**Proof.**
We will check hypotheses (a), (b), and (c) of Lemma 52.16.10. 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 51.12.1.

Observe that for each $p \geq 0$ the limit topology on $\mathop{\mathrm{lim}}\nolimits H^ p(U, \mathcal{F}_ n)$ is the $I$-adic topology by Lemma 52.4.5. In particular, hypothesis (b) holds.

We know that $M = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}_ n)$ 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 10.96.12. In particular hypothesis (c) holds.

For each $n \geq 0$ let us write $Ob_ n = \mathop{\mathrm{lim}}\nolimits _ N H^1(U, I^ n\mathcal{F}_ N)$. A special case is $Ob = Ob_0 = \mathop{\mathrm{lim}}\nolimits _ N H^1(U, \mathcal{F}_ N)$. 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.)

We set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$, we pick generators $f_1, \ldots , f_ r$ of $I$, we pick $c \geq 1$, and we choose $\Phi _\mathcal {F}$ as in Lemma 52.4.4. We will use the results of Lemma 52.2.4 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 $I^ n\mathcal{F} = \mathop{\mathrm{lim}}\nolimits I^ n\mathcal{F}_ N$. 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

\[ Ob'_ n = \mathop{\mathrm{Im}}\left( \bigoplus \nolimits _{e_1 + \ldots + e_ r = n} Ob \cdot T_1^{e_1} \ldots T_ r^{e_ r} \to Ob_ n \right) \]

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 Cohomology, Lemma 20.35.1 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 $\bigoplus _{n \geq 0} \mathop{\mathrm{Im}}(M_ n \to Ob'_ n)$. 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 $\mathop{\mathrm{Im}}(M_ n \to M_{n - k}) = \mathop{\mathrm{Im}}(M_{n'} \to M_{n - k})$ for all $n' \geq n$.

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.
$\square$

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