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Proof. To see what the statement means, please look at Derived Categories, Definitions 13.6.1 and 13.36.3. It was shown in Lemmas 15.74.4 and 15.74.5 that $D_{perf}(R) \subset D(R)$ is a strictly full, saturated, triangulated subcategory of $D(R)$. Of course $R \in D_{perf}(R)$.

Recall that $\langle R \rangle = \bigcup \langle R \rangle _ n$. To finish the proof we will show that if $M \in D_{perf}(R)$ is represented by

with $M^ i$ finite projective, then $M \in \langle R \rangle _{b - a + 1}$. The proof is by induction on $b - a$. By definition $\langle R \rangle _1$ contains any finite projective $R$-module placed in any degree; this deals with the base case $b - a = 0$ of the induction. In general, we consider the distinguished triangle

By induction the truncated complex $\sigma _{\leq b - 1}M^\bullet $ is in $\langle R \rangle _{b - a}$ and $M_ b[-b]$ is in $\langle R \rangle _1$. Hence $M^\bullet \in \langle R \rangle _{b - a + 1}$ by definition. $\square$