TITLE: Does convergence of Fourier transforms imply convergence of measures?
QUESTION [5 upvotes]: Let $\{\sigma_n\}$ be a sequence of measures on the complex unit circle $\mathbb{T}$ and let $\sigma$ also be such a measure. Suppose that $\hat{\sigma_n}(k) \rightarrow \hat{\sigma}(k)$ as $n\rightarrow\infty$ for every $k$ in $\mathbb{Z}$. Does that imply a convergence of the sequence $\{\sigma_n\}$ to $\sigma$? if so, in what sense?

REPLY [2 votes]: This is used in proofs of the central limit theorem in probability theory.  If $X_1,X_2,X_3,\ldots$ is a sequence of independent identically distributed real-valued random variables and $\mathbb{E}(X_1^2)<\infty$, then the probability distribution of the random variable
$$
\frac{(X_1+\cdots+X_n)-n\mathbb{E}(X_1)}{\sqrt{n\operatorname{var}(X_1)}}
$$
approaches the standard normal distribution, whose probability density function is
$$
x\mapsto \frac{1}{\sqrt{2\pi}} e^{-x^2/2},\tag{1}
$$
as $n\to\infty$.
Proof: Show that the sequence of Fourier transforms of this sequence of probability distributions approaches the Fourier transform of the stated limit.
This of course relies on an affirmative answer to the question posed here, except that it deals with probability measures on the line rather than on the compact circle.  I don't think it's hard to deduce the case for the circle from the case for the line, since as a measure space, the circle is merely a subset of finite measure, of the line.
In what sense?  The sense of convergence used in the statement of this theorem in probability theory is usually stated something like this: Let $F$ be the cumulative probability distribution function $F(x) = \Pr(X_1\le x)$.  Then the sequence of cumulative probability distribution functions whose convergence to $F$ is assserted, converges pointwise to $F$ on the set of points of continuity of $F$.  Now notice that the cumulative distribution function corresponding to $(1)$ is continuous everywhere, so that last phrase doesn't matter.
(The reason for not requiring convergence at points where $F$ is not continuous can be seen by thinking about the probability measure assigning probability $1$ to every set containing $1/n$ as a member and $0$ to every set disjoint from $\{1/n\}$.  Under this definition, this approaches the measure concentrating all probability at $0$, as $n\to\infty$.  One wants the definition to imply that.)
See the Wikipedia article: Lévy's continuity theorem.