Is the set $\{ (X_n)_{n \in \mathbb{N}} \text{ has a nondecreasing subsequence} \}$ measurable?
In general, the set described need not be measurable, but it will always be universally measurable. Universally measurable sets are those subsets of $\Omega$ which lie in the completion of $\mathcal{A}$ with respect to every sigma-finite measure. That is, for a sigma-finite measure $\mu$ on $(\Omega,\mathcal{A})$, let $\mathcal{A}_\mu$ be the completion. This is the sigma-algebra of sets of the form $B\cup C$ where $B\in\mathcal{A}$ and $C$ is contained in a set in $\mathcal{A}$ of zero $\mu$-measure. The universal completion of $\mathcal{A}$ is $$ \overline{\mathcal{A}}=\bigcap_\mu\mathcal{A}_\mu $$ where the intersection is over all sigma-finite measures on $(\Omega,\mathcal{A})$. The set $A$ in the question can be shown to be in $\mathcal{A}_\mu$ for each such $\mu$ and, in particular, is in $\overline{\mathcal{A}}$. This means that it has a well-defined measure with respect to any sigma-finite measure $\mu$. However, it need not be in $\mathcal{A}$.
We can exactly classify the possible sets $A$ in terms of analytic sets. There are many equivalent definitions of analytic sets, but I will take the following for now (which makes sense for any sigma-algebra $\mathcal{A}$).
A set $A\subseteq\Omega$ is $\mathcal{A}$-analytic if and only if it is the projection of an $\mathcal{A}\otimes\mathcal{B}(\mathbb{R})$ measurable subset of $\Omega\times\mathbb{R}$ onto $\Omega$. i.e., $$ A = \left\{x\in\Omega\colon(x,y)\in S{\rm\ for\ some\ }y\in\mathbb{R}\right\} $$ for some $S\in\mathcal{A}\otimes\mathcal{B}(\mathbb{R}).$
The definition given by Wikipedia is for the case where $\Omega$ is a Polish space and $\mathcal{A}$ its Borel sigma-algebra, but this definition can be applied to any measurable space $(\Omega,\mathcal{A})$. A nice introduction to analytic sets is given in appendix A5 of Stochastic Integration with Jumps by Klaus Bichteler, available free online on his homepage.
Useful well-known properties of analytic sets are the following.
- Every $\mathcal{A}$-analytic set is in the universal completion $\overline{\mathcal{A}}$.
- If $\Omega$ is an uncountable Polish space with Borel sigma-algebra $\mathcal{A}$, (for example, $\Omega=\mathbb{R}$, $\mathcal{A}=\mathcal{B}(\mathbb{R})$) then there exists $\mathcal{A}$-analytic sets which are not in $\mathcal{A}$.
The following classifies the sets $A$ in the question in terms of analytic sets.
Lemma: The following are equivalent.
- There is a sequence of real-valued random variables $X_n$ on $(\Omega,\mathcal{A})$ such that $$A=\left\{\omega\in\Omega\colon n\mapsto X_n(\omega){\rm\ has\ a\ nondecreasing\ subsequence}\right\}\qquad{\rm(1)}$$
- $A$ is $\mathcal{A}$-analytic.
Once we have proven this lemma, then the statements above about the measurability of $A$ follow.
To prove that the first statement of the lemma implies the second, consider $A$ given by (1). Define $S\subseteq\Omega\times(0,\infty]$ by $$ S=\left\{(\omega,x)\colon X_n(\omega){\rm\ has\ a\ nondecreasing\ subsequence\ tending\ to\ }x\right\}. $$ This can be shown to be in $\mathcal{A}\otimes\mathcal{B}((0,\infty])$ and its projection onto $\Omega$ is $A$. So, $A$ is analytic.
It just remains to prove that the second statement of the lemma implies the first. For this, I will use the alternative definition of an analytic set $A$ as given by the Suslin operation. This can be shown to be equivalent to the definition above. Let $\omega=\{0,1,2,\ldots\}$ denote the natural numbers, $\omega^{\lt\omega}=\bigcup_{n=1}^\infty\omega^n$ denote the nonempty finite sequences in $\omega$, and $\omega^\omega$ denote the infinite sequences in $\omega$. For each $x\in\omega^\omega$ and positive integer $n$, write $x\vert_n\in\omega^{\lt\omega}$ for the initial sequence of length $n$ from $x$, $$ x\vert_n = (x_0,x_1,\ldots,x_{n-1}). $$ A Suslin scheme $P$ is a collection of sets $P_x\in\mathcal{A}$ over $x\in\omega^{<\omega}$ and the Suslin operation maps this to $$ A=\bigcup_{x\in\omega^\omega}\bigcap_{n=1}^\infty P_{x\vert_n}. $$ Every $\mathcal{A}$-analytic set can be expressed in this form. We can express $A$ as a projection of a reasonably nice subset of $\Omega\times\mathbb{R}$. Start by choosing a collection of intervals $U_x=[a_x,b_x)\subseteq[0,1)$ over $x\in\omega^{\lt\omega}$ satisfying the properties
- $\bigcap_{n=1}^\infty U_{x\vert_n}\not=\emptyset$ for all $x\in\omega^\omega$.
