Let $X^n:= X_1^n,...,X_n^n$ be a sequence of $E^n$ valued random variables defined on a probability space $(\Omega,\mathcal{F})$. Define $Y^n$ as a selection on $X^n$. Meaning $\mathbb{P}(Y^n=X_i^n|X^n=x)=p_i^n(x)$ with $\sum_i p_i^n(x) =1$. Is it immediate that $Y^n$ is $\mathcal{F}^n=\sigma(X^n)$ measurable? Is it the case for all the $Y^n$ defined like this? for all the $Y^n$ absolutely continuous wrt $X^n$? How can i prove it formally?
EDIT (updating progessively): @justt gave me some more insight. The selection process just needs to be independant from $X^n$. Turning the focus of the question on how to accomplish that.
Context: I want to select a random (following a given law) entry in a $n$ sized random tab. And I want to show that the selection process is measurable wrt to the law of the tab. I don't really know where to start, usually measurability is swept under the table...
If you want your selection procedure to be independent of the sequence $X^n$, which is very likely the case, you need to introduce additional randomness (the random index $I^n$). So no $Y^n$ is not going to be $\sigma(X^n)$-measurable, but rather $\sigma(X^n,I^n)$-measurable.
Your answer makes it look like you are conflating the notion of measurability of a random variable wrt another, and absolute continuity between the law of a random variable and the law of another. These notions have nothing in common and are completely orthogonal (neither implies the other).