Actually it is quicker to sketch the proof than checking a reference. 

Assume  $u:\mathbb{R}^n\to \mathbb{R}$ is measurable and periodic wrto $x_i$ with period $b_i - a_i$, for $1\le i\le n$. Then by a linear change of variables $\|u\| _ {p,M}=\|u _ \nu \| _ {p, M} $ for $1\le p\le\infty$. So if $u\in L^p_{loc}$ the sequence $\{u _ \nu \} _ {\nu\in\mathbb{N} } $ is bounded in  $L^p(M)$. As a general fact, when checking the weak (or weak*) convergence of a bounded sequence in a Banach space $E$ (resp., in its dual), a norm-dense set of test element of $E^*$ (resp. $E$) is sufficient. Here the thesis easily follows using as test functions 
continuous functions on $M$ , for which it holds
$$\Bigg | \int_M u _ \nu(x) \phi(x)dx  -  \int_M \tilde u \phi(x)dx \Bigg | \leq \|u\|_{1,M}\\ \omega( {\delta}/{\nu})\\ ,$$

where $\omega$ is a modulus of continuity for the uniformly continuous  function $\phi$ on $M$ and $\delta$ is the diameter of $M$ (add and subtract a term $\int_M u _ \nu(x) \hat \phi(x)dx$ with a suitable discretization  $\hat \phi$ of $\phi$). Note that, of course, the argument wouldn't work for $p=1$, as continuous functions are not dense in $L^\infty$.

**More details.** Let $u\in L^1 _ {loc}$  with (multi)period $M$, $\phi \in C^0(M)$, and $\nu \in \mathbb{N} _ +$. First observe that $M$ is partitioned into $\nu^{n}$ smaller parallelepipeds, say $\{ M_j\} _ {1\le j \le \nu^n}$, which are translated copies of $\nu ^{-1} M$, and have in particular diameter $\delta / \nu$, where as above $\delta:=\mathrm{diam}(M)\\ $ .  Define a simple function $\hat\phi$ that takes a constant value on each of these $M _ j$, namely $$\hat \phi _ {|M _ j}:= \frac{1}{|M _ j|}\int _ { M _ j}\phi (x)dx,\quad \mathrm{for\\ any\\ \\ } 1\le j \le \nu^n .$$ Note that by continuity of $\phi$,  $\hat\phi _ {|M _ j}=\phi(\xi _ j)$ for some $\xi _ j\in M _ j $, whence  $\| \hat \phi - \phi\| _ {\infty, M}\le \omega (\delta / \nu)$ . Moreover, since $u$ has period $M$, the function $u _ \nu$ has period $\nu ^{-1} M$, and in particular its integral on each $M _ j$ is $\int _ {M _ j} u _ \nu (x) dx =\nu ^{-n} \int _ M u(x) dx=\nu ^{-n} |M|\\ \tilde u $. Then, since  $\hat\phi$ is constant on each $M _ j$ we have
$$ \int_M u _ \nu(x) \hat \phi(x)dx = \sum _ {j=1} ^{\nu\\, ^n} \int _ {M _ j} u _ \nu(x) \hat \phi(x)dx= \sum _ {j=1} ^{\nu\\, ^n}\frac{1}{|M _ j|}\int _ { M _ j}\phi (x)dx \int _ {M _ j} u _ \nu(x) dx=$$ 
$$= \sum _ {j=1} ^{\nu\\, ^n}\\ \tilde u\\,  \int _ { M _ j}\phi (x)dx  = \tilde u \int _ M \phi(x) dx\\ .$$
Thus, 
$$\Bigg | \int _ M u _ \nu(x) \phi(x)dx  -  \int _ M \tilde u \phi(x)dx \Bigg| = \Bigg | \int _ M u _ \nu(x) \big( \phi(x) - \hat \phi(x)\big)dx \Bigg| \le \|u\|_{1,M}\\ \omega( {\delta}/{\nu})\\ .$$

This is interpretated: if $u\in L^1 _ {loc}$,  then $u _ \nu$ converges to $\tilde u$ in the weak* sense of measures on $M$. Moreover, as remarked above, it follows immediately that if $u\in L^p _ {loc}$ for some $1 < p < \infty$, the convergence is  weak  $L^{p}(M)$, and if $u\in L^\infty _ {loc}$ the convergence is weak* $L^\infty (M)$.