Does $\sup_{t \leq T} |M_{n_k}(t)-M_{m_k}(t)|\to 0$ imply $\lim_k M_{n_k}(t)$ exists and is continuous? This came up in proving that $\mathcal{M}^2_c$ is a complete metric space using the invariant metric induced by
$$
||M|| = \sum_k \frac{||M(k)||_2\wedge 1}{2^n}.
$$
Suppose $M_n(t)$ is a sequence of continuous martingales such that
$$
E \sup_{t \leq T} |M_n(t)-M_m(t)|^2 \to 0
$$
as $m,n \to \infty$.
I would like to say "then $M_n$ is uniformly Cauchy along a subsequence" and hence converges to a continuous limit, but I stumble when I actually try to put to words why.
Certainly, for any $n_k,m_k \to \infty$ I may choose a subsequence of indices (WLOG the whole sequence) such that $\sup_{t \leq T} |M_{n_k}(t)-M_{m_k}(t)|\to 0$ a.s. Is this enough to conclude that the a.s. limit $\lim_k M_{n_k}(t)$ is continuous? I feel like there should be some simple way to show that $N_k := M_{n_k}$ is actually uniformly Cauchy which would finish the proof.
Edit: Obviously if $m_k=n_k$ then we can't conclude anything. Assume $m_k=n_k+1$ for all $k$.
 A: By assumption,
$$\mathbb{E}\left( \sup_{t \leq T} |M_n(t)-M_m(t)|^2 \right) \to 0$$
as $m,n \to \infty$ which means that $(M_n)_n$ is a Cauchy sequence in $L^2(\Omega,\mathbb{P};C[0,T])$, i.e. the space of functions
$$\Omega \ni \omega \mapsto f(\omega) \in C[0,T]$$
endowed with the norm
$$\|f\|_{L^2}^2 = \int \|f(\omega)\|_{\infty}^2 \, d\mathbb{P}(\omega) = \int \sup_{t \leq T} |f(\omega,t)|^2 \, d\mathbb{P}(\omega) \tag{1}$$
where $\|\cdot\|_{\infty}$ denotes the supremum norm. Since this space is complete (see Lemma below), we conclude that there exists $M \in L^2(\Omega,\mathbb{P};C[0,T])$ such that $M_n \to M$ in $L^2(\Omega,\mathbb{P};C[0,T])$, i.e.
$$\mathbb{E}\left( \sup_{t \leq T} |M_n(t)-M(t)|^2 \right) \to 0$$
as $n \to \infty$. Consequently, there exists a subsequence such that
$$\sup_{t \leq T} |M_{n(k)}(t)-M(t)| \to 0 \quad (k \to \infty)$$
This means that $M$ is a uniform limit of continuous functions, thus continuous.

Lemma $L^2(\Omega,\mathbb{P};C[0,T])$ endowed with the norm defined in $(1)$ is complete.
Proof: Note that the mapping defined in $(1)$ is indeed a norm. It suffices to show that for any sequence $(f_n)_n \subseteq L^2(\Omega,\mathbb{P};C[0,T])$ such that $$\sum_{k \geq 1} \|f_k\|_{L^2} < \infty$$ the limit $\lim_{n \to \infty} \sum_{k=1}^n f_k$ exists.  By triangle inequality, we have
$$\left( \int \left( \sum_{j=1}^n \|f_j\|_{\infty} \right)^2 \, d\mathbb{P} \right)^{\frac{1}{2}} = \left\| \sum_{j=1}^n \|f_j\|_{\infty} \right\|_{L^2(\mathbb{P})} \leq \sum_{j=1}^n \bigg\| \|f_j\|_{\infty} \bigg\|_{L^2(\mathbb{P})} \leq \sum_{j \geq 1} \|f_j\|_{L^2}$$
By the monotone convergence theorem, we conclude that
$$\int \bigg(\underbrace{\sum_{j \geq 1} \|f_j\|_{\infty}}_{=:g}\bigg)^2 \, d\mathbb{P}< \infty$$
in particular $g<\infty$ a.s.. Therefore,
$$f(\omega,t) := \sum_{j \geq 1} f_j(\omega,t) \qquad (\omega \in \Omega, t \leq T)$$
exists a.s. and the series converges a.s. uniformly in $t$. Since
$$\left\| f(\omega)- \sum_{j=1}^n f_j(\omega) \right\|_{\infty} \leq \sum_{j \geq n+1} \|f_j(\omega)\|_{\infty} \leq g(\omega) \in L^2(\mathbb{P})$$
the claim follows by the dominated convergence theorem.
