Does weak convergence of measures preserve absolute continuity?

Let $\{ \sigma_n \}$ be a sequence of positive measures on the complex unit circle $\mathbb{T}$ with its borel sets, and Suppose that $\{ \sigma_n \}$ converges weakly to $\sigma$ which is also such a measure. Suppose that $\mu$ is another positive measure on $\mathbb{T}$ such that $\sigma_n\ll\mu$ for every $n$. Does this imply that $\sigma\ll\mu$? If so, can we tell anything about $d\sigma/d\mu$?

Solution 1:

No, it doesn't. Let $\lambda$ be the arglength measure and $\phi_n \ge 0$ a continuous function on $\mathbb T$ with $\int_{\mathbb T} \phi_n\, d\lambda = 1$ and $\phi_n(x) = 0$ if $|x-1| \ge \frac 1n$. Then for each continuous function $f\colon \mathbb T \to \mathbb R$ we have $\int_{\mathbb T} f\phi_n d\lambda \to f(1)$, that is $\phi_n \lambda \to \delta_1$ weakly. But $\delta_1$ is not $\lambda$-continuous.