Why $\hat{\mu}\in L^2(\mathbb{R}^n)$ implies $\mu$ absolutely continuous?

Question

Let $\mu$ be a compactly supported probability measure on $\mathbb{R}^n$. Suppose that $\hat{\mu}\in L^2(\mathbb{R}^n)$, then $\mu$ must be absolutely continuous with respect the Lebesgue measure. How to prove ?

The Fourier transform $\hat{\mu}$ is given by $$ \hat{\mu}(\xi)=\int e^{-2\pi i\langle\xi,x\rangle}d\mu(x). $$

Thank you and sorry if my English wasn't correct.

Answer 1

If you provide the definition of absolutely continuous measures, you will be able to prove it.

Answer 2

Hint

Prove that $\mu$ has a $L^2$ Lebesgue density. Then conclusion is trivial since

$$\mu(E) = \int_E f_\mu \ \mathrm d\lambda$$