I don't think the definition of $\mu$ is relevant to your problem, you only need that $\int\phi d\mu =0 $ if $\phi$ is $2\pi$-periodic.

The reason is that the measure
$$
d\nu(x) = \sum_{n\in\mathbb{Z}^2}d\mu(x-2\pi n)
$$
(which is the convolution of the indicator function of the lattice $\chi_{2\pi\mathbb{Z}^2}$ with $d\mu$) is defined to make integration of $\phi$ against $d\nu$ the same as integration of the periodization of the $\phi$ against $d\mu$:
$$
\int \phi(x)d\nu(x) =\int\phi(x)\left(\sum_{n\in\mathbb{Z}^2}d\mu(x-2\pi n)\right) = \int\left(\sum_{n\in\mathbb{Z}^2}\phi(y+2\pi n)\right)d\mu(y)
$$
The periodization
$$
\sum_{n\in\mathbb{Z}^2}\phi(y+2\pi n)
$$
is $2\pi$-periodic, so the second integral is always zero.

(Throughout, let's assume that $\phi$ has enough decay to make all the integrals converge. Also, we need to exchange the order of the integral and the sum to make the change of variables $x\to y+2\pi n$, then change the order back.)

other questionit seems that these questions relate to sciencedirect.com/science/article/pii/S0007449710000953 $\endgroup$