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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 $d\sigma/d\mu$?

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You can use \ll for $\ll$. – martini Sep 19 '12 at 12:17
In fact, the usual way for physicists to talk about "the delta function" is to approximate this singular measure as the weak limit of a sequence of absolutely continuous measures. – GEdgar Sep 19 '12 at 13:14
up vote 5 down vote accepted

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.

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What if all the functions $d\sigma_n/d\mu$ are non-negative and uniformly bounded (there is some $M>0$ such that for every $n$ we have $d\sigma_n/d\mu <M$)? – user25640 Sep 20 '12 at 21:32

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