What is the solution to POTW #281 - Oct 30, 2018?

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Here is this week's POTW:

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Given a complex Borel measure $\mu$ on the torus $\Bbb T^1$, define the Fourier coefficients of $\mu$ by $\hat{\mu}(n) := \int_{\Bbb T} e^{-2\pi i nx}\, d\mu(x)$, $n\in \Bbb Z$. Show that if the sequence $(\hat{\mu}(n))\in \ell^1(\Bbb Z)$, then $\mu$ has a Radon-Nikyodym derivative with respect to the Lebesgue measure on $\Bbb T$.

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No one answered this week’s problem. You can read my solution below.

Spoiler

Define
$$F(x) = \sum_{n \in \Bbb Z} \hat{\mu}(n) e^{2\pi i n x}\quad (x\in \Bbb T)$$
Since $\sum \lvert \hat{\mu}(n)\rvert < \infty$, by the Weierstrass $M$-test it follows that $F$ is a continuous function on $\Bbb T$, and
$$\hat{F}(m) = \sum_{n = -\infty}^\infty \hat{\mu}(n) \int_{\Bbb T} e^{2\pi i (n-m)x}\, dx = \hat{\mu}(m)$$By linearity, $$\int_{\Bbb T} g(x)F(x)\, dx = \int_{\Bbb T} g(x)\, d\mu(x)$$whenver $g$ is a trigonometric polynomial. By density of trig polynomials in $C(T)$, this integral equation holds whenever $g$ is continuous. Finally, since simple functions can be uniformly approximated by continuous functions, the integral equation holds whenever $g$ is a simple function. Therefore, for all measurable subsets $A\subset \Bbb T$, $$\mu(A) = \int_{\Bbb T} 1_A(x)\, d\mu(x) = \int_{\Bbb T} 1_A(x)F(x)\, dx = \int_A F(x)\, dx$$ This shows that $\dfrac{d\mu}{dm} = F$.