Is the space of continuous local martingales equipped with the topology of uniform convergence on compact sets complete? Let


*

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$

*$M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ denote the set of continuous local $\mathcal F$-martingales on $(\Omega,\mathcal A,\operatorname P)$


Is $M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ equipped with the topology of uniform convergence on compact sets$^1$ complete?

$^1$ i.e. If $(M^n)_{n\in\mathbb N}\subseteq M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ and $M\in M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$, then $M_n\xrightarrow{n\to\infty}M$ in $M_{c,\:\text{loc}}(\mathcal F,\operatorname P)$ if and only if $$\sup_{0\le s\le t}\left|M^n_s-M_s\right|\xrightarrow{n\to\infty}0\;\;\;\text{in probability for all }t\ge 0\tag1.$$
 A: Yes, the space of continuous local martingales equipped with ucp convergence is complete.
Let $(M^n)_{n \in \mathbb{N}} \subseteq M_{c,\text{loc}}(\mathcal{F},P)$ be a Cauchy sequence with respect to the ucp convergence. Then we can choose for any fixed $T>0$ numbers $n_k \in \mathbb{N}$, $n_1 < n_2 < \ldots$, such that
$$\mathbb{P} \left( \sup_{t \leq T} |M^{n_k}_s-M_s^{n_{k-1}}|>2^{-k} \right) \leq 2^{-k}.$$
By the Borel Cantelli lemma, this implies that
$$\mathbb{P} \left( \sup_{t \leq T} |M^{n_k}_t-M_t^{n_{k-1}}|>2^{-k} \, \, \text{for infinitely many $k$} \right)=0.$$
Using that
$$M_t^n - M_t^m = \sum_{k=m+1}^n (M_t^k-M_t^{k-1}), \qquad m<n,$$
it is not difficult to see that this implies that $(M_t^{n_k}(\omega))_{t \leq T}$ is a Cauchy sequence in $C([0,T],\|\cdot\|_{\infty})$ for $\mathbb{P}$-almost all $\omega$. By the completeness of $C[0,T]$, we find that
$$M_t(\omega) := \lim_{k \to \infty} M_t^{n_k}(\omega)$$
exists almost surely and defines a process with continuous sample paths. Since $(M^n)_{n \in \mathbb{N}}$ is a Cauchy sequence with respec to the ucp convergence and $M^{n_k} \to M$ uniformly on $[0,T]$ with probability $1$, it follows that
$$\forall \epsilon>0: \quad \lim_{n \to \infty} \mathbb{P} \left( \sup_{t \leq T} |M_t^n-M_t|> \epsilon \right)=0,$$
and so $M^n \to M$ in ucp. Consequently, $M$ is the ucp limit of a sequence of continuous local martingales, and so $M$ is itself a local martingale (see Theorem 5 here).
