Let $\mu\_n,n\in \mathbb N$ be a random probability measures and let $\mu$ be a deterministic probability measure on $\mathbb R$. That is to say, that the $\mu\_n$ are measurable maps from a probability space $(\Omega,\mathcal{T},\mathbf{P})$ to the space of $M\_1(\mathbb R)$ equipped with the Borel-$\sigma$-algebra generated by the topology of weak convergence. Assume that the expected measures $\nu\_n:=\mathbf E\mu\_n$, defined via duality as $\int f~d\nu\_n:=\mathbf E\int f~d\mu\_n$ for all $f\in C\_b(\mathbb R)$, converge weakly to $\mu$, i.e. that for all $f\in C\_b(\mathbb R)$ we have the convergence of $\mathbf E\int f d\mu\_n$ to $\int f d\mu$. By Levy's continuity Theorem this is equivalent to the statement that the characteristic function $\phi\_{\nu\_n}(t):=\int e^{itx}~d\nu\_n(x)$ converges pointwise to the characteristic function $\phi\_\mu$ of $\mu$. Is a similar statement also true for *weak convergence in probability* the random measures, i.e. is there a connection between the following statements?


* For all $\epsilon>0$ the sequence $\mathbf{P}(d\_{BL}(\mu\_n,\mu)>\epsilon)$ converges to zero, where $d\_{BL}$ is the bounded Lipschitz metric $$d\_{BL}(\mu,\nu)=\sup\left\{\left\lvert\int f~d\mu-\int f~d\nu\right\lvert~:~f\colon\mathbb{R}\to\mathbb{R}\text{ is 1-Lipschitz and }\lVert f\lVert\_\infty\leq1\right\}$$ (which completely metrizes the topology of weak convergence)
* For all $f\in C\_b(\mathbb R)$ it holds that $\int f~d\mu\_n$ converges in probability to $\int f~d\mu$.
* For all $t\in \mathbb R$, $\phi\_{\mu\_n}(t)$ converges to $\phi\_\mu(t)$ in probability.
* On some small interval $[0,\epsilon]$, the sequence $\sup\_{0\leq t\leq \epsilon}\lvert \phi\_{\mu\_n}(t)-\phi\_\mu(t)\lvert$ converges in probability to zero.
* For all $k\in\mathbb N$, the sequence of moments $\int x^k~d\mu\_n$ converges in probability to $\int x^k d\mu$.
* For all $k\in\mathbb N$, the sequence of expected moments $\mathbf E\int x^k~d\mu\_n$ converges to $\int x^k d\mu$ and the variances $\mathbf{Var}\int x^k~d\mu\_n$ converge to zero.


Clearly, for the last two statements to be equivalent to any of the above, we would need some condition on the determinacy of $\mu$ via its moments. Therefore, for simplicity assume that $\mu$ is subgaussian, and that the $\mu\_n$ are uniformly subgaussian in the sense that $\mathbf E\mu\_n([-R,R]^c)\leq C e^{-C R^2}$ for some $C$ and all $n,R$.



Weak convergence of random measures
I found a way to prove the implication I was mostly interested in (the last statement implies the second): Below my argument, for the sake of completeness.


Fix a function $f$. A sequence of random variables $X\_n$ converges to a random variable $X$ in probability if and only if any subsequence $X\_{n\_m}$ has a further subsequence $X\_{n\_{m\_l}}$ which converges to $X$ almost surely. Apply this to the sequence in question and pick the subsequence $n\_{m\_l}$ in such a way that $\mathbf{Var}\int x~d\mu\_{n\_{m\_l}}$ is summable. By Borel-Cantelli this implies that $\int x~d\mu\_{n\_{m\_l}}$ converges almost surely to $\int x~d\mu$. Continue inductively and choose a diagonal sequence (for convenience still named the same) such that the sequences $\int x^k~d\mu\_{n\_{m\_l}}$ converge to $\int x^k~d\mu$ almost surely for all $k\in\mathbb N$. If the limiting distribution is uniquely determined by its moment this implies that with full probability $\mu\_{n\_{m\_l}}$ converges weakly to $\mu$, which, in particular, implies that the integral against $f$ converges almost surely.


