I would like to know if there is a measurable set $M \subset \Bbb{R}$ such that

$M$ has finite Lebesgue measure $0 < \lambda(M) < \infty$,

$M$ is unbounded in the sense that $\lambda(M \setminus [-r,r]) > 0$ for all $r > 0$.

The indicator function of $M$ is an $L^p$-Fourier multiplier for some $p \neq 2$, i.e., the operator $L^2 \to L^2, f \mapsto \mathcal{F}^{-1}(\widehat{f} \cdot 1_M)$ can be extended to a bounded linear operator on $L^p(\Bbb{R})$ for some $p \neq 2$.

We have $\mathcal{F}^{-1} 1_M \in \bigcap_{1 < q < \infty} L^q (\Bbb{R})$.

As a first step, it would also be fine to only have an example where the first three properties are satisfied.

Some observations:

By a result of Lebedev and Olevskii (https://eudml.org/doc/58174) any set satisfying condition 3 from above needs to be an open set, at least after modifying $M$ on a null-set.

Having finite measure is a necessary condition for condition 4, since if $\mathcal{F}^{-1} 1_M \in L^2$, then $1_M \in L^2$, so that $M$ needs to have finite measure.

As noted in a paper by Mockenhaupt and Ricker , the class of all sets $M$ such that $1_M$ is an $L^p$-Fourier multiplier forms an algebra of sets that

a) is closed under scalings and translations,

b) contains all (bounded or unbounded) intervals, since the Hilbert transform is bounded on $L^p$,

c) contains all unions of dyadic intervals $[2^j, 2^{j+1})$ for $j \in \Bbb{Z}$, essentially by Littlewood-Paley theory.

But I was unable to use these properties to either construct a set $M$ as I would like to have, or to show that no such set can exist.

Any help would be appreciated.

exactlyat the powers of $2$, so you can just shift a bit and subtract the standard partition). $\endgroup$