Show martingale space is a Banach space 
Let $\mathcal{H}^1 = \{M \in \mathcal{M}; E[\sup_{t\geq 0} |M_t|] < \infty\}$, where $\mathcal{M}$ is the space of right continuous with left limits martingales. Show that $\mathcal{H}^1$ is complete.

So far I have used the completeness of $L^1$ for each $t$ to get a candidate process for $\mathcal{H}^1$. I had to take a right continuous version in order for this to make sense. But I am having trouble showing that this process is in $\mathcal{H}^1$. Any help would be appreciated. Thanks.
 A: Lemma: On the space $\tilde{\Omega} := \{f: [0,\infty) \times \Omega \to \mathbb{R}; f$ càdlàg$\}$ we define a norm by
$$\|f\|_{L^1} := \mathbb{E}(\|f\|_{\infty}) := \mathbb{E} \left( \sup_{t \geq 0} |f(t)| \right).$$
Then $$L^1(\Omega; D[0,\infty)) := \{f \in \tilde{\Omega}; \|f\|_{L^1}<\infty\}$$ is a complete normed space.
Proof: As a composition of two norms, $\|\cdot\|_{L^1}$ defines a norm. In order to show completeness, it suffices to prove that for any sequence $(f_n)_n \subseteq L^1(\Omega,D[0,\infty))$ such that $$\sum_{k \geq 1} \|f_k\|_{L^1} < \infty$$ the limit $\lim_{n \to \infty} \sum_{k=1}^n f_k$ exists.  By triangle inequality, we have
$$\int \left( \sum_{j=1}^n \|f_j\|_{\infty} \right) \, d\mathbb{P} = \left\| \sum_{j=1}^n \|f_j\|_{\infty} \right\|_{L^1(\mathbb{P})} \leq \sum_{j=1}^n \bigg\| \|f_j\|_{\infty} \bigg\|_{L^1(\mathbb{P})} \leq \sum_{j \geq 1} \|f_j\|_{L^1}$$
By the monotone convergence theorem, we conclude that
$$\int \bigg(\underbrace{\sum_{j \geq 1} \|f_j\|_{\infty}}_{=:g}\bigg) \, d\mathbb{P}< \infty$$
in particular $g<\infty$ a.s.. Therefore,
$$f(\omega,t) := \sum_{j \geq 1} f_j(\omega,t) \qquad (\omega \in \Omega, t \geq 0)$$
exists a.s. and the series converges a.s. uniformly in $t$. In particular, $f$ has càdlàg sample paths. Since
$$\left\| f(\omega)- \sum_{j=1}^n f_j(\omega) \right\|_{\infty} \leq \sum_{j \geq n+1} \|f_j(\omega)\|_{\infty} \leq g(\omega) \in L^1(\mathbb{P})$$
the claim follows by the dominated convergence theorem.

Now let $(M^n)_{n \in \mathbb{N}} \subseteq \mathcal{H}^1$ be a Cauchy sequence. Then $(M^n)_{n \in \mathbb{N}} \subseteq L^1(\Omega, D[0,\infty))$ and therefore there  exists by the above lemma $M \in L^1(\Omega, D[0,\infty))$ such that $\|M^n-M\|_{L^1} \to 0$ as $n \to \infty$. It remains to show that $M$ is a martingale. This follows directly from the fact that $\|M^n-M\|_{L^1} \to 0$ implies $\|M^n(t)-M(t)\|_{L^1(\mathbb{P})} \to 0$ for all $t \geq 0$.
