Let $D$ be the set of Dirac measures. In fact, $D$ would constitut a reasonable event, regarding probabilistic considerations (since we want to talk about random measures). However, in general it is not an event.
Consider, e.g., $X = [0,1]$ with the cocountable $\sigma$-algebra $\mathcal{B} = \sigma (\{\{x\} : x \in \mathcal{X})$ and note that the $\sigma$-algebra $\mathcal{M}_X$ is countably determined. Then, there is no countable set system $\mathcal{E} \subset \mathcal{B}$ such that $\text{pr}_{\mathcal{E}}^{-1}\big(\text{pr}_{\mathcal{E}}(D)\big) = D$, where $\text{pr}_{\mathcal{E}} : M_X \to [0,\infty]^{\mathcal{E}}$ is the projection of the set functions on $\mathcal{E}.$ This is due to the fact that we can define a measure $\mu$ by $\mu(B) = 1,$ if $B^c$ is countable and else $\mu(B) = 0,$ where $B \in \mathcal{B}.$ Note, $\mu$ is in $\text{pr}_{\mathcal{E}}^{-1}\big(\text{pr}_{\mathcal{E}}(D)\big)$ since $\mu(B) \in \{0,1\}$ for all $B,$ but $\mu$ is not in $D$ since there is no $x\in X$ such that $\mu = \delta_x.$ Hence, $D$ is not $\mathcal{M}_X$-measurable.
Let $(X,\mathcal{T})$ be a locally compact second countable Hausdorff space, then the set of all locally finite measures on $(X,\mathcal{B})$ is $\mathcal{M}_X$-measurable.
To prove this, we first note that there is a countable basis $\mathcal{U}$ of $\mathcal{T}$ and a sequence $(G_k)_{k\in\mathbb{N}}\in\mathcal{T}^\mathbb{N}$ of relatively compact sets such that $G_k \uparrow X$ for $k \to \infty.$ Then, $\mathcal{E} :=\big\{ \bigcap \mathcal{O} : \mathcal{O} \subset \mathcal{U} \cup G(\mathbb{N}),\ |\mathcal{O}| < \infty \big\}$ is a countable and intersection stable generator of $\mathcal{B}.$ According to the measure uniqueness theorem, a locally finite measure $\mu$ is uniquely determined by $\mu|_{\mathcal{E}}$. Consequently, the set of all locally finite measures is countably determined, that is, $\mathcal{M}_X$-measurable.