Let $(X,\mathcal{F}_X)$ be a meausurable space and let $\textbf{P}(X)$ the set of all probability meausures over $(X,\mathcal{F}_X)$.
For $A\in\mathcal{F}_X$ consider de evaluation map $$ {i_A}:{\textbf{P}(X)}\rightarrow{[0,1]}$$ $$ {\mathbb{P}_X}\mapsto{i_A(\mathbb{P}_X) = \mathbb{P}_X(A).}$$ equate P (X) by the smallest $ \sigma- $ algebra such that the family of functions $ \{i_ {A} \} _ {A \in \mathcal {F} _X} $ is measurable.
Then $(P(X),\sigma(\{i_ {A}\})=\sigma\left( \{i_{F_X}^{-1}(B): F_X\in\mathcal{F}_X\; \text{ y } B\in\mathcal{B}(\mathbb{R})\} \right))$ is a meausurable space.
Consider a function defined by:
$${\alpha_{(X,\mathcal{F}_X)}}:{(X,\mathcal{F}_X)}\rightarrow{(\textbf{P}(X),\sigma(\{i_{A}\}))} $$
$$ x\mapsto{\alpha_X(x) = \delta_x,} $$
where $\delta_x$ is the is the measure of Dirac at point $x\in X$.
Then $\alpha_{(X,\mathcal{F}_X)}$ is it is a measurable function.
In fact, let $F\in \{i_{F_X}^{-1}(B): F_X\in\mathcal{F}_X\; \text{ y } B\in\mathcal{B}(\mathbb{R})\} $, thanks to Proving measurability of a function only by checking generating sets it is enough to prove that $$ \alpha_{(X,\mathcal{F}_X)}^{-1}(F)\in\mathcal{F}_X $$.
We have $A=i^{-1}_{F_X}(B)$ with $F_X\in\mathcal{F}_X$ and $B\in\mathcal{B}[0,1]$, then
\begin{align} \alpha^{-1}(A) & = \alpha^{-1}(i^{-1}_{F_X}(B)), \\ & = \{x\in X: \alpha(x)\in i^{-1}_{F_X}(B) \} , \\ & = \{x\in X: \alpha_{x}(F_X) \in B \}. \end{align}
We know that $\alpha_{x}(F_X) = 1$ if $x\in F_X$ or $\alpha_{x}(F_X) = 0$ if $x\in F_X^C$.
With this,
- If $\{1\}\in B$, then $\alpha^{-1}(A) = F_X$,
- If $\{0\}\in B$, then $\alpha^{-1}(A) = F_X^C$,
- If $\{1\}\notin B$ and If $\{0\}\notin B$, then $\alpha^{-1}(A) = \emptyset$
- If $\{1\} \in B$ and If $\{0\}\in B$, then $\alpha^{-1}(A) = X$.
¿Is this correct?
Therefore, in all cases we have to $\alpha^{-1}(A)\in\mathcal{F}_X$ and then $\alpha$ is a meausurable function.