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.
Best Answer
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*}