Let $\omega_A^\bullet$ be a dualizing complex. Let $\delta$ be its dimension function, see Dualizing Complexes, Section \ref{dualizing-section-dimension-function}. An important role will be played by the finite $A$-modules $$ E^i = \Ext_A^i(M, \omega_A^\bullet) $$ For $\mathfrak p \subset A$ we will write $H^i_\mathfrak p$ to denote the local cohomology of an $A_\mathfrak p$-module with respect to $\mathfrak pA_\mathfrak p$. Then we see that the $\mathfrak pA_\mathfrak p$-adic completion of $$ (E^i)_\mathfrak p = \Ext^{\delta(\mathfrak p) + i}_{A_\mathfrak p}(M_\mathfrak p, (\omega_A^\bullet)_\mathfrak p[-\delta(\mathfrak p)]) $$ is Matlis dual to $$ H^{-\delta(\mathfrak p) - i}_{\mathfrak p}(M_\mathfrak p) $$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-special-case-local-duality}. In particular we deduce from this the following fact: an ideal $J \subset A$ annihilates $(E^i)_\mathfrak p$ if and only if $J$ annihilates $H^{-\delta(\mathfrak p) - i}_{\mathfrak p}(M_\mathfrak p)$. \medskip\noindent Set $T_n = \{\mathfrak p \in T \mid \delta(\mathfrak p) \leq n\}$. As $\delta$ is a bounded function, we see that $T_a = \emptyset$ for $a \ll 0$ and $T_b = T$ for $b \gg 0$. \medskip\noindent Assume (2). Let us prove the existence of $J$ as in (1). We will use a double induction to do this. For $i \leq s$ consider the induction hypothesis $IH_i$: $H^a_T(M)$ is annihilated by some $J \subset A$ with $V(J) \subset T'$ for $0 \leq a \leq i$. The case $IH_0$ is trivial because $H^0_T(M)$ is a submodule of $M$ and hence finite and hence is annihilated by some ideal $J$ with $V(J) \subset T$. \medskip\noindent Induction step. Assume $IH_{i - 1}$ holds for some $0 < i \leq s$. Pick $J'$ with $V(J') \subset T'$ annihilating $H^a_T(M)$ for $0 \leq a \leq i - 1$ (the induction hypothesis guarantees we can do this). We will show by descending induction on $n$ that there exists an ideal $J$ with $V(J) \subset T'$ such that the associated primes of $J H^i_T(M)$ are in $T_n$. For $n \ll 0$ this implies $JH^i_T(M) = 0$  (Algebra, Lemma \ref{algebra-lemma-ass-zero}) and hence $IH_i$ will hold. The base case $n \gg 0$ is trivial because $T = T_n$ in this case and all associated primes of $H^i_T(M)$ are in $T$. \medskip\noindent Thus we assume given $J$ with the property for $n$. Let $\mathfrak q \in T_n$. Let $T_\mathfrak q \subset \Spec(A_\mathfrak q)$ be the inverse image of $T$. We have $H^j_T(M)_\mathfrak q = H^j_{T_\mathfrak q}(M_\mathfrak q)$ by Lemma \ref{lemma-torsion-change-rings}. Consider the spectral sequence $$ H_\mathfrak q^p(H^q_{T_\mathfrak q}(M_\mathfrak q)) \Rightarrow H^{p + q}_\mathfrak q(M_\mathfrak q) $$ of Lemma \ref{lemma-local-cohomology-ss}. Below we will find an ideal $J'' \subset A$ with $V(J'') \subset T'$ such that $H^i_\mathfrak q(M_\mathfrak q)$ is annihilated by $J''$ for all $\mathfrak q \in T_n \setminus T_{n - 1}$. Claim: $J (J')^i J''$ will work for $n - 1$. Namely, let $\mathfrak q \in T_n \setminus T_{n - 1}$. The spectral sequence above defines a filtration $$ E_\infty^{0, i} = E_{i + 2}^{0, i} \subset \ldots \subset E_3^{0, i} \subset