We are reading J. E. Björk's book: Rings of Differential Operators and we don't understand one step at Lemma 3.4:
Let $\Gamma$ and $\Omega$ be two filtrations on the left $A_n(K)$-module $M$ and assume $\Gamma$ is good (meaning the associated graded module is finitely generated over the graded ring $gr(A_n(K))$). Then there exists an integer $w$ such that $\Gamma_w \subset \Omega_{v + w}$ for all $v$.
In the proof we put: $R_v = \mathscr{T}_v\Gamma_0 + \dots + \mathscr{T}_{v-v_0}\Gamma_{v_0}$. And the argument goes: as $v$ is increasing, the sets $R_v$ are also increasing. But how can it be that $R_v \supset \Gamma_v$ , since every summand in $R_v$ is in $\Gamma_v$ (by the definition of this kind of filtration)?
Thanks in advance!
You've probably figured this out by now, but for future readers, I think your observation that $R_v\subseteq\Gamma_v$ is correct, and in fact, given the proof in Lemma 3.4 that $R_v\supseteq\Gamma_v$, equality holds $R_v = \Gamma_v,\forall v\ge v_0$.
cf. Proposition 8.3.1 in S. C. Coutinho's A Primer of Algebraic D-Modules, where Coutinho shows that for a good filtration $\Gamma_{i+k}\subseteq B_i\Gamma_k$ (the reverse containment, $\Gamma_{i+k}\supseteq B_i\Gamma_k$, being automatic as part of the definition of a filtration w.r.t $\mathcal B$),
3.1 PROPOSITION. Let $M$ be a left $A_n$-module. A filtration $\Gamma$ of $M$ with respect to $\mathcal B$ (the Bernstein filtration) is good if and only if there exists $k_0$ such that $\Gamma_{i+k} = B_i\Gamma_k$ for all $k > k_0$.