Is there way to choose $R$, $r$ so that the Lebesgue measure $m(A_{R,r}) \to 0$ as $\varepsilon\to 0$?

Question

Let an integrable function $f:\mathbb{R}\to\mathbb{R^+}$ such that its Fourier transform $\widehat f(\lambda)\ne 0$ for almost $\lambda\in\mathbb{R}$. Define ${A_{R,r}} = \left\{ {\lambda:\left| \lambda \right| \leqslant R,\left| {\hat f\left( \lambda \right)} \right| \leqslant r} \right\}$ ($R>0$, $r>0$). Is there way to choose $R_{\varepsilon}$, $r_{\varepsilon}$ so that $R_\varepsilon\to\infty$, $r_\varepsilon\to 0$ and the Lebesgue measure $m\left( {{A_{{R_\varepsilon },{r_\varepsilon }}}} \right) \to 0$ as $\varepsilon\to 0$?

Answer 1

For each fixed $R > 0$, we have $m(A_{R,r}) \leq m([-R,R]) < \infty$ and $$\bigcap_{n \in \Bbb{N}} A_{R,1/n} = \{\lambda \, \mid |\hat{f}(\lambda)| = 0\},$$ where the last set has measure zero by assumption on $f$.

Furthermore, $A_{R,r} \subset A_{R,s}$ for $r < s$.

Using continuity of the measure $m$ from above, we conclude

$$ 0 = m\bigg(\bigcap_n A_{R, 1/n}\bigg) = \lim_n m(A_{R,1/n}). $$

Hence, for each $\varepsilon > 0$, we can choose $R_\varepsilon = 1/\varepsilon$ and some $r_\varepsilon = 1/n_\varepsilon$ with $n_\varepsilon > 1/\varepsilon$ and $m(A_{R_\varepsilon, r_\varepsilon}) = m(A_{R_\varepsilon, 1/n_\varepsilon}) < \varepsilon$.

It is now easy to check that all your requirements are fulfilled.