This question was posted yesterday but seems like it was since deleted by the OP.
The question: Let $f \in L^2(\mathbb{R})$, then show that $\{f(x-k)\}_{k \in \mathbb{Z}}$ forms an orthonormal set in $L^2(\mathbb{R})$ if and only if $\sum_{k \in \mathbb{Z}} |\hat{f}(\xi - k)|^2 = 1$ almost surely.
One of the proposed ideas for the forwards direction was to consider the periodization $F(\xi) = \sum_{k \in \mathbb{Z}} |\hat{f}(\xi - k)|^2$. Then computing its Fourier coefficients, we see that $$\hat{F}(n) = \int_0^1 \sum_{k \in \mathbb{Z}} |\hat{f}(\xi - k)|^2 e^{-2 \pi i n\xi}\, d\xi =\int_0^1 \sum_{k \in \mathbb{Z}} |\hat{f}(\xi - k)|^2 e^{-2 \pi i n(\xi - k)}\, d\xi.$$ From here I would like to interchange the sum and integral to get an integral over $\mathbb{R}$, but I am unsure if this is justified. Assuming this can be done we can wrtie $\hat{F}(n) = \langle \hat{f}(\xi -k), e^{2\pi i n(\xi - k)}\hat{f}(\xi-k)\rangle$. Is this in the right direction? For the other direction I am not too sure ... Thank you for any input!
Having $\{f(x-k)\}_{k \in \mathbb{Z}}$ form an orthonormal set is equivalent to $$\int_{\mathbb R}\overline {f(x-k)}f(x-l)dx=\int_{\mathbb R}\overline {f(x)}f(x-l+k)dx=\begin{cases} 1, & \text{if }\ k=l \\ 0, & \text{otherwise} \end{cases}$$ for all $k,l$ in $\mathbb Z$. Thus the orthonormality is equivalent to $$ \int_{\mathbb R}\overline {f(x)}f(x-n)dx = \begin{cases} 1, & \text{if }\ n=0 \\ 0, & \text{otherwise} \end{cases}$$
But since
$$\begin{split} \int_{\mathbb R}\overline {f(x)}f(x-n)dx &=\int_{\mathbb R} |\hat f(\xi)|^2e^{-2i\pi n\xi}d\xi &\,\,\,\,\,\,\, (1)\\ &=\sum_{k\in\mathbb Z}\int_{-k}^{-k+1} |\hat f(\xi)|^2e^{-2i\pi n\xi}d\xi &\,\,\,\,\,\,\, (2)\\ &= \sum_{k\in\mathbb Z}\int_{0}^{1} |\hat f(\xi-k)|^2e^{-2i\pi n(\xi - k)}d\xi &\,\,\,\,\,\,\, (3)\\ &= \int_{0}^{1} \sum_{k\in\mathbb Z}|\hat f(\xi-k)|^2e^{-2i\pi n(\xi - k)}d\xi&\,\,\,\,\,\,\, (4)\\ &= \int_{0}^{1} \sum_{k\in\mathbb Z}|\hat f(\xi-k)|^2e^{-2i\pi n\xi}d\xi&\,\,\,\,\,\,\, (5) \end{split}$$ where $$\begin{split} (1) &\ \text{by the Plancherel-Parseval theorem} \\ &\ \text{and the fact that the Fourier transform of $f(\cdot - n)$ is $\xi\mapsto \hat f (\xi)e^{-2i\pi n \xi}$}\\ (2) &\ \text{by splitting the real line into intervals of length 1}\\ (3) &\ \text{by the change of variable }\xi \rightarrow \xi-k\\ (4) &\ \text{by the Fubini theorem for changing the order of sums/integrals}\\ (5) &\ \text{by }e^{2i\pi nk}=1. \end{split} $$ Note that, to your question, $(4)$ is justified by the fact that $$\sum_{k\in\mathbb Z}\int_{0}^{1} \left||\hat f(\xi-k)|^2e^{-2i\pi n(\xi - k)}\right|d\xi = \sum_{k\in\mathbb Z}\int_{0}^{1} \hat f(\xi-k)|^2d\xi=\int_{\mathbb R}|\hat f(\xi)|^2d\xi<+\infty$$ This is just another form of the Lebesgue dominated convergence theorem.
To recap, the orthonormality of the shifted functions is equivalent to $$\int_{0}^{1} \sum_{k\in\mathbb Z}|\hat f(\xi-k)|^2e^{-2i\pi n\xi}d\xi=\begin{cases} 1, & \text{if }\ n=0 \\ 0, & \text{otherwise} \end{cases}\tag{6}$$ Now, by the same Fubini theorem, $$\int_0^1\sum_{k\in\mathbb Z}|\hat f(\xi-k)|^2d\xi=\sum_{k\in\mathbb Z}\int_0^1 |\hat f(\xi-k)|^2d\xi = \int_{\mathbb R}|\hat f(\xi)|^2<+\infty$$ thus $g:\xi\mapsto \sum_{k\in\mathbb Z}|\hat f(\xi-k)|^2 \in L^1([0,1])$. And by $(6)$, that function's Fourier coefficients are all $0$ except for the constant term (equal to 1). This is equivalent to that function being equal to $1$ almost everywhere.