Suppose that $\{X^i\_t\}\_{1\leq i\leq n;\,t\in [0,T]}$ is an interacting particle system prescribed by certain SDEs, with each $X^i \in \mathcal{X}$ (the state space). Define the associated empirical measure as $$\mu^n\_t = \frac{1}{n}\sum\_{i=1}^n \delta\_{X^i\_t},\quad t\in [0,T].$$ It is mentioned in page 7 of [Domingo-Enrich, Jelassi, Mensch, Rotskoff, and Bruna - A mean-field analysis of two-player zero-sum games](https://arxiv.org/abs/2002.06277) that $\{\mu^n\_t\}\_{t\in [0,T]} \in \mathcal{C}\left([0,T],\mathcal{P}(\mathcal{X})\right)$. However, in page 30 it is also stated that
$\{\mu^n\_t\}\_{t\in [0,T]}$ is a $\mathcal{P}\left(\mathcal{C}\left([0,T],\mathcal{X}\right)\right)$-valued random element. I am a bit confused on the aforementioned two spaces, and from my intuition the two spaces are different. Can anyone shed some light on what is going here?



Distinction between $\mathcal{C}\left([0,T],\mathcal{P}(\mathcal{X})\right)$ and $\mathcal{P}\left(\mathcal{C}\left([0,T],\mathcal{X}\right)\right)$
$\newcommand\om{\omega}\newcommand\Om{\Omega}\newcommand\C{\mathcal C}\newcommand\X{\mathcal X}\newcommand\Y{\mathcal Y}\newcommand\P{\mathcal P}$Apparently, here $\P(\Y)$ means the set of all probability measures on (a certain $\sigma$-algebra over) a set $\Y$.


First of all, it is not true that
$\mu^n:=(\mu^n\_t)\_{t\in[0,T]}\in\C([0,T],\P(\X))$. Rather, $\mu^n$ is a random element of $\C([0,T],\P(\X))$, assuming that $\mu^n\_t$ is continuous in $t$ in an appropriate sense.


Next, for each $i$, consider the maps $X^i$ defined by the formula
$$[0,T]\ni t\mapsto X^i(t):=X^i\_t.$$
Then $X^i$ is a random element of $C([0,T],\X)$, provided that $X^i\_t$ is continuous in $t$.


Further let
$$\widetilde{\mu^n}:=\frac1n\,\sum\_{i=1}^n\delta\_{X^i},$$
so that $\widetilde{\mu^n}$ is a random element of $\P(\C([0,T],\X))$.


Knowing the random probability measure $\widetilde{\mu^n}$ over $\C([0,T],\X)$, one can restore the family $\mu^n=(\mu^n\_t)\_{t\in[0,T]}$ of random probability measures over $\X$. Indeed, for any $A\subseteq\X$ we have
$$\mu^n\_t(A)=\widetilde{\mu^n}(B\_{t,A}),$$
where $B\_{t,A}:=\{f\in\C([0,T],\X)\colon f(t)\in A\}$. This allows one to identify the family $\mu^n$ of random elements of $\P(\X)$ with the random element $\widetilde{\mu^n}$ of $\P(\C([0,T],\X))$.




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N.B.1: One should not use $\{\mu^n\_t\}\_{t\in[0,T]}$ for $(\mu^n\_t)\_{t\in[0,T]}$, since curly brackets are reserved for sets, whereas parentheses are used for families of elements of a set $S$ (i.e., for functions with values in $S$).


N.B.2: Here one should not write $X^i\in\X$. Indeed, for some set $\Om$ and for each pair $(t,\om)\in[0,T]\times\Om$, it is, not $X^i$, but $X^i\_t(\om)$ that is an element of $\X$.


