 Solution

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