Does the (distributional) support of the Fourier transform of an $L^p$-function with $p<\infty$ have positive measure? Suppose that $f \in L^p(\mathbb R^n)$ such that $1\leq p < \infty$. Let $\hat f$ be the Fourier transform of $f$. Clearly, if $p=1$ or $p=2$ then the support of $\hat f$ has positive Lebesgue measure, provided that $f \neq 0$. Using Hausdorff-Young this also holds if $p \in [1,2]$. On the other hand, if $p=\infty$ then this statement is not true as the Delta distribution shows.
What about $p \in (2,\infty)$?
That is, let $f \in L^p(\mathbb R^n)$ such that $f \neq 0$ and $p \in (2,\infty)$. Does the support of $\hat f$ (in the sense of distributions) have positive Lebesgue measure? Or does it at least contain a limit point?
 A: No. In dimension $d\ge 2$, the surface measure of the sphere gives a counterexample: its Fourier transform has power decay.
For $d=1$, there are also singular measures whose Fourier transform has power decay. One can consider a closed set of positive Fourier dimension.
As for the last part of your question, the support of $\widehat{f}$ cannot be a bounded discrete (= finite) set since then $\widehat{f}$ is a combination of $\delta$'s and their derivatives, so $f$ is a linear combination of functions of the type $x^n e^{iax}$, which cannot be in $L^p$.
In the case of an infinite discrete set it's clear that a measure supported by such a set will no longer work as a counterexample. In fact, if $\mu$ is any measure with a point part, then, by Wiener's theorem, $f=\widehat{\mu}$ satisfies
$$
T\lesssim \int_{-T}^T |f|^2\, dt \le \left( \int_{-T}^T|f|^p\, dt\right)^{2/p} (2T)^{1-2/p} ,
$$
so $f\notin L^p$.
A: I can prove that when $p\in (2,\infty)$, then the Fourier transform need not be represented by a locally integrable function. The proof presented below is not constructive. For an explicit example of a function whose Fourier transform has support of Lebesgue measure zero, see the example of Christian Remling.

Theorem.
If $p\in (2,\infty)$, then there is
$f\in L^p(\mathbb{R}^n)$ such that the distributional Fourier transform
$\hat{f}$ is not given by a locally integrable function.

In the proof we will need the following result.

Lemma.
Fix $z\in\mathbb{C}$ with ${\rm re}\, z>0$ and choose $z^{1/2}$ to have
${\rm re}\, z^{1/2}>0$. Then the Fourier transform of
$$
f(x)=z^{-1/2} e^{-\pi x^2/z},
\quad
x\in\mathbb{R}
$$
equals
$$
\hat{f}(\xi) = e^{-\pi z\xi^2},
\quad
\xi\in\mathbb{R}\, .
$$
In particular for $\delta>0$ we have
\begin{equation}
\label{a}
\left(\frac{e^{-\pi x^2/(1+i\delta)}}{(1+i\delta)^{1/2}}\right)^\wedge(\xi)=
e^{-\pi(1+i\delta)\xi^2}\, .
\end{equation}

Proof.
If $f(x)=e^{-4\pi^2tx^2}$, $t>0$, then
$\hat{f}(\xi)=(4\pi t)^{-1/2} e^{-\xi^2/(4t)}$. Hence also
$$
\left( (4\pi t)^{-1/2} e^{-x^2/(4t)}\right)^\wedge(\xi)=
e^{-4\pi^2 t\xi^2}\, .
$$
Taking $4\pi t=a$ we have
\begin{equation}
\left(\frac{e^{-\pi x^2/a}}{a^{1/2}}\right)^\wedge(\xi)=
e^{-\pi\xi^2 a}
\quad
\mbox{for $a>0$.}
\tag1
\end{equation}
Let
$$
f_z(x)= z^{-1/2} e^{-\pi x^2/z},
\quad
{\rm re}\, z>0\, .
$$
Fix $\xi\in\mathbb{R}$. It is easy to see that the function
$$
z\mapsto \hat{f}_z(\xi),
\quad
{\rm re}\, z>0
$$
is holomorphic. This easily follows from the integral formula
that defines $\hat{f}_z(\xi)$ and the fact that we can differentiate
this formula with respect to $z$ under the sign of the integral.
The function
$$
z\mapsto g_z(\xi)=e^{-\pi z\xi^2},
\quad
{\rm re}\, z>0
$$
is holomorphic (on $\mathbb{C}$).
Since $\hat{f}_z(\xi)=g_z(\xi)$ on the half line
${\rm re}\, z>0$, ${\rm im}\, z=0$ by (1), we conclude that
$\hat{f}_z(\xi)=g_z(\xi)$ for all ${\rm re}\, z>0$.
$\Box$
Proof of the theorem.
Let $p>2$. Suppose that for every $f\in L^p(\mathbb{R}^n)$, $\hat{f}$ is represented
by a locally integrable function. let $\eta\in C_0^\infty(\mathbb{R}^n)$,
$\eta(x)=1$ for $|x|\leq 1$. Then the operator
$$
L^p(\mathbb{R}^n)\ni f\stackrel{T}{\mapsto} \eta\hat{f}\in L^1(\mathbb{R}^n)
$$
is linear. We will prove that $T$ is bounded. Recall that according to the
closed graph theorem in order to prove that
a linear operator between Banach spaces $T:X\to Y$ is bounded it suffices to
show the implication
$$
x_k\to x,
\quad
Tx_k\to y
\qquad
\Rightarrow
\qquad
Tx=y
$$
which is easily equivalent to the implication
$$
x_k\to 0,
\quad
Tx_k\to y
\qquad
\Rightarrow
\qquad
y=0.
$$
Suppose that $f_n\to 0$ in $L^p$ and $Tf_n\to g$ in $L^1$. Then for every $\psi\in\mathscr{S}_n$
we have
$$
Tf_n[\psi]\to \int_{\mathbb{R}^n} g(x)\psi(x)\, dx\, .
$$
On the other hand
$$
Tf_n[\psi] = \eta\hat{f}_n[\psi] =
f_n[(\eta\psi)\,\hat{}\,]=\int_{\mathbb{R}^n} f_n(x)(\eta\psi)\,\hat{}\,(x)\, dx\to 0\, ,
$$
so
$$
\int_{\mathbb{R}^n} g(x)\psi(x)\, dx =0
$$
for all $\psi\in \mathscr{S}_n$ and hence $g=0$ a.e. This proves boundedness of $T$.
In particular
$$
\int_{B(0,1)} |\hat{f}| \leq\int_{\mathbb{R}^n}|\eta\hat{f}|\leq
\Vert f\Vert_p\, .
$$
Let now
$$
f_\delta(x)=\frac{e^{-\pi x^2/(1+i\delta)}}{(1+i\delta)^{1/2}}\, .
$$
Then according to the lemma
$$
\hat{f}_\delta(\xi) = e^{-\pi(1+i\delta)\xi^2}
$$
and according to our estimate
$$
\int_{-1}^1 |\hat{f}_\delta|\leq C\Vert f_\delta\Vert_p\, .
$$
We have
$$
\int_{-1}^1 |\hat{f}_\delta| = \int_{-1}^1 e^{-\pi\xi^2}\, d\xi = A >0\, ,
$$
where $A$ is a constant independent of $\delta$. Hence
$$
A^p\leq C^p\int_\mathbb{R} |f_\delta|^p =
C\int_\mathbb{R} \frac{e^{-\pi x^2p/(1+\delta^2)}}{(1+\delta^2)^{p/2}}\, dx \to
0
\quad
\mbox{as $\delta\to 0$}
$$
which is a contradiction.
$\Box$
