= Solution
Write $m(dx)=dx/(2\pi)$ for the normalized <Haar measure>. The identity $\sin^2(kx)=\tfrac12-\tfrac12\cos(2kx)$ gives $\mu_k=\tfrac12m-\tfrac12\cos(2kx)m$. For any fixed <trigonometric polynomial>, <orthogonality of complex exponentials> makes the oscillatory integral zero once $2k$ exceeds its largest frequency. The norms satisfy $\|\mu_k\|=\mu_k(\mathbb T)=1/2$, so the dense-test criterion applies. Therefore
$$
\boxed{\mu_k\overset{*}{\rightharpoonup}\tfrac12m,\qquad d\mu=\frac{dx}{4\pi}.}
$$
In particular the limit has mass one half; treating these <measures> as probabilities would introduce a factor-of-two error.
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