Is $Y=Y(\omega) = \inf_{0 \leq t \leq 1}X_t(\omega) = 1_A(\omega)$ not measurable if $A \notin \mathcal{B}[0,1]?$ Consider the probability space $([0,1], \mathcal{B}[0,1],\lambda)$ where $\lambda$ is the Lebesgue measure. Let $A \subset [0,1] $ and define
$ X_t(\omega) = 
\begin{cases} 0 \;\; if \; \; t = \omega \;, \; \omega \notin A \\
1 \; \; \; otherwise 
\end{cases} $
I would like to argue that all $X_t, t \in [0,1]$ are random variables, but that $Y=Y(\omega) = \inf_{0 \leq t \leq 1} X_t(\omega) = 1_A(\omega)$ is not measurable if $A \notin \mathcal{B}[0,1].$
I can't quite figure out how to solve this. Is $A$ open in $[0,1]$? If so, then $X_t(\omega)$ is not right continuous. For example, if $A = (0.5,0.6)$ then $X_{0.5}(\omega) = 0$ but $X_{0.5}+(\omega) = 1$ which violates $\lim_{y \to x, y > x} F(y) = F(x) $.

Observation $1$. Let $(X, \mathfrak A) $ be a measurable space and let $A\subset X. $ Then $\mathbf 1_A$ is $\mathfrak A$-measurable if and only if $A\in \mathfrak A. $
The proof is straightforward application of the definition of measurability of function. The conspicuous thing to notice is the set
$$\mathsf T:= \{x\in X:\mathbf 1_A(x)\leq \alpha\}\in \mathfrak A~~\forall~\alpha\in \mathbb R. \tag 1$$ Take $\alpha=0.5;$ then $\mathsf T = A^\complement\in \mathfrak A. $ This means $\mathsf T^\complement = A \in \mathfrak A. $
OP now can use this observation according to the necessity of the problem in hand.

Reference:
$\rm [I]$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific Publishing, $2014;$ chapter $1, $ pp. $72-73.$
