**Proof.**
Let us consider the set

\[ B = \{ \mathfrak p \not\in V(I),\ \mathfrak p \in T,\text{ and } \text{depth}(M_\mathfrak p) \leq s\} \]

Choose $J_0 \subset J$ such that $V(J_0)$ is the closure of $B \cup V(J)$.

Claim I: $V(J_0) \cap V(I) = V(J) \cap V(I)$.

Proof of Claim I. The inclusion $\supset $ holds by construction. Let $\mathfrak p$ be a minimal prime of $V(J_0)$. If $\mathfrak p \in B \cup V(J)$, then either $\mathfrak p \in T$ or $\mathfrak p \in V(J)$ and in both cases $V(\mathfrak p) \cap V(I) \subset V(J) \cap V(I)$ as desired. If $\mathfrak p \not\in B \cup V(J)$, then $V(\mathfrak p) \cap B$ is dense, hence infinite, and we conclude that $\text{depth}(M_\mathfrak p) < s$ by Local Cohomology, Lemma 51.9.2. In fact, let $V(\mathfrak p) \cap B = \{ \mathfrak p_\lambda \} _{\lambda \in \Lambda }$. Pick $\mathfrak q_\lambda \in V(\mathfrak p_\lambda ) \cap V(J) \cap V(I)$ as in (3). Let $\delta : \mathop{\mathrm{Spec}}(A) \to \mathbf{Z}$ be the dimension function associated to a dualizing complex $\omega _ A^\bullet $ for $A$. Since $\Lambda $ is infinite and $\delta $ is bounded, there exists an infinite subset $\Lambda ' \subset \Lambda $ on which $\delta (\mathfrak q_\lambda )$ is constant. For $\lambda \in \Lambda '$ we have

\[ \text{depth}(M_{\mathfrak p_\lambda }) + \delta (\mathfrak p_\lambda ) - \delta (\mathfrak q_\lambda ) = \text{depth}(M_{\mathfrak p_\lambda }) + \dim ((A/\mathfrak p_\lambda )_{\mathfrak q_\lambda }) \leq d + s \]

by (3) and the definition of $B$. By the semi-continuity of the function $\text{depth} + \delta $ proved in Duality for Schemes, Lemma 48.2.8 we conclude that

\[ \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_{\mathfrak q_\lambda }) = \text{depth}(M_\mathfrak p) + \delta (\mathfrak p) - \delta (\mathfrak q_\lambda ) \leq d + s \]

Since also $\mathfrak p \not\in V(I)$ we read off from (4) that $\mathfrak p \in T$, i.e., $V(\mathfrak p) \cap V(I) \subset V(J) \cap V(I)$. This finishes the proof of Claim I.

Claim II: $H^ i_{J_0}(M) \to H^ i_ J(M)$ is an isomorphism for $i \leq s$ and $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$.

Proof of claim II. Choose $\mathfrak p \in V(J')$ not in $V(J_0)$. It suffices to show that $H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for $i \leq s$, see Local Cohomology, Lemma 51.2.6. Observe that $\mathfrak p \in T$. Hence since $\mathfrak p$ is not in $B$ we see that $\text{depth}(M_\mathfrak p) > s$ and the groups vanish by Dualizing Complexes, Lemma 47.11.1.

Claim III. The final statement of the lemma is true.

By Claim II for $i \leq s$ we have

\[ H^ i_ T(M) = H^ i_{J_0}(M) = H^ i_{J'}(M) \]

for all ideals $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. See Local Cohomology, Lemma 51.5.3. Let us check the hypotheses of Local Cohomology, Proposition 51.10.1 for the subsets $T \subset T \cup V(I)$, the module $M$, and the integer $s$. We have to show that given $\mathfrak p \subset \mathfrak q$ with $\mathfrak p \not\in T \cup V(I)$ and $\mathfrak q \in T$ we have

\[ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > s \]

If $\text{depth}(M_\mathfrak p) \geq s$, then this is true because the dimension of $(A/\mathfrak p)_\mathfrak q$ is at least $1$. Thus we may assume $\text{depth}(M_\mathfrak p) < s$. If $\mathfrak q \in V(I)$, then $\mathfrak q \in V(J) \cap V(I)$ and the inequality holds by (4). If $\mathfrak q \not\in V(I)$, then we can use (3) to pick $\mathfrak q' \in V(\mathfrak q) \cap V(J) \cap V(I)$ with $\dim ((A/\mathfrak q)_{\mathfrak q'}) \leq d$. Then assumption (4) gives

\[ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_{\mathfrak q'}) > s + d \]

Since $A$ is catenary this implies the inequality we want. Applying Local Cohomology, Proposition 51.10.1 we find $J'' \subset A$ with $V(J'') \subset T \cup V(I)$ such that $J''$ annihilates $H^ i_ T(M)$ for $i \leq s$. Then we can write $V(J'') \cup V(J_0) \cup V(I) = V(J'I)$ for some $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Replacing $J_0$ by $J'$ the proof is complete.
$\square$

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