Fourier coefficients of a (finite, regular, positive) measure are absolutely summable => the measure has a density

Question

Let $\mu$ be a finite, regular, positive measure on $[0,1)$ such that $\sum_{n\in\mathbb{Z}} |\hat{\mu}(n)| < \infty$. How can I prove that there exists $f(x)$ such that $\mu(dx) = f(x)dx$? Actually, I am quite sure it should be the case that $f(x) = \sum_{n\in\mathbb{Z}} e^{inx}\hat{\mu}(n)$ since if $\mu(dx) = f(x)dx$, then $\hat{\mu}(n) = \int_0^1 e^{-inx}\,d\mu(dx) = \int e^{-inx}f(x)\,dx = \hat{f}(n)$. Assuming this is true, I want to verify that $\int_A f(x)\,dx = \int_A \,d\mu$ for any Borel subset of $[0,1)$. So far I have reduced the problem to verifying that $\int_0^1 \phi_n f(x)\,dx = \int_0^1 \phi_n\,d\mu$ since I can approximate the indicator function of $A$ by a sequence of continuous functions, $\phi_n$. Unfortunately, I cannot see how to verify this is true.

Answer 1

Your approach is correct: since $f$ has the same Fourier coefficients as $\mu$, it suffices to show that the equality of coefficients implies that the two measures ($ f(x)\,dx$ and $\mu$) agree.

You also correctly observed that it suffices to show $$\int_0^1 \phi f(x)\,dx = \int_0^1 \phi\,d\mu\tag{1}$$ for every continuous $\phi$.

In the context of Fourier series it is natural to consider $[0,1)$ as $\mathbb R/\mathbb Z$, namely, a circle. You can approximate the indicator functions by continuous functions on the circle, which means the additional condition $\phi(0)=\phi(1-)$. The functions $e^{2\pi i n t}$, $n\in\mathbb Z$, form an algebra that separates the points of the circle, and is closed under conjugation. By the Stone-Weierstrass theorem, every continuous function on the circle is a uniform limit of the elements of this algebra. Passing to the limit under the integrals, you will get (1).