If $f$ has compact support, choose $x_j \to 0$ so that the functions $f(x - x_j)$ are independent

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Let $f : \mathbb{R}^n \to \mathbb{C}$ be a continuous function. Let $A$ denote the support of $f$, and suppose that $A$ is compact and that $m(A) > 0$, where $m$ denotes Lebesgue measure.

I would like to should that there exists a sequence $\{x_j\}_{j=1}^\infty \subseteq \mathbb{R}^n$ with $x_j \to 0$ such that the functions $f_j = f(x - x_j)$ are linearly independent outside a set of measure zero.

This construction is made in the proof of the Theorem 7.12 in Teschl's Mathematical Methods in Quantum Mechanics, and is part of an alternate strategy (avoiding complex analysis) to show that if a function $f$ and its Fourier transform $\hat{f}$ are both compactly supported, then $f \equiv 0$.

The support of the function $f_j$ is $A + x_j$, but I need to find a suitable way of choosing the $x_j$ so that the sets $A_j = A + x_j$ are sufficiently separated to ensure the linear independence.

Hints or solutions are greatly appreciated!

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If I understood correctly the question, you can choose $x_j=a_j e$ with $e$ any non zero vector and $a_j\searrow 0$, and you can remove "outside a set of measure zero".

Consider the affine hyperplane orthogonal to $e$ touching $A$ from outside, that is $H_t=\{x:\langle x, e\rangle =t\}$ with $t=\sup_{y\in A} \langle y,e\rangle$, and pick any point $x_A$ in the intersection $H_t\cap A$. Then for any given finite set $\{a_{i_1},\ldots, a_{i_k}\}$ the function among $f_{j}$ with biggest $a_j$ is the only one not vanishing near its "touching point" $x_A+x_{j}$. This forces its coefficient in the linear combination to be zero, and you can proceed inductively.