Recall that $\textit{Coh}(U, I\mathcal{O}_U)$ is an abelian category, see Cohomology of Schemes, Lemma \ref{coherent-lemma-inverse-systems-abelian}. Over affine opens of $U$ the object $(\mathcal{F}_n)$ corresponds to a finite module over a Noetherian ring (Cohomology of Schemes, Lemma \ref{coherent-lemma-inverse-systems-affine}). Thus the kernels of the maps $f^N : (\mathcal{F}_n) \to (\mathcal{F}_n)$ stabilize for $N$ large enough. By Lemmas \ref{lemma-map-kernel-cokernel-on-closed} and \ref{lemma-essential-image-completion} in order to prove the lemma we may replace $(\mathcal{F}_n)$ by the image of such a map. Thus we may assume $f$ is injective on $(\mathcal{F}_n)$. After this replacement the equivalent conditions of Lemma \ref{lemma-equivalent-f-good} hold for the inverse system $(\mathcal{F}_n)$ on $U$. We will use this without further mention in the rest of the proof. \medskip\noindent We will check hypotheses (a), (b), and (c) of Lemma \ref{lemma-when-done}. Hypothesis (b) holds by Lemma \ref{lemma-topology-I-adic-f}. \medskip\noindent Pick a inverse system of modules $\{M_n\}$ as in Lemma \ref{lemma-system-of-modules}. We may assume $H^0_\mathfrak m(M_n) = 0$ by replacing $M_n$ by $M_n/H^0_\mathfrak m(M_n)$ if necessary. Then we obtain short exact sequences $$ 0 \to M_n \to H^0(U, \mathcal{F}_n) \to H^1_\mathfrak m(M_n) \to 0 $$ for all $n$. Let $E$ be an injective hull of the residue field of $A$. By Lemma \ref{lemma-helper-algebraize} and our current assumption (4) we can choose, an integer $m \geq 0$, finite $A$-modules $N_1$ and $N_2$ annihilated by $f^c$ for some $c \geq 0$ and compatible systems of maps $$ H^i_\mathfrak m(M_n) \to \Hom_A(N_i, E), \quad i = 1, 2 $$ for $n \geq m$ with the properties stated in the lemma. \medskip\noindent We know that $M = \lim H^0(U, \mathcal{F}_n)$ is an $A$-module whose limit topology is the $f$-adic topology. Thus, given $n$, the module $M/f^nM$ is a subquotient of $H^0(U, \mathcal{F}_N)$ for some $N \gg n$. Looking at the information obtained above we see that $f^cM/f^nM$ is a finite $A$-module. Since $f$ is a nonzerodivisor on $M$ we conclude that $M/f^{n - c}M$ is a finite $A$-module. In this way we see that hypothesis (c) of Lemma \ref{lemma-when-done} holds. \medskip\noindent Next, we study the module $$ Ob = \lim H^1(U, \mathcal{F}_n) = \lim H^2_\mathfrak m(M_n) $$ For $n \geq m$ let $K_n$ be the kernel of the map $H^2_\mathfrak m(M_n) \to \Hom_A(N_2, E)$. Set $K = \lim K_n$. We obtain an exact sequence $$ 0 \to K \to Ob \to \Hom_A(N_2, E) $$ By the above the limit topology on $Ob = \lim H^2_\mathfrak m(M_n)$ is the $f$-adic topology. Since $N_2$ is annihilated by $f^c$ we conclude the same is true for the limit topology on $K = \lim K_n$. Thus $K/fK$ is a subquotient of $K_n$ for $n \gg 1$. However, since $\{K_n\}$ is pro-isomorphic to a inverse system of finite length $A$-modules (by the conclusion of Lemma \ref{lemma-helper-algebraize}) we conclude that $K/fK$ is a subquotient of a finite length $A$-module. It follows that $K$ is a finite $A$-module, see Algebra, Lemma \ref{algebra-lemma-finite-over-complete-ring}. (In fact, we even see that $\dim(\text{Supp}(K)) = 1$ but we will not need this.) \medskip\noindent Given $n \geq 1$ consider the boundary map $$ \delta_n : H^0(U, \mathcal{F}_n) \longrightarrow \lim_N H^1(U, f^n\mathcal{F}_N) \xrightarrow{f^{-n}} Ob $$ (the second map is an isomorphism) coming from the short exact sequences $$ 0 \to f^n\mathcal{F}_N \to \mathcal{F}_N \to \mathcal{F}_n \to 0 $$ For each $n$ set $$ P_n = \Im(H^0(U, \mathcal{F}_{n + m}) \to H^0(U, \mathcal{F}_n)) $$ where $m$ is as above. Observe that $\{P_n\}$ is an inverse system and that the map $f : \mathcal{F}_n \to \mathcal{F}_{n + 1}$ on global sections maps $P_n$ into $P_{n + 1}$. If $p \in P_n$, then $\delta_n(p) \in K \subset Ob$ because $\delta_n(p)$ maps to zero in $H^1(U, f^n\mathcal{F}_{n + m}) = H^2_\mathfrak m(M_m)$ and the composition of $\delta_n$ and $Ob \to \Hom_A(N_2, E)$ factors through $H^2_\mathfrak m(M_m)$ by our choice of $m$. Hence $$ \bigoplus\nolimits_{n \geq 0} \Im(P_n \to Ob) $$ is a finite graded $A[T]$-module where $T$ acts via multiplication by $f$. Namely, it is a graded submodule of $K[T]$ and $K$ is finite over $A$. Arguing as in the proof of Lemma \ref{lemma-ML-general}\footnote{Choose homogeneous generators of the form $\delta_{n_j}(p_j)$ for the displayed module. Then if $k = \max(n_j)$ we find that for $n \geq k$ and any $p \in P_n$ we can find $a_j \in A$ such that $p - \sum a_j f^{n - n_j} p_j$ is in the kernel of $\delta_n$ and hence in the image of $P_{n'}$ for all $n' \geq n$. Thus $\Im(P_n \to P_{n - k}) = \Im(P_{n'} \to P_{n - k})$ for all $n' \geq n$.} we find that the inverse system $\{P_n\}$ satisfies ML. Since $\{P_n\}$ is pro-isomorphic to $\{H^0(U, \mathcal{F}_n)\}$ we conclude that $\{H^0(U, \mathcal{F}_n)\}$ has ML. Thus hypothesis (a) of Lemma \ref{lemma-when-done} holds and the proof is complete.