# What is the source of this simple lemma in harmonic analysis over compact groups?

I recently needed this simple fact of harmonic analysis:

Let $G$ be a discrete abelian group and $X\subset G$ be a finite subset. Then for any $\epsilon>0$ the set $\Gamma_\epsilon=\{ \gamma\in\widehat{G} : \lvert \widehat{X}(\gamma)\rvert \geq (1-\epsilon)\lvert X\rvert\}$ is a closed subset of $\widehat{G}$ with strictly positive Haar measure; $\mu(\Gamma_\epsilon)>0$.

The point is the positivity of the Haar measure; this fails for $\epsilon=0$ or $X$ infinite, for example.

The proof is simple: observe that this set contains $\Gamma_\epsilon'-\Gamma_\epsilon'$ if $\Gamma_\epsilon'$ has the shape $$\{ \gamma : \lvert \theta_x - \gamma(x) \rvert \leq \epsilon/2\textrm{ for all }x\in X\}$$ for any $\theta_x$, and then note that $\widehat{G}$ is covered by at most $(2\epsilon^{-1})^{\lvert X\rvert}$ many such sets.

I was surprised that I couldn't find a reference or more discussion in Rudin's `Fourier analysis on groups', and Google didn't turn up anything either (at least, nothing I could recognise as this fact). It feels like this fact is a simple corollary of some deeper fact of harmonic analysis.

Hence my questions are:

What is a reference for this fact? What is the correct generalisation to general locally compact abelian groups?

An even vaguer question: what is the iceberg of which this is a tip?

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Does $\widehat{X}$ for a subset $X$ mean the Fourier transform of the characteristic function of that subset? If so, the inequality defining your $\Gamma_\varepsilon$ appears in the proof of Corollary 4 of math.uconn.edu/~kconrad/blurbs/gradnumthy/loccptascoli.pdf with $G$ being any locally compact abelian group and $X$ being any compact subset of $G$ (denoted as $K$ rather than $X$). Maybe Theorem 2 or Corollary 4 there is the iceberg you're looking for. The positivity of Haar measure of a measurable subset of a locally compact abelian group is true if it contains an open set. – KConrad Jun 4 '14 at 12:17
Yes, it does. Thanks! That's the kind of statement I was looking for. – Thomas Bloom Jun 4 '14 at 12:30