Related : How do we formally define "j-th smallest element"?
Let $F_j:\mathbb{R}^n\rightarrow \mathbb{R}$ be a $j$-th smallest element picking function.
How do I prove that $F_j$ is Borel measurable?
I know how to prove it for the case $j=1$ and $j=n$, but I am not sure how to prove this for the case $1<j<n$.
Thank you in advance!
Fleshing out the comment so that the question won't get stuck in the unanswered list.
Let $\{f_1,f_2,\ldots,f_n\}$ be the collection of measurable functions, $f_i:\Bbb{R}^n\to\Bbb{R}$. Let $S$ be the collection of subsets of $\{1,2,\ldots,n\}$ that have exactly $j$ elements. Observe that $|S|=\binom nj$, so $S$ is a finite collection.
To each set $M\in S$ we can define the maximum $$ f_M:\Bbb{R}^n\to\Bbb{R}, x\mapsto\max\{f_i(x)\mid i\in M\}. $$ As a maximum of a finite set of measurable functions we can conclude that $f_M$ is measurable for each $M\in S$.
Your claim then follows from the observation that the function $F_j(x)$ picking $j$th smallest from the list $(f_1(x),f_2(x),\ldots,f_n(x))$ for all $x$, can be written as $$ F_j=\min\{f_M\mid M \in S\}. $$
As the minimum of a finite collection of measurable functions this is immediately seen to be measurable.