Let $\mathbb T:=[0,1]/(0\sim 1)$. Its easy to see that if $f:\mathbb T\to \mathbb R$ and $k\in\mathbb Z$ is not zero, then $f_k(x):=f(kx)$ (1) defines a map $ f_k:\mathbb T\to \mathbb R$, and $$ \int_{\mathbb T} f_k(x) dx = \int_{\mathbb T} f(x)dx$$ Indeed: identifying with 1-periodic functions ...