# An equivalent condition for a measure on $\mathbb{T}$ to be symmetric

Is it correct that a Borel probability measure $\sigma$ on the complex unit circle $\mathbb{T}$ is symmetric (i.e. $\sigma(A)=\sigma(\overline{A})$ for every Borel set) iff $\hat{\sigma}(n)=\hat{\sigma}(-n)$ for every $n\in\mathbb{Z}$?

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If $\mu(A)=\mu(\overline A)$ for all $A$, then $\widehat \mu(n)=\widehat \mu(-n)$ for any $n\in\Bbb Z$, as $\mu=\widetilde \mu$ where $\widetilde \mu(B):=\mu(\overline B)$.
Conversely, assume that for any $n\in\Bbb Z$, $\widehat \mu(n)=\widehat \mu(-n)$. Then by Stone-Weierstrass theorem, we have for any real-valued continuous function on $\Bbb T$ that $$\int_{\Bbb T}f(z)d\mu(z)=\int_{\Bbb T}f(\overline z)d\mu(z).$$ As any characteristic function of a closed set can be approached pointwise by continuous function, we deduce that $\mu(A)=\mu(\overline A)$ for any closed set $A$. Using $\pi$-$\lambda$ systems, we get that this relationship holds for any Borel-measurable set.