How can we show that the $k$th order statistic is measurable?

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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(E,\le)$ be a partially ordered set and $\mathcal E$ be a $\sigma$-algebra on $E$
  • $n\in\mathbb N$
  • $X_1,\ldots,X_n:\Omega\to E$ be $(\mathcal A,\mathcal E)$-measurable, $$X_{1:n}:=\min\left\{X_1,\ldots,X_n\right\}$$ and $$X_{i:n}:=\min\left\{X_j:X_j>X_{i-1:n}\right\}\;\;\;\text{for }j\in\left\{2,\ldots,n\right\}.\tag1$$

Question 1: The definition $(1)$ is broken, unless at least almost surely $X_i\ne X_j$ for all $i\ne j$. Can we fix the definition such that $X_{1:n}\le\cdots\le X_{n:n}$ even when $X_1=\cdots=X_n$, where a tie is broken by declaring that $X_i$ is smaller than $X_j$ whenever $X_i=X_j$ and $i<j$?

Question 2: Given the fixed definition, are we able to show that each $X_{i:n}$ is $(\Omega,\mathcal E)$-measurable? This is clear to me when $X_i\ne X_j$ for all $i\ne j$ and $(E,\mathcal E)=(\mathbb R,\mathcal B(\mathbb R))$ in which case the minimum is a continuous function. So, maybe we need to impose further conditions on $\mathcal E$ and its relation to $\le$.