Convergence of random measure Suppose that $S$ is a separable metric space or Polish. Let $μ_{n},n∈N $ be a random probability measures and let μ be a deterministic probability measure on $S$. That is to say, that the $ μ_{n}$ are measurable maps from a probability space $(Ω,T,P)$ to the space of $M_1(S) $ equipped with the Borel-σ-algebra generated by the topology of weak convergence. Assume that the expected measures $ν_{n}:=Eμ_{n}$, defined via duality as $∫f dν_{n}:=E∫f dμ_{n}$ for all $f∈C_{b}(S)$, converge weakly to μ, i.e. that for all $f∈C_{b}(S)$ we have the convergence of $E∫fdμ_{n}$ to $∫fdμ$.


*

*For all ϵ>0 the sequence $P(d_{BL}(μ_{n},μ)>ϵ)$ converges to zero, where $d_{BL}$ is the bounded Lipschitz metric 
$$
d_{BL}(μ,ν)=\sup\Big\{\Big\vert∫f dμ−∫f dν\Big|: f:S\to\mathbb{R} \; \mbox{is 1-Lipschitz and $\Vert f\Vert_\infty\leq 1$}\Big\}
$$
which completely metrizes the topology of weak convergence)

*For all $f∈C_{b}(S)$ it holds that $∫f dμ_{n}$ converges in probability to $∫f dμ$. 
please i want to know if 1) implies 2) or if 2) implies 1) or if 1) is equivalent to 2) 
and in the case where $μ_{n}$ is not necessary a probability measure but just a  random measure, do the same results remain true?
 A: In your setting, the two are equivalent. Let's first show that your assumptions imply that the $\mu_n$, viewed as random variables in $M_1(S)$, are tight. Since $S$ is Polish and the $\nu_n$ converge weakly to a limit, we know from Prokhorov that the sequence $\nu_n$ is tight so, for every $\varepsilon > 0$ there exists a compact set $K_\varepsilon \subset S$ such that $\nu_n(K_\varepsilon) > 1-\varepsilon $, uniformly in $n$. A simple application of Markov's inequality then shows that $P(\mu_n(K_\varepsilon) > 1-\sqrt \varepsilon) > 1-\sqrt \varepsilon$.
Define now $$\mathcal{K}_\delta = \{\mu \in M_1(S)\,:\, \mu(K_{2^{-2m}\delta^2}) > 1- 2^{-m} \delta \quad \forall m \in \mathbb{N}\}.$$
The set $\mathcal{K}_\delta$ is tight and therefore precompact in $M_1(S)$. Furthermore, we know from above that $P(\mu_n(K_{2^{-2m}\delta^2}) > 1-2^{-m} \delta) > 1-2^{-m} \delta$ so that, summing all of these probabilities, one has $P(\mu_n \in \mathcal{K}_\delta) > 1-\delta$. Now that we know that the $\mu_n$ are tight, we know that there are (possibly random) accumulation points $\hat \mu$.
The claim now is that both of your conditions are equivalent to the statement that $\hat \mu = \mu$ almost surely. It is clear that this statement implies your second claim by continuity of the maps $\eta \mapsto \int f\,d\eta$. It also implies the first claim since $d_{BL}$ is a metric for $M_1(S)$ and convergence in law to a deterministic element is the same as convergence in probability. The fact that your claims both imply that $\hat \mu$ cannot be anything other than $\mu$ is pretty much immediate.
I haven't checked the details, but I guess that if the $\mu_n$ are allowed to be random positive measures, then the claim still holds. (Testing with $f=1$ gives an a priori bound on the mass of the measure and identifies the mass of the limit.)
A: The equivalence of 1. and 2. is essentially a special case of [1] van der Vaart/Wellner (1996), Weak Convergence and Empirical Processes - With Applications to Statistics. (Unfortunately this book mainly covers nonseparable metric spaces, so it may be hard to read.)
That  1. implies 2. even if $\mu$ is arbitrary, is an easy consequence of [1], Lemma 1.9.2 (ii), rewritten to the separable case:
$\mu_n \to \mu$ in probability iff every subsequence $(\mu_{n'})$ has a further subsequence $(\mu_{n''})$ with $\mu_{n''} \to \mu$ almost surely.
The metric space $\mathbb{D}$ in [1] here are $(M_1(S),d_{BL})$ resp. $\mathbb{R}$. These spaces are separable.
That 2. implies 1., now for constant $\mu$, is a direct consequnce of [1], Lemma 1.10.2 (iii). 
