Let $f\colon\mathbb{R}\to\mathbb{R}$ 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.
$\large\text{My attempt}$
Let $k\in\mathbb{Z}$ be an integer such that $$a-\pi \le 2\pi k< a+\pi,$$ then
\begin{align} \Large\int_{I_a}f\;d\lambda=\int_{[a-\pi, a+\pi]} f\;d\lambda= \Large &\int_{[a-\pi,2\pi k]}f\;d\lambda +\int_{[2\pi k,a+\pi]}f\;d\lambda\\ \Large\color{blue}{=}&\int_{[a+\pi, 2\pi k+2\pi]} f\;d\lambda+\int_{[2\pi k, a+\pi]}f\;d\lambda\\ \Large=&\int_{[2\pi k, 2\pi k+2\pi]}f\;d\lambda\\ =&\int_{[0.2\pi]}f\;d\lambda \end{align}
I can't prove equality in blue, ie that $$\Large \int_{[a-\pi,2\pi k]}f\;d\lambda=\int_{[a+\pi,2\pi k+2\pi]}f\;d\lambda$$
Question. The fact that $f$ is $2\pi-$periodic comes into play in this equality, but what role does the invariance under translations of the Lebesgue measure play?
If $f=1$ the equality is evident.
But already if $f$ is a simple function the only thing I can say is that the two intervals have the same Lebesgue measure, but how can this help us to conclude that the integrals are equal?
