statement without this formula.
The hint I have is to use the sequential criterion for integrability, namely: $f$ is integrable on $[a,b]$ if and only if there is a sequence of partitions $(P_n)$ of $[a,b]$ such that $\lim U(f, P_n)-L(f, P_n) =0$, in which case $\lim U(f, P_n)=\lim L(f, P_n)=\int_{a}^{b}f$.
I don't see how to apply this, however.
I'd appreciate any help.
Assuming $f$ is $2\pi$-periodic, it should be clear $f$ is Riemann integrable (RI) on every $[\pi,3\pi], [3\pi,5\pi],\dots$ because the upper minus lower sums on each such interval correspond exactly to those on $[-\pi,\pi],$ where we know $f$ is RI. Same thing for $[-3\pi,-\pi], [-5\pi,-3\pi], \dots$
Recall two easily proved results: i) if $f$ is RI on both $[a,b]$ and $[b,c],$ then $f$ is RI on $[a,c];$ ii) if $f$ is RI on $[a,b]$ and $a\le a'<b'\le b,$ then $f$ is RI on $[a',b'].$ From this and the above it follows that $f$ is RI on any closed bounded interval.