I'm trying to understand the proof of Lemma 9 from Sprindzhuk's book Metric Theory of Diophantine approximation. Here's what it says.
Lemma 9. Fix integers $m$ and $n$ and consider the tori $\mathbb{T}^{mn} = \mathbb{R}^{mn}/\mathbb{Z}^{mn}$ and $\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$. Given two independent rows $q_1, q_2 \in \mathbb{Z}^m$, consider the maps $S_{1}, S_{2}:\mathbb{T}^{mn} \to \mathbb{T}^n$ induced by $$ S_{1} (X) = {q_1}X,\ S_{2}(X) = {q_2}X, $$ where we write $mn$-vectors as $m\times n$ matrices $X$. If $A_1, A_2$ denote two measurable subsets in $\mathbb{T}^n$. Then we have $$ \left|S_{1}^{-1}(A_1) \cap S_{2}^{-1}(A_2)\right| = |A_1|| A_2|, $$ where the $|\cdot|$ denote Haar measures on each of the tori.
Proof. Let $\chi_1, \chi_2$ denote the characteristic functions of $A_1, A_2$ in $\mathbb{T}^n$ respectively. For each $k\in \mathbb{Z}^n$, we have Fourier coefficients $$ \widehat{\chi_j}(k) = \int \chi_j(t)e^{-2i\pi k\cdot t}dt. $$ And thus we have convergence (I guess of functions in $L^2(\mathbb{T}^n, |\cdot|)$) \begin{equation}\label{1}\tag{1} \sum_{k \in \mathbb{Z}^n} \widehat{\chi_j}(k)e^{2i\pi k\cdot(\cdot)} \to \chi_j(\cdot). \end{equation} Now, we can compute \begin{equation}\label{2}\tag{2} \begin{split} \left|S_1^{-1}(A_1) \cap S_2^{-1}(A_2)\right| &= \int \chi_1(S_1(X))\chi_2(S_2(X))dX \\ &= \sum_{k\in \mathbb{Z}^n} \sum_{l \in \mathbb{Z}^n} \widehat{\chi_1}(k)\widehat{\chi_2}(l) \int \exp\left\lbrace2i\pi \left(k\cdot S_1(X) + l\cdot S_2(X)\right)\right\rbrace dX. \end{split} \end{equation} Question. What kind of convergence is being used to obtain the last equality above? If only $L^2$-convergence were being used, then surely we would get an expression involving the Fourier coefficients $\widehat{\chi_j \circ S_j}(M)$ where $M \in \mathbb{Z}^{mn}$?
What am I misunderstanding?
As Mason points out, I'm being stupid and just $L^2$-convergence suffices. However I want to also point out that there seems to be a proof that avoids Fourier series.
Consider the map $S:\mathbb{T}^{mn} \to \mathbb{T}^n \times \mathbb{T}^n$ induced by $S(X) = (S_1(X), S_2(X))$. The independence of $q_1, q_2$ easily implies that the map $S$ is onto. And of course, this map is a group homomorphism.
If $\mu$ is the probability Haar measure on $\mathbb{T}^{mn}$ then I claim that $S_*\mu$ is also the probability Haar measure on $\mathbb{T}^n \times \mathbb{T}^n$. For $$ (S_*\mu)(S(X)+U) = \mu (S^{-1}(S(X)+U)) = \mu (X+S^{-1}(U)) = (S_*\mu)(U). $$ Thus $$|S_1^{-1}(A_1) \cap S_2^{-1}(A_2)| = \mu(S^{-1}(A_1\times A_2)) = S_*\mu(A_1\times A_2)=|A_1||A_2|.$$ So we're done?