Fix $m$. Let $\mathfrak q \in T$. By Lemmas \ref{lemma-bootstrap-inherited} and \ref{lemma-algebraize-local-cohomology-bis} we see that $$ H^i_\mathfrak q(M_\mathfrak q) \longrightarrow \lim H^i_\mathfrak q(M_\mathfrak q/I^nM_\mathfrak q) $$ is an isomorphism for $i \leq s$. The inverse systems $\{H^i_\mathfrak q(I^nM_\mathfrak q)\}_{n \geq 0}$ and $\{H^i_\mathfrak q(M/I^nM)\}_{n \geq 0}$ satisfy the Mittag-Leffler condition for all $i$, see Lemma \ref{lemma-ML-local}. Thus looking at the inverse system of long exact sequences $$ 0 \to H^0_\mathfrak q(I^nM_\mathfrak q) \to H^0_\mathfrak q(M_\mathfrak q) \to H^0_\mathfrak q(M_\mathfrak q/I^nM_\mathfrak q) \to H^1_\mathfrak q(I^nM_\mathfrak q) \to H^1_\mathfrak q(M_\mathfrak q) \to \ldots $$ we conclude (some details omitted) that there exists an integer $m'(m, \mathfrak q) \geq m$ such that for all $k \geq m'(m, \mathfrak q)$ the map $H^i_\mathfrak q(I^kM_\mathfrak q) \to H^i_\mathfrak q(I^mM_\mathfrak q)$ is zero for $i \leq s$ and the image of $H^{s + 1}_\mathfrak q(I^kM_\mathfrak q) \to H^{s + 1}_\mathfrak q(I^mM_\mathfrak q)$ is independent of $k \geq m'(m, \mathfrak q)$ and maps injectively into $H^{s + 1}_\mathfrak q(M_\mathfrak q)$. \medskip\noindent Suppose we can show that $m'(m, \mathfrak q)$ can be chosen independently of $\mathfrak q \in T$. Then the lemma follows immediately from Local Cohomology, Lemmas \ref{local-cohomology-lemma-zero} and \ref{local-cohomology-lemma-essential-image}. \medskip\noindent Let $\omega_A^\bullet$ be a dualizing complex. Let $\delta : \Spec(A) \to \mathbf{Z}$ be the corresponding dimension function. Recall that $\delta$ attains only a finite number of values, see Dualizing Complexes, Lemma \ref{dualizing-lemma-universally-catenary}. Claim: for each $d \in \mathbf{Z}$ the integer $m'(m, \mathfrak q)$ can be chosen independently of $\mathfrak q \in T$ with $\delta(\mathfrak q) = d$. Clearly the claim implies the lemma by what we said above. \medskip\noindent Pick $\mathfrak q \in T$ with $\delta(\mathfrak q) = d$. Consider the ext modules $$ E(n, j) = \text{Ext}^j_A(I^nM, \omega_A^\bullet) $$ A key feature we will use is that these are finite $A$-modules. Recall that $(\omega_A^\bullet)_\mathfrak q[-d]$ is a normalized dualizing complex for $A_\mathfrak q$ by definition of the dimension function associated to a dualizing complex, see Dualizing Complexes, Section \ref{dualizing-section-dimension-function}. The local duality theorem (Dualizing Complexes, Lemma \ref{dualizing-lemma-special-case-local-duality}) tells us that the $\mathfrak qA_\mathfrak q$-adic completion of $E(n, -d - i)_\mathfrak q$ is Matlis dual to $H^i_\mathfrak q(I^nM_\mathfrak q)$. Thus the choice of $m'(m, \mathfrak q)$ for $i \leq s$ in the first paragraph tells us that for $k \geq m'(m, \mathfrak q)$ and $j \geq -d - s$ the map $$ E(m, j)_\mathfrak q \to E(k, j)_\mathfrak q $$ is zero. Since these modules are finite and nonzero only for a finite number of possible $j$ (small detail omitted), we can find an open neighbourhood $W \subset \Spec(A)$ of $\mathfrak q$ such that $$ E(m, j)_{\mathfrak q'} \to E(m'(m, \mathfrak q), j)_{\mathfrak q'} $$ is zero for $j \geq -d - s$ for all $\mathfrak q' \in W$. Then of course the maps $E(m, j)_{\mathfrak q'} \to E(k, j)_{\mathfrak q'}$ for $k \geq m'(m, \mathfrak q)$ are zero as well. \medskip\noindent For $i = s + 1$ corresponding to $j = - d - s - 1$ we obtain from local duality and the results of the first paragraph that $$ K_{k, \mathfrak q} = \Ker(E(m, -d - s - 1)_\mathfrak q \to E(k, -d - s - 1)_\mathfrak q) $$ is independent of $k \geq m'(m, \mathfrak q)$ and that $$ E(0, -d - s - 1)_\mathfrak q \to E(m, -d - s - 1)_\mathfrak q/K_{m'(m, \mathfrak q), \mathfrak q} $$ is surjective. For $k \geq m'(m, \mathfrak q)$ set $$ K_k = \Ker(E(m, -d - s - 1) \to E(k, -d - s - 1)) $$ Since $K_k$ is an increasing sequence of submodules of the finite module $E(m, -d - s - 1)$ we see that, at the cost of increasing $m'(m, \mathfrak q)$ a little bit, we may assume $K_{m'(m, \mathfrak q)} = K_k$ for $k \geq m'(m, \mathfrak q)$. After shrinking $W$ further if necessary, we may also assume that $$ E(0, -d - s - 1)_{\mathfrak q'} \to E(m, -d - s - 1)_{\mathfrak q'}/K_{m'(m, \mathfrak q), \mathfrak q'} $$ is surjective for all $\mathfrak q' \in W$ (as before use that these modules are finite and that the map is surjective after localization at $\mathfrak q$). \medskip\noindent Any subset, in particular $T_d = \{\mathfrak q \in T \text{ with }\delta(\mathfrak q) = d\}$, of the Noetherian topological space $\Spec(A)$ with the endowed topology is Noetherian and hence quasi-compact. Above we have seen that for every $\mathfrak q \in T_d$ there is an open neighbourhood $W$ where $m'(m, \mathfrak q)$ works for all $\mathfrak q' \in T_d \cap W$. We conclude that we can find an integer $m'(m, d)$ such that for all $\mathfrak q \in T_d$ we have $$ E(m, j)_\mathfrak q \to E(m'(m, d), j)_\mathfrak q $$ is zero for $j \geq -d - s$ and with $K_{m'(m, d)} = \Ker(E(m, -d - s - 1) \to E(m'(m, d), -d - s - 1))$ we have $$ K_{m'(m, d), \mathfrak q} = \Ker(E(m, -d - s - 1)_{\mathfrak q} \to E(k, -d - s - 1)_{\mathfrak q}) $$ for all $k \geq m'(m, d)$ and the map $$ E(0, -d - s - 1)_\mathfrak q \to E(m, -d - s - 1)_\mathfrak q/K_{m'(m, d), \mathfrak q} $$ is surjective. Using the local duality theorem again (in the opposite direction) we conclude that the claim is correct. This finishes the proof.