We will check hypotheses (a), (b), and (c) of Lemma \ref{lemma-when-done}. Before we start, let us point out that the modules $H^0(U, \mathcal{F}''_n)$ and $H^1(U, \mathcal{F}''_n)$ are finite $A$-modules for all $n$ by Local Cohomology, Lemma \ref{local-cohomology-lemma-finiteness-Rjstar}. \medskip\noindent Observe that for each $p \geq 0$ the limit topology on $\lim H^p(U, \mathcal{F}_n)$ is the $I$-adic topology by Lemma \ref{lemma-topology-I-adic}. In particular, hypothesis (b) holds. \medskip\noindent We know that $M = \lim H^0(U, \mathcal{F}_n)$ is an $A$-module whose limit topology is the $I$-adic topology. Thus, given $n$, the module $M/I^nM$ is a subquotient of $H^0(U, \mathcal{F}_N)$ for some $N \gg n$. Since the inverse system $\{H^0(U, \mathcal{F}_N)\}$ is pro-isomorphic to an inverse system of finite $A$-modules, namely $\{H^0(U, \mathcal{F}''_N)\}$, we conclude that $M/I^nM$ is finite. It follows that $M$ is finite, see Algebra, Lemma \ref{algebra-lemma-finite-over-complete-ring}. In particular hypothesis (c) holds. \medskip\noindent For each $n \geq 0$ let us write $Ob_n = \lim_N H^1(U, I^n\mathcal{F}_N)$. A special case is $Ob = Ob_0 = \lim_N H^1(U, \mathcal{F}_N)$. Arguing exactly as in the previous paragraph we find that $Ob$ is a finite $A$-module. (In fact, we also know that $Ob/I Ob$ is annihilated by a power of $\mathfrak a$, but it seems somewhat difficult to use this.) \medskip\noindent We set $\mathcal{F} = \lim \mathcal{F}_n$, we pick generators $f_1, \ldots, f_r$ of $I$, we pick $c \geq 1$, and we choose $\Phi_\mathcal{F}$ as in Lemma \ref{lemma-cd-is-one-for-system}. We will use the results of Lemma \ref{lemma-properties-system} without further mention. In particular, for each $n \geq 1$ there are maps $$ \delta_n : H^0(U, \mathcal{F}_n) \longrightarrow H^1(U, I^n\mathcal{F}) \longrightarrow Ob_n $$ The first comes from the short exact sequence $0 \to I^n\mathcal{F} \to \mathcal{F} \to \mathcal{F}_n \to 0$ and the second from $I^n\mathcal{F} = \lim I^n\mathcal{F}_N$. We will later use that if $\delta_n(s) = 0$ for $s \in H^0(U, \mathcal{F}_n)$ then we can for each $n' \geq n$ find $s' \in H^0(U, \mathcal{F}_{n'})$ mapping to $s$. Observe that there are commutative diagrams $$ \xymatrix{ H^0(U, \mathcal{F}_{nc}) \ar[r] \ar[dd] & H^1(U, I^{nc}\mathcal{F}) \ar[dd] \ar[rd]^{\Phi_\mathcal{F}} \\ & & \bigoplus_{e_1 + \ldots + e_r = n} H^1(U, \mathcal{F}) \cdot T_1^{e_1} \ldots T_r^{e_r} \ar[ld] \\ H^0(U, \mathcal{F}_n) \ar[r] & H^1(U, I^n\mathcal{F}) } $$ We conclude that the obstruction map $H^0(U, \mathcal{F}_n) \to Ob_n$ sends the image of $H^0(U, \mathcal{F}_{nc}) \to H^0(U, \mathcal{F}_n)$ into the submodule $$ Ob'_n = \Im\left( \bigoplus\nolimits_{e_1 + \ldots + e_r = n} Ob \cdot T_1^{e_1} \ldots T_r^{e_r} \to Ob_n \right) $$ where on the summand $Ob \cdot T_1^{e_1} \ldots T_r^{e_r}$ we use the map on cohomology coming from the reductions modulo powers of $I$ of the multiplication map $f_1^{e_1} \ldots f_r^{e_r} : \mathcal{F} \to I^n\mathcal{F}$. By construction $$ \bigoplus\nolimits_{n \geq 0} Ob'_n $$ is a finite graded module over the Rees algebra $\bigoplus_{n \geq 0} I^n$. For each $n$ we set $$ M_n = \{s \in H^0(U, \mathcal{F}_n) \mid \delta_n(s) \in Ob'_n\} $$ Observe that $\{M_n\}$ is an inverse system and that $f_j : \mathcal{F}_n \to \mathcal{F}_{n + 1}$ on global sections maps $M_n$ into $M_{n + 1}$. By exactly the same argument as in the proof of Lemma \ref{lemma-ML-general} we find that $\{M_n\}$ is ML. Namely, because the Rees algebra is Noetherian we can choose a finite number of homogeneous generators of the form $\delta_{n_j}(z_j)$ with $z_j \in M_{n_j}$ for the graded submodule $\bigoplus_{n \geq 0} \Im(M_n \to Ob'_n)$. Then if $k = \max(n_j)$ we find that for $n \geq k$ and any $z \in M_n$ we can find $a_j \in I^{n - n_j}$ such that $z - \sum a_j z_j$ is in the kernel of $\delta_n$ and hence in the image of $M_{n'}$ for all $n' \geq n$ (because the vanishing of $\delta_n$ means that we can lift $z - \sum a_j z_j$ to an element $z' \in H^0(U, \mathcal{F}_{n'c})$ for all $n' \ge n$ and then the image of $z'$ in $H^0(U, \mathcal{F}_{n'})$ is in $M_{n'}$ by what we proved above). Thus $\Im(M_n \to M_{n - k}) = \Im(M_{n'} \to M_{n - k})$ for all $n' \geq n$. \medskip\noindent Choose $n$. By the Mittag-Leffler property of $\{M_n\}$ we just established we can find an $n' \geq n$ such that the image of $M_{n'} \to M_n$ is the same as the image of $M' \to M_n$. By the above we see that the image of $M' \to M_n$ contains the image of $H^0(U, \mathcal{F}_{n'c}) \to H^0(U, \mathcal{F}_n)$. Thus we see that $\{M_n\}$ and $\{H^0(U, \mathcal{F}_n)\}$ are pro-isomorphic. Therefore $\{H^0(U, \mathcal{F}_n)\}$ has ML and we finally conclude that hypothesis (a) holds. This concludes the proof.