- $U_x\cap U_y=\emptyset$ unless $x=z\vert_m$ and $y=z\vert_n$ for some $z\in\omega^\omega$ and positive integers $m,n$.
For example, we can take $$ a_{x_0,\ldots,x_n}=\sum_{k=0}^n2^{-k-x_0-\ldots-x_{k-1}}(1-2^{-x_k}) $$ and $b_{x_0,\ldots,x_n}=a_{x_0,\ldots,x_n+1}$.
Define $S\subseteq\Omega\times\mathbb{R}$ by $$ S=\bigcap_{n=1}^\infty\bigcup_{x\in\omega^n}P_x\times U_x. $$ Then, $A$ is the projection of $S$ onto $\Omega$. Set $$ S_n=\bigcap_{k=1}^n\bigcup_{x\in\omega^k}P_x\times U_x. $$ Then, $S_n$ decreases to $S$ and the paths $X_n(\omega,t)=1_{\{(\omega,t)\in S_n\}}$ are right-continuous. The set $A$ is precisely the set of $\omega\in\Omega$ such that the process $\sum_{n=1}^\infty 2^{-n}X_n(\omega,t)$ hits $1$. This relates the question to measurability of hitting times of right-continuous processes as mentioned in my comment to the original question.
For each positive integer $n$, define a sequence $Y_{n,m}$ of random variables over $m=0,1,2,\ldots$ by \begin{align} Y_{n,0}(\omega)&=1.\\ Y_{n,m+1}(\omega)&=\sup\{x\in[0,Y_{n,m}(\omega)-1/n]\colon (\{\omega\}\times[x,Y_{n,m}(\omega)])\cap S_n\not=\emptyset\} \end{align}
Then $m\mapsto Y_{n,m}$ are decreasing sequences of random variables and, using the convention $\sup\emptyset=-\infty$, each sequence is constant at $-\infty$ after a finite number of steps. It can be seen that a point $(\omega,t)\in S$ if and only if there is a subsequence $Y_{n_k,m_k}(\omega)$ strictly decreasing to $t$.
Finally, let $X_k(\omega)$ be the sequence of random variables $Y_{n,m}(\omega)$ in order of increasing $m$ and then increasing $n$, after each value of $-\infty$ and each repeated value is removed. In case this terminates, we can set $X_k(\omega)=k$ once there are no remaining values left in the sequence. Then, a point $(\omega,t)$ is in $S$ if and only if $X_k(\omega)$ has a subsequence decreasing to $t$, and $A$ is precisely the set of $\omega\in\Omega$ for which $-X_k(\omega)$ has a nondecreasing subsequence.
Finally, I'll point out that although I showed above that there are cases where $A$ is not in $\mathcal{A}$, the construction above would give rather contrived examples. However, the existence of any single counterexample implies that any sufficiently `generic' construction of the space $(\Omega,\mathcal{A})$ will also give $A\not\in\mathcal{A}$.
For example, consider the following standard construction of a space with an infinite sequence of random variables. Let $\Omega=\mathbb{R}^{\mathbb{N}}$ be the space of infinite sequences $\omega=(\omega_1,\omega_2,\ldots)$ of real numbers. Then, for each positive integer $n\in\mathbb{N}$, define $X_n\colon\Omega\to\mathbb{R}$ by $X_n(\omega)=\omega_n$. Finally, let $\mathcal{A}$ be the sigma-algebra generated by the $X_n$. i.e., $\mathcal{A}$ is generated by the sets $X_n^{-1}(S)$ for $n\in\mathbb{N}$ and $S\in\mathcal{B}(\mathbb{R})$. Then, $$ A = \left\{\omega\in\Omega\colon X_n(\omega){\rm\ has\ a\ nondecreasing\ subsequence}\right\} $$ is not in $\mathcal{A}$.
To see this, consider any counterexample $(\tilde\Omega,\mathcal{\tilde A})$ with random variables $\tilde X_n\colon\tilde\Omega\to\mathbb{R}$ such that the set $\tilde A$ on which $\tilde X_n$ has a nondecreasing subsequence is not measurable. Define the map $f\colon\tilde\Omega\to\Omega$ by $f(\tilde\omega)=\omega$ where $\omega_n=\tilde X_n(\tilde\omega)$. Then, $f$ is measurable and $\tilde A=f^{-1}(A)$. If $A$ was in $\mathcal{A}$ then this implies that $\tilde A\in\mathcal{\tilde A}$, a contradiction.