**Proof.**
Part (1) follows as $E_0^{p, n - p} = \text{gr}^ p K^ n$. Part (2) is clear from Equation (12.24.5.1). We will use Lemma 12.24.10 to prove that the spectral sequence weakly converges. Fix $p, n \in \mathbf{Z}$. The right hand side of (12.24.6.1) is equal to $F^ pK^ n \cap \mathop{\mathrm{Ker}}(d) + F^{p + 1}K^ n$ because $F^{p + r}K^ n = 0$ for $r \gg 0$. Thus (12.24.6.1) is an equality. The left hand side of (12.24.6.2) is equal to $F^ pK^ n \cap \mathop{\mathrm{Im}}(d) + F^{p + 1}K^ n$ because $F^{p - r + 1}K^{n - 1} = K^{n - 1}$ for $r \gg 0$. Thus (12.24.6.2) is an equality. Since the filtration on $H^ n(K^\bullet )$ is finite by (2) we see that we have abutment. To prove we have convergence we have to show the spectral sequence is regular which follows as it is bounded (Lemma 12.24.8) and we have to show that $H^ n(K^\bullet ) = \mathop{\mathrm{lim}}\nolimits _ p H^ n(K^\bullet )/F^ pH^ n(K^\bullet )$ which follows from the fact that the filtration on $H^*(K^\bullet )$ is finite proved in part (2).

Proof of (4). Assume that for some $r \geq 0$ we have $E_ r^{p, q} \in \mathcal{C}$ for some weak Serre subcategory $\mathcal{C}$ of $\mathcal{A}$. Then $E_{r + 1}^{p, q}$ is in $\mathcal{C}$ as well, see Lemma 12.10.3. By boundedness proved above (which implies that the spectral sequence is both regular and coregular, see Lemma 12.24.8) we can find an $r' \geq r$ such that $E_\infty ^{p, q} = E_{r'}^{p, q}$ for all $p, q$ with $p + q = n$. Thus $H^ n(K^\bullet )$ is an object of $\mathcal{A}$ which has a finite filtration whose graded pieces are in $\mathcal{C}$. This implies that $H^ n(K^\bullet )$ is in $\mathcal{C}$ by Lemma 12.10.3.
$\square$

## Comments (2)

Comment #7055 by Xiaolong Liu on

Comment #7245 by Johan on