An argument for getting $L^2$ convergence from $L^1$ convergence and boundedness in $L^2$ This is a theorem proof from Protter's Stochastic Integration and Differential Equations on the completeness of the space $\mathcal{H}^2$ of semimartingales.
Setting: Let $A$ be a (predictable) integrable variation (IV) process with $A_0 = 0$, and with the $L^2$ norm on a probability space we have $$\Vert \int_0^\infty |dA_s| \vert_{L^2} < \infty.$$
We would like to show that this norm space is complete.
Suppose $(A^n)$ is a Cauchy sequence of (predictable) IV processes in $\Vert \cdot \Vert_2$ where $\Vert A \Vert_p = \Vert \int_0^\infty |dA_s| \Vert_{L^p}, p \ge 1$. To show $(A^n)$ converges it suffices to show a subsequence converges. Therefore, without loss of generality we can assume $\sum_n \Vert A^n \Vert_2 < \infty$.
Then $\sum A^n$ converges in $\Vert \cdot \Vert_1$ to a limit $A$. Moreover $$\lim_{m \to \infty} \sum_{n \ge m} \int_0^\infty |dA_s^n| = 0$$ in $L^1$ and is dominated in $L^2$ by $\sum_n \int_0^\infty |dA_s^n|.$ Therefore $\sum A^n$ converges to the limit $A$ in $\Vert \cdot \Vert_2$ as well, and there is a subsequence converging almost surely.
I cannot follow the arguments in bold above. Firstly, how do we get that $\sum A^n$ converges in $\Vert \cdot \Vert_1$ given $\sum_n \Vert A^n \Vert_2 < \infty$? And why do we get that $\sum_{n \ge m} \int_0^\infty |dA_s^n| \to 0$ as $m\to \infty$ in $L^1$?
Finally, how do we get convergence in $\Vert \cdot \Vert_2$ from these results?
 A: There's a lot wrong with this. First of all, that $A^n$ is Cauchy does not imply that we can have $\sum_{n}\|A^n\|_{2} < \infty$ after passing to a subsequence. Consider a constant sequence $A^n = A$ as a counterexample. Secondly, we will not be able to conclude the convergence of $\sum_{n\in\mathbb{N}} A^n$ in any norm unless we have $A^n \to 0$.
What we can do is find a subsequence such that $\sum_{n\in\mathbb{N}} \|A^{n+1}-A^{n}\|_{2} < \infty$. We may then write
$$ A^{n} = A^{1} + \sum_{i=1}^{n-1} (A^{i+1} - A^{i}).$$
Note that for any $t > 0$ we have
$$ |A_t^{i+1}-A_t^{i}|=\left|\int_{0}^{t} dA^{i+1}_s - \int_{0}^{t} dA^{i}_s\right| \leq \int_{0}^{t} |d(A^{i+1}-A^{i})_s|.$$
It follows that
$$ \left\|\sum_{i=1}^{\infty} \sup_{t\geq 0} |A^{i+1}-A^{i}| \right\|_{L^2} \leq \left\|\sum_{i=1}^{\infty} \int_{0}^{\infty} |d(A^{i+1}-A^{i})_s| \right\|_{L^2} \leq \sum_{i=1}^{\infty}\|A^{i+1} - A^{i}\|_2.$$
This shows that $\sum_{i=1}^{n-1} (A^{i+1}-A^{i})$ converges almost surely uniformly as $n\to\infty$ to some predictable $\bar{A}$ with $\|\bar{A}\|_{2} < \infty$. Letting $A := \bar{A}-A^1$, we have that $A^n \to A$ almost surely uniformly on $[0,\infty)$. By Fatou's lemma we obtain
$$ \|A^{n} - A\|_2 \leq \liminf_{m\to\infty} \|A^{n} - A^{m}\|_{2} \leq \liminf_{m\to\infty} \|\sum_{i=n}^{m} (A^{i+1} - A^{i})\|_2 \leq \sum_{i=n}^{\infty} \|A^{i+1}-A^{i}\|_2,$$
which vanishes as $n\to \infty$.
