Why is the hydrodynamic limit a convergence in probability? The hydrodynamic limit is a matter that involves several notation and definitions and so I will  will assume that whoever reads this question has basic knowledge about the notation and understands it (let me know if you need any clarifications)...

*

*$\mathbb{T}=[0,1)$

*$\mathbb{T}_n = \mathbb{Z}/n\mathbb{Z}=\{0,1,\dots,n-1\}$ is the discrete torus with $n$ points

*$\{0,1\}^{\mathbb{T}_n}=\{\eta:\mathbb{T}_n \to \{0,1\} \ | \ \eta \text{ is a function}\}$

*$\langle{\mu},{G}\rangle := \int_{\mathbb{T}} G(u) \ \mu(du)$
The hydrodynamic limit states that
Hydrodynamic limit
Consider a process $\{\eta_t\}_{t \geq 0}$. Fix a continuous initial profile $\rho_0: \mathbb{T} \to [0,1]$ and let $\{\mu_n\}_{n \in \mathbb{N}}$ be a sequence of probability measures associated to $\rho_0$. Then, for any $t \in [0,T]$, for every $\delta > 0$ and every $G \in C(\mathbb{T})$, it holds that
\begin{equation}
\begin{split}
& \lim_{n \to \infty} \mathbb{P}_{\mu_n}\left(\eta_{.} \in D_{\Omega_n}[0,T] : \left| \frac1n \sum_{x \in \mathbb{T}_n} G\left(\frac{x}{n}\right)\eta_t(x)-\int_{\mathbb{T}} G(u) \rho(t,u) \ du \right| > \delta \right)\\
=& \lim_{n \to \infty} \mathbb{P}_{\mu_n}\left(\eta_{.} \in D_{\Omega_n}[0,T] : \left| \langle\pi^n_t, G \rangle - \langle\pi_t, G \rangle\right| > \delta \right)
=0
\end{split}
\end{equation}
where $\pi_t$ is an absolutely continuous measure (with respect to the Lebesgue measure) and whose density $\rho_t$ (i.e. $\pi_t(du) = \rho(t,u) du$) is a solution of some PDE with initial condition $\rho_0$.
However, the hydrodynamic limit is sometimes stated as the converge of the random measures $\pi^n_t$ to the absolutely continuous measure $\pi_t$ with respect to the probability $\mathbb{P}_{\mu_n}$.
But what it would make sense to me would be to say that $\langle\pi^n_t, G \rangle$ converges in probability to $\langle\pi_t, G \rangle$ (and not from $\pi^n_t$ to $\pi_t$).
Can anyone help me understand why is the hydrodynamic limit equivalent to the convergence in probability of $\pi^n_t$ to $\pi_t$?
 A: Edit: I should have mentioned here that I am assuming that these random measures are all coupled together on a single probability space. This will typically be the case by construction in practice. If not and if your $\pi_t$ is a deterministic limit measure (which I assume is the case, since otherwise you need to specify how you couple $\pi_t$ to $\pi_t^n$ anyway) then you can construct such a coupling by working with a countable product space in which $\mu_n$ corresponds to the coordinate projection to the $n^{th}$ level and the different coordinate projections are independent.
First, notice that you can view $\mu_t^n$ as a sub-probability measure on $\mathbb{T}$ by setting $\mu_t^n = \frac{1}{n}\sum_{x \in \mathbb{T}_n} \delta_{x/n} \eta_t(x)$. The claim is that if $\langle \pi_t^n, G \rangle \to \langle \pi_t, G \rangle$ in probability for each $G \in C(\mathbb{T})$, then we actually have $\pi_t^n \to \pi_t$ in probability as measures.
It suffices to prove that each subsequence $\pi_t^{n_k}$ has a further subsequence which converges almost surely to $\pi_t$. Pass to an arbitrary subsequence $n_k$. Fix a countable dense subset $\{G_\alpha : \alpha \in \mathbb{N}\}$ of $\{ G \in C(\mathbb{T}) : 0 \leq G \leq 1\}$. After diagonalizing, we may pass to a further subsequence $n_{k_j}$ along which we have the almost sure convergence $\langle \pi_t^{n_{k_j}}, G_\alpha \rangle \to \langle \pi_t, G_\alpha \rangle$ for all $\alpha \in \mathbb{N}$ as $j \to \infty$. We work on the full probability event where this convergence holds. Because $\mathbb{T}$ is compact, in each realization of the environment, we may pass to a weakly convergent subsequence $\pi_t^{n_{k_{j_\ell}}}$. Call the limit point $\gamma$. We have $\langle \pi_t^{n_{k_{j_\ell}}}, G_\alpha \rangle \to \langle \pi_t, G_\alpha \rangle$. Since the family $\{G_\alpha : \alpha \in \mathbb{N}\}$ is dense in $C(\mathbb{T})$, it follows that $\gamma = \pi_t$. Therefore $\pi_t^{n_{k_j}} \to \pi_t$ as measures almost surely, which implies the result.
