Let $(\Omega , \mathcal F , \mathbb P )$ be a probability space and $E$ polish space. Now assume that for every $x\in E$ there an event $A_x \in \mathcal F$ is given. Further let $X:\Omega \to E$ be a random variable such that $\sigma (X) $ is independent from $A_x$ for every $x\in E$. Assume that there is a set $N \subset E$ such that $\mathbb P (A_x) = 1$ for all $x \in E \setminus N$ and $\mathbb P (X \in N ) =0 $.
Furthermore, we can assume that $\omega \mapsto 1_{A_{X(\omega)}}(\omega ) $ is measurable.
Can we deduce that $\mathbb P (A_X) = 1 ?$
Of course, Fubini's theorem comes into mind. One is tempted to write $$\mathbb P (A_X) = \int_{E} \mathbb P (A_x) \mathbb{P} (X \in d x)=1,$$ but the measurability of $x\mapsto \mathbb P(A_x) $ was not a given assumption.
I will try to give an example for an application: Assume that $X = (X_i)_{i\in \mathbb N}$ is a sequence of i.i.d. random variables with values in $\mathbb R$ and $(Z_i)_{i\in\mathbb N}$ is a sequence of i.i.d. standard normal distributed random variables independent from $(X_i)_{i\in\mathbb N}$. Thus $E = \mathbb R ^\mathbb N$ and now define for every $x\in E$ the events $$A_x = \{ L_x := \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n Z_i 1_{(-\infty , 0]}(x_i) + Z_i^2 1_{(0,\infty)}(x_i) \text{ exists and is equal to } \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n 1_{(0,\infty)}(x_i) \} $$
$1_{A_X}$ is measurable and on the set $E\setminus N := \{x\in E: \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n 1_{(0,\infty)}(x_i) \text{ exists}\}$ we have $\mathbb P(A_x) = 1$. In this case we could argue that $x\mapsto 1_{A_x}$ is measurable, but can we conclude without it that $\mathbb P (A_X)= 1$?