= Solution
Read the coefficient inside the limit as $\widehat{\mu_k}(n)$; the printed omission of its index would otherwise make that condition independent of the sequence. Weak-star convergence immediately implies convergence of every coefficient by testing $e^{-inx}$. Conversely coefficient convergence gives convergence against every <trigonometric polynomial>. Since $\|\mu_k\|=1$, the dense-test argument yields \b[weak-star convergence to $\mu$].
The limit is necessarily a <probability>. Testing the constant function gives $\mu(\mathbb T)=1$. For every nonnegative continuous $f$, $\int f\,d\mu=\lim_k\int f\,d\mu_k\geq0$. A <Radon measure> whose integrals against all such functions are nonnegative is positive, by the positive case of the <Riesz representation theorem> (or continuous approximation of Borel indicators). Thus \b[$\mu$ is nonnegative and has total mass one].
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