We know that given a $T$-periodic function $f(x)$ on $\mathbb{R}$,
$$\int_{[s_0,s_0+T]}f(x)dx=\int_{[s_1,s_1+T]}f(x)dx$$
for any $s_0$ and $s_1$. This can be proved easily by change of variable. It is also true if the domain is 'splitted' properly. For example,
$$\int_{[s_0,s_0+T]}f(x)dx=\int_{[s_0,t_0]\cup[t_0+T,s_0+2T]}f(x)dx$$
where $s_0<t_0<s_0+T$. For this specific and simple domain, it can be proved easily as well by shifting by $T$ the domain pieces on the right-hand side and merging the pieces into a whole one same as that on the left-hand side. We can conclude informally that the integrals are equal if one domain can be transformed by shifting and merging to the other one.
My question, for a general measure space $(X, \mu)$ and $T$-periodic function $f(x)$, is there any similar conditions on the domains $\Omega_0$ and $\Omega_1$ such that
$$\int_{\Omega_0}f(x)dx=\int_{\Omega_1}f(x)dx.$$
My conjecture is that one condition is $\mu(\Omega_0)=\mu(\Omega_1)$ so that the new merged domain from one is of the same size as the other one. This is of course not sufficient. We need to characterize the shifting property which I have no idea about.