In fact, this is true for every $\mu$ and every $1 \le p < \infty$. It's a pretty standard "textbook" fact, which is probably why you're not finding papers that discuss it.

One possible way to prove it is following Aleksei Kulikov's hint. Suppose they were not dense. Then by Hahn-Banach and the $L^p$-$L^q$ duality, there would exist some nonzero $f \in L^q(\mu)$, where $\frac{1}{p}+\frac{1}{q}=1$, such that $\int f(x) e^{isx}\,\mu(dx) = 0$ for all $s$. Now let $d\nu = f\,d\mu$, noting that $f \in L^q \subset L^1$ so that $\nu$ is a finite signed measure, and we have $\int e^{isx}\,\nu(dx) = 0$ for all $s$, which is to say that the Fourier transform of $\nu$ is identically zero. But a measure (or distribution) with a vanishing Fourier transform can only be zero; this fact will be in any basic Fourier theory text. Thus the complex exponentials are indeed dense in $L^p(\mu)$, and we assumed nothing about $\mu$.

This is generally false for $p=\infty$, since the uniform closure of the exponentials only contains continuous functions. Off the top of my head, I think it will be true if and only if the support of $\mu$ is a discrete set, with no limit points, which in particular means it must be a finite or countable sum of point masses.

In fact, this is true for every $\mu$ and every $1 \le p < \infty$. It's a pretty standard "textbook" fact, which is probably why you're not finding papers that discuss it.

One possible way to prove it is following Aleksei Kulikov's hint. Suppose they were not dense. Then by Hahn-Banach and the $L^p$-$L^q$ duality, there would exist some nonzero $f \in L^q(\mu)$, where $\frac{1}{p}+\frac{1}{q}=1$, such that $\int f(x) e^{isx}\,\mu(dx) = 0$ for all $s$. Now let $d\nu = f\,d\mu$, noting that $f \in L^q \subset L^1$ so that $\nu$ is a finite signed measure, and we have $\int e^{isx}\,\nu(dx) = 0$ for all $s$, which is to say that the Fourier transform of $\nu$ is identically zero. But a measure (or distribution) with a vanishing Fourier transform can only be zero; this fact will be in any basic Fourier theory text. Thus the complex exponentials are indeed dense in $L^p(\mu)$, and we assumed nothing about $\mu$.

When $p=\infty$, this is false for practically all interesting measures $\mu$. For instance:

• It is false if the support of $\mu$ is a set with a limit point. For if so, then we may find a sequence $x_n$ in the support of $\mu$ which converges to some $x$. We can then inductively choose open sets $U_n$ which are pairwise disjoint, have $x_n \in U_n$, and have diameters shrinking to 0. Consider the function $f = \sum_k 1_{U_{2k}}$ which takes the value 1 on all the even-numbered $U_n$ and is 0 elsewhere. If $g$ is any function with $\|f-g\|_{L^\infty(\mu)} < 1/4$, then since each $U_n$ has positive measure, it must contain a point $y_n$ with $|f(y_n) - g(y_n)| < 1/4$, which is to say $g(y_n) > 3/4$ for even $n$, and $g(y_n) < 1/4$ for odd $n$. But the $y_n$ converge to $x$, so such $g$ cannot be continuous, and in particular it cannot be a finite sum of complex exponentials.

• It is false if the support of $\mu$ is a countable set whose pairwise distances are not bounded away from 0 (e.g. the set $\{\sqrt{n} : n \in \mathbb{N}\}$). Any $\mu$ with countable support is atomic, so this means we can find distinct points $(x_n, y_n)$, all having positive measure, and with $|x_n - y_n| \to 0$. Let $f$ take the value 1 on all the $x_n$ and 0 on all the $y_n$. Now if $g$ is a finite sum of complex exponentials, then it is Lipschitz with some constant $M$, and so if $n > 2M$, we have $|g(x_n) - g(y_n)| < M/n < 1/2$. This forces either $|f(x_n) - g(x_n)| > 1/2$ or $|f(y_n) - g(y_n)| > 1/2$, and so $\|f-g\|_{L^\infty(\mu)} > 1/2$.

This leaves measures whose support set is something like $\mathbb{N}$ (it is obviously true for measures whose support is finite). For instance, are the complex exponentials are dense in $\ell^\infty$? I don't know the answer to that question.