Let $f\colon\mathbb{R}\to\mathbb{C}$ be $2\pi$-periodic function. For all $a\in\mathbb{R}$ we consider the interval $I_a=[a-\pi, a+\pi]$. If $f\in L^1(I_0)$, then $$\int_{I_0}f\;d\lambda=\int_{I_a}\;f\;d\lambda,$$ where $\lambda$ is the Lebesgue measure.
I know the this result follows from the fact that the function $f$ is $2\pi$- periodic and traslation invariance of Lebesgue measure, but I don't know how to prove it explicitly.
Could anyone give me a suggestion? Thanks!
Hint: Write $I_a$ as the (disjoint) union $[a - \pi, (2k+1)\pi]\cup [(2k+1)\pi, a + \pi]$ for some $k\in\mathbb Z$. Can you now separately "shift" these two parts onto (disjoint) subsets of $I_0$ using the periodicity of $f$?