Sufficient condition for u.c.p. convergence of processes

Question

Assume that all processes are cadlag. I am trying to prove the following claim:

Let a sequence of processes $X_n$ be given. Assume that for all $s$ in a dense subset of $\mathbb R^+$

$ X_n(s) \overset{\mathbb P}{\rightarrow} X(s). $

If the paths of $X_n$ are increasing for all $n$ a.s. and the paths of $X$ are continuous a.s., then we also have, for all $t>0,\epsilon>0$, ucp convergence:

$ \lim_{n\to \infty} \mathbb P \left( \sup_{s\leq t} |X_n(s)-X(s)|>\epsilon\right)=0. $

My attempt was to use that if a sequence of increasing functions converges pointwise to a continuous function, then convergence is uniform. To do so I thought that if the dense subset is countable, you can find a subsequence $n_k$ for which, for all $s$, $X_{n_k}(s) \to X(s)$ a.s.

Then applying the theorem you get ucp convergence along the subsequence. But how can you now conclude that it holds for the original sequence?

Answer 1

Recall that for a sequence $(Y_n)_n$ of random variables the following statements are equivalent:

• $Y_n \to Y$ in probability.
• There exists $Y$ such that for any subsequence of $(Y_n)_{n \in \mathbb{N}}$, we can find a subsequence which converges almost surely to $Y$.

If we set $$Y_n := \sup_{s \leq t} |X_n(s)-X(s)|,$$ then your claim is equivalent to $Y_n \to 0$ in probability.

Consider an arbitrary subsequence $(Y_{n(k)})_k$ of $(Y_n)_n$. Using the considerations mentioned in your question, we can find a subsequence which converges to $0$ almost surely. By the above statement, it follows that $Y_n \to 0$ in probability.