E_2^{0, i} = H^0_\mathfrak q(H^i_{T_\mathfrak q}(M_\mathfrak q)) $$ The module $E_\infty^{0, i}$ is annihilated by $J''$. The subquotients $E_j^{0, i}/E_{j + 1}^{0, i}$ for $i + 1 \geq j \geq 2$ are annihilated by $J'$ because the target of $d_j^{0, i}$ is a subquotient of $$ H^j_\mathfrak q(H^{i - j + 1}_{T_\mathfrak q}(M_\mathfrak q)) = H^j_\mathfrak q(H^{i - j + 1}_T(M)_\mathfrak q) $$ and $H^{i - j + 1}_T(M)_\mathfrak q$ is annihilated by $J'$ by choice of $J'$. Finally, by our choice of $J$ we have $J H^i_T(M)_\mathfrak q \subset H^0_\mathfrak q(H^i_T(M)_\mathfrak q)$ since the non-closed points of $\Spec(A_\mathfrak q)$ have higher $\delta$ values. Thus $\mathfrak q$ cannot be an associated prime of $J(J')^iJ'' H^i_T(M)$ as desired. \medskip\noindent By our initial remarks we see that $J''$ should annihilate $$ (E^{-\delta(\mathfrak q) - i})_\mathfrak q = (E^{-n - i})_\mathfrak q $$ for all $\mathfrak q \in T_n \setminus T_{n - 1}$. But if $J''$ works for one $\mathfrak q$, then it works for all $\mathfrak q$ in an open neighbourhood of $\mathfrak q$ as the modules $E^{-n - i}$ are finite. Since every subset of $\Spec(A)$ is Noetherian with the induced topology (Topology, Lemma \ref{topology-lemma-Noetherian}), we conclude that it suffices to prove the existence of $J''$ for one $\mathfrak q$. \medskip\noindent Since the ext modules are finite the existence of $J''$ is equivalent to $$ \text{Supp}(E^{-n - i}) \cap \Spec(A_\mathfrak q) \subset T'. $$ This is equivalent to showing the localization of $E^{-n - i}$ at every $\mathfrak p \subset \mathfrak q$, $\mathfrak p \not \in T'$ is zero. Using local duality over $A_\mathfrak p$ we find that we need to prove that $$ H^{i + n - \delta(\mathfrak p)}_\mathfrak p(M_\mathfrak p) = H^{i - \dim((A/\mathfrak p)_\mathfrak q)}_\mathfrak p(M_\mathfrak p) $$ is zero (this uses that $\delta$ is a dimension function). This vanishes by the assumption in the lemma and $i \leq s$ and Dualizing Complexes, Lemma \ref{dualizing-lemma-depth}. \medskip\noindent To prove the converse implication we assume (2) does not hold and we work backwards through the arguments above. First, we pick a $\mathfrak q \in T$, $\mathfrak p \subset \mathfrak q$ with $\mathfrak p \not \in T'$ such that $$ i = \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim((A/\mathfrak p)_\mathfrak q) \leq s $$ is minimal. Then $H^{i - \dim((A/\mathfrak p)_\mathfrak q)}_\mathfrak p(M_\mathfrak p)$ is nonzero by the nonvanishing in Dualizing Complexes, Lemma \ref{dualizing-lemma-depth}. Set $n = \delta(\mathfrak q)$. Then there does not exist an ideal $J \subset A$ with $V(J) \subset T'$ such that $J(E^{-n - i})_\mathfrak q = 0$. Thus $H^i_\mathfrak q(M_\mathfrak q)$ is not annihilated by an ideal $J \subset A$ with $V(J) \subset T'$. By minimality of $i$ it follows from the spectral sequence displayed above that the module $H^i_T(M)_\mathfrak q$ is not annihilated by an ideal $J \subset A$ with $V(J) \subset T'$. Thus $H^i_T(M)$ is not annihilated by an ideal $J \subset A$ with $V(J) \subset T'$. This finishes the proof of the proposition.