Weak star limit of Carleson measure I am studying some problems concerning approximation of inner functions and I ask myself this question:
Is the weak star limit of Carleson measure still a Carleson measure?
I suspect that the answer is no, but I cannot find any counterexample.
Any reference would be appreciated.
 A: Tl;dr The answer is indeed no as you suspected, because any finite measure with compact support on $\mathbb{D}$ is a Carleson measure, however, any non- Carleson measure can be written as a weak$^{*}$ limit of a sequence of finite measures with finite support. A more detailed explanation follows below.
Recall that a measure $\mu$ on $\mathbb{D}$ is Carleson if for any Carleson box $Q_{\ell,\theta_{0}}$, where:
$$
Q_{\ell, \theta}:=\{z=re^{it}: 1-\ell<r<1, \hspace{2pt} \theta<t<\theta+\ell\}
$$
we have that:
$$
\mu(Q_{\ell,\theta_{0}})\leq{C\ell}
$$
for some constant $C>0$ independent of $\ell$ and $\theta$. In particular, notice that any finite measure with finite support on $\mathbb{D}$ is Carleson.
Since $\mathbb{D}$ is LCH (locally compact and Hausdorff), the Reisz representation theorem tells us that $C_{0}(\mathbb{D})^{*}\cong{\mathcal{M}(\mathbb{D})}$ where $\mathcal{M}(\mathbb{D})$ is the set of Radon measures on $\mathbb{D}$. Thus, given measures $(\mu_{n})_{n\geq{1}}$ on $\mathbb{D}$ we say that $\mu_{n}\xrightarrow{w^{*}}\mu$ for some measure $\mu$ on $\mathbb{D}$ if for every $f\in{C_{0}(\mathbb{D})}$,
$$
\int_{\mathbb{D}}f(x)d\mu_{n}(x)\rightarrow{\int_{\mathbb{D}}f(x)d\mu(x)} \hspace{5pt} \text{as $n\rightarrow{\infty}$}
$$
Let $\nu$ be any non- Carleson probability measure on $\mathbb{D}$ and let $X_{1}, X_{2}, X_{3}, ...$ be i.i.d. samples from $\nu$. Let $\Omega$ be the probability space on which $X_{1}, X_{2}, .... $ are defined. Then the random measure $\nu_{n}(\omega, \cdot)$ is given by:
$$
\nu_{n}(\omega, \cdot)=\frac{1}{n}\sum_{k=1}^{n}\delta_{X_{k}(\omega)}(\cdot)
$$
By the law of large numbers, for any fixed $f\in{C_{0}(\mathbb{D})}$,
$$
\int_{\mathbb{D}}f(x)d\nu_{n}(\omega, x)=\frac{1}{n}\sum_{k=1}^{n}f(X_{k}(\omega))\rightarrow{\mathbb{E}f(X_{1}(\omega))}=\int_{\mathbb{D}}f(x)d\nu(x) \hspace{5pt} \text{as $n\rightarrow{\infty}$}
$$
Since the countable intersection of events of full measure has full measure, the above convergence can be made to hold for countably many functions in $C_{0}(\mathbb{D})$ simultaneously. However, $C_{0}(\mathbb{D})$ is separable and so, by density, this is enough to ensure that:
$$
\int_{\mathbb{D}}f(x)d\nu_{n}(\omega, x)\rightarrow{\int_{\mathbb{D}}f(x)d\nu(x)} \hspace{5pt} \text{as $n\rightarrow{\infty}$ for all $f\in{C_{0}(\mathbb{D})}$, a.s.}
$$
In other words, for a.e. $\omega\in{\Omega}$, the Carleson measures $\nu_{n}(\omega, \cdot)$ converge in weak$^{*}$ to $\nu$ which is not Carleson.
