**Proof.**
Let $K$ be in $D_\mathcal {A}(\mathcal{O})$. By Lemma 21.25.2 we have $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$. By Lemma 21.23.3 we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*(\tau _{\geq -n}K)$. Let $V' \in \mathcal{B}'$ and let $\{ V'_ i \to V'\} $ be an element of $\text{Cov}_{V'}$. Then we consider

\[ H^ j(V'_ i, Rf_*K) = H^ j(u(V'_ i), K) \quad \text{and}\quad H^ j(V'_ i, Rf_*(\tau _{\geq -n}K)) = H^ j(u(V'_ i), \tau _{\geq -n}K) \]

The assumption in Situation 21.25.5 implies that the last group is independent of $n$ for $n$ large enough depending on $j$ and $d_{V'}$. Some details omitted. We apply this for $j$ and $j - 1$ and via Lemma 21.23.2 this gives that

\[ H^ j(V'_ i, Rf_*K) = \mathop{\mathrm{lim}}\nolimits H^ j(V'_ i, Rf_*(\tau _{\geq -n} K)) \]

and the system on the right is constant for $n$ larger than a constant depending only on $d_{V'}$ and $j$. Thus Lemma 21.23.6 implies that

\[ H^ j(Rf_*K)(V') \longrightarrow \left(\mathop{\mathrm{lim}}\nolimits H^ j(Rf_*(\tau _{\geq -n}K))\right)(V') \]

is injective. Since the elements $V' \in \mathcal{B}'$ cover every object of $\mathcal{C}'$ we conclude that the map $H^ j(Rf_*K) \to \mathop{\mathrm{lim}}\nolimits H^ j(Rf_*(\tau _{\geq -n}K))$ is injective. The spectral sequence

\[ E_2^{p, q} = R^ pf_*H^ q(\tau _{\geq -n}K) \]

converging to $H^{p + q}(Rf_*(\tau _{\geq -n}K))$ (Derived Categories, Lemma 13.21.3) and assumption (3) show that $H^ j(Rf_*(\tau _{\geq -n}K))$ is constant for $n \geq N - j$. Hence $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is injective for $j \geq N - n$.

Thus we proved the lemma with “isomorphism” in the last line of the lemma replaced by “injective”. However, now choose $j$ and $n$ with $j \geq N - n$. Then consider the distinguished triangle

\[ \tau _{\leq -n - 1}K \to K \to \tau _{\geq -n}K \to (\tau _{\leq -n - 1}K)[1] \]

See Derived Categories, Remark 13.12.4. Since $\tau _{\geq -n}\tau _{\leq -n -1}K = 0$, the injectivity already proven for $\tau _{-n - 1}K$ implies

\[ 0 = H^ j(Rf_*(\tau _{\leq -n - 1}K)) = H^{j + 1}(Rf_*(\tau _{\leq -n - 1}K)) = H^{j + 2}(Rf_*(\tau _{\leq -n - 1}K)) = \ldots \]

By the long exact cohomology sequence associated to the distinguished triangle

\[ Rf_*(\tau _{\leq -n - 1}K) \to Rf_*K \to Rf_*(\tau _{\geq -n}K) \to Rf_*(\tau _{\leq -n - 1}K)[1] \]

this implies that $H^ j(Rf_*K) \to H^ j(Rf_*(\tau _{\geq -n}K))$ is an isomorphism.
$\square$

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