Define the Fourier dimension of a set $S\subset\mathbb{R}^n$ to be the the supremum of $s\in [0,n]$ such that there exists a Borel measure $m$ such that 1.) $m(S)\in (0,\infty)$, 2.) $m$ has a compact support contained in $S$ and 3.) $\forall x\in\mathbb{R}^n:|\hat{m}(x)|\leq |x|^{-s/2}$ (for the aforementioned $s$), when
$$\hat{m}(x) \equiv \int_{\mathbb{R}^n}e^{-2\pi i \xi \cdot x}dm(\xi)$$
I was quite surprised that I did not find anything on the following question, so there is a chance that is is completely trivial: If we are given a closed interval $S := [a, b]\subset\mathbb{R}$, do we know that $S$ has a Fourier dimension equal to one? If so, what should be our choice of measure? What if $S = \bigcup_{i=1}^K[a_i, b_i]$ for some $K\in\mathbb{N}$ with $[a_i, b_i]\cap [a_j, b_j] = \varnothing$ iff $i\neq j$.
I initially thought that the regular Lebesgue measure restricted to the interval in question should do the trick: $x = 0$ is no issue if we interpret the condition on $|\hat{m}|$ as a right-sided limit at the origin. But when $x\neq 0$ the following example gave me some doubt: Take $b = 4, a = -2$. Then if $x\neq 0$ we get the general formula
$$|\hat{m}(x)|^2 = \left|\frac{i}{2\pi x}\left(e^{-2\pi i xb} - e^{-2\pi ixa}\right)\right|^2 = \frac{1 - \cos\left(2\pi x(b - a)\right)}{2\pi^2x^2}$$
giving us that $|\hat{m}(x)|^2 \leq |x|^{-s} \Longleftrightarrow 1 - \cos\left(2\pi x(b - a)\right) \leq 2\pi^2|x|^{2 - s}$.
Then, if the Fourier dimension were to be $1$, we'd get in general
$$1 - \cos\left(2\pi x(b - a)\right) \leq 2\pi^2|x|$$
which in our case reduces to
$$1 - \cos\left(12\pi x\right) \leq 2\pi^2|x|$$
But for e.g. $x = 0.05$ the LHS is equal to $\approx 1.309$ while the RHS is equal to $\approx 0.987$. Hence the Lebesgue measure restricted to $[a, b]$ does not always work (it does e.g. when $a = 0, b = 1$). What should the choice of measure be in this case and when one works with union of intervals?