Integral of a $2\pi$-periodic function over a $2\pi$-length interval Suppose $f:\mathbb{R}\to \mathbb{R}$ is a $2\pi$-periodic function such that $f$ is Lebesgue integrable over the fundamental interval $[-\pi,\pi]$. 
I have to prove that for every $s\in \mathbb{R}$ $f$ is Lebesgue integrable over $[s-\pi,s+\pi]$ and 

$$\int_{s-\pi}^{s+\pi}f(x)dx=\int_{-\pi}^{\pi}f(x)dx.$$

I tried writing $s=\overline{x}+2\overline{k}\pi$, with $x\in[-\pi,\pi)$ and $\overline{k}\in \mathbb{Z}$, and then using change of variable formula for $C^1$-diffeomorphism, combined with the fact that $f(x+2k\pi)=f(x)$ for every $k\in \mathbb{Z}$, but still seems not working at all. Please note that I'm not referring to Riemann integral.
Any hint would be really appreciated.

What you are missing is the identity
$$\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx \quad\text{if $a \le b \le c$}
$$
I would suggest choosing $k$ to be the unique integer such that
$$s-\pi \le 2 \pi k < s+\pi
$$
and then
\begin{align*}
\int_{s-\pi}^{s+\pi} f(x) dx & = \int_{s-\pi}^{2\pi k} f(x) dx + \int_{2 \pi k}^{s+\pi} f(x) dx \\
&= \int_{s+\pi}^{2\pi k + 2 \pi} f(x) \, dx + \int_{2 \pi k}^{s+\pi} f(x) dx\\
&= \int_{2 \pi k}^{2 \pi k + 2 \pi} f(x) dx \\
&= \int_0^{2\pi} f(x) dx
\end{align*}
