I am reading the proof of the uniqueness of the solution to an SDE given locally Lipschitz conditions. Here, the proof proceeds by cutting the drift and diffusion coefficients on $|x|\le R$. So we have $X^R$ as a solution on this restriction and for $S>R$, $X^S$ is a solution for the restriction on $|x| \le S$.
The proof claims that once we have $$\lim_{R \to \infty}P(\sup_{t \le T}|X_t^R - X_t^S|>0) \to 0$$ shows that $X_t:= \limsup_{R \to \infty} X_t^R$ exists uniformly for $t \in [0,T]$.
More generally, I have seen from texts in Stochastic Calculus use the fact that if $P(\sup_{0 \le t \le T} |X_t^n - X_t^m|>c) \le 0$ as $n,m\to \infty$ for every $c>0$, then there exists $X=X_t$ such that $\sup_{0\le t \le T} |X_t^n - X_t| \to 0$ in probability.
I am trying to work out the details for this. How do we get from uniform Cauchy in probability a uniform limit in probability? I can get a pointwise limit for each $t$, but I can't figure out how to show that this limit will be uniform over compact sets.
Let me give a proof of the more general claim. Write $|f|_\infty=\sup_{0\leq t\leq T}|f(t)|$ and remember that this norm renders the space of continuous functions $[0,T]\to\mathbb{R}^d$ complete. Since $P(|X^n-X^m|_\infty>c)\to 0$ as $m,n\to\infty$ for any $c>0$, we can find a subsequence such that $P(|X^{n_{k+1}}-X^{n_k}|_\infty>2^{-k})\leq 2^{-k}$ for all $k\in\mathbb{N}$. By Borel-Cantelli, on a set of probability $1$, $X^{n_k}$ defines a Cauchy sequence in a Banach space, whence on this set the limit $X$ exists. Finally note that $$ P(|X^n-X|_\infty>c)\leq P(|X^n-X^{n_k}|_\infty>c)+P(|X^{n_k}-X|_\infty>c) $$ which vanishes by taking $k,n\to\infty$.