**Proof.**
We first make a preliminary remark. Suppose that $(S', s') \to (S, s)$ is an elementary étale neighbourhood. Denote $\mathcal{F}'$ the pullback of $\mathcal{F}$ to $X' = X \times _ S S'$. By the discussion following Definition 38.16.1 we see that $\mathcal{F}'$ is pure along $X'_{s'}$. Moreover, $\mathcal{F}'$ is flat over $S'$ along $X'_{s'}$. Then it suffices to prove that $\mathcal{F}'$ is universally pure along $X'_{s'}$. Namely, given any morphism $(T, t) \to (S, s)$ of pointed schemes the fibre product $(T', t') = (T \times _ S S', (t, s'))$ is flat over $(T, t)$ and hence if $\mathcal{F}_{T'}$ is pure along $X_{t'}$ then $\mathcal{F}_ T$ is pure along $X_ t$ by Lemma 38.16.6. Thus during the proof we may always replace $(s, S)$ by an elementary étale neighbourhood. We may also replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ due to the local nature of the problem.

Choose an elementary étale neighbourhood $(S', s') \to (S, s)$ and a commutative diagram

\[ \xymatrix{ X \ar[d] & X' \ar[l]^ g \ar[d] \\ S & \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}) \ar[l] } \]

such that $X' \to X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ is étale, $X_ s = g((X')_{s'})$, the scheme $X'$ is affine, and such that $\Gamma (X', g^*\mathcal{F})$ is a free $\mathcal{O}_{S', s'}$-module, see Lemma 38.12.11. Note that $X' \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ is of finite type (as a quasi-compact morphism which is the composition of an étale morphism and the base change of a finite type morphism). By our preliminary remarks in the first paragraph of the proof we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$. Hence we may assume there exists a commutative diagram

\[ \xymatrix{ X \ar[dr] & & X' \ar[ll]^ g \ar[ld] \\ & S & } \]

of schemes of finite type over $S$, where $g$ is étale, $X_ s \subset g(X')$, with $S$ local with closed point $s$, with $X'$ affine, and with $\Gamma (X', g^*\mathcal{F})$ a free $\Gamma (S, \mathcal{O}_ S)$-module. Note that in this case $g^*\mathcal{F}$ is universally pure over $S$, see Lemma 38.17.4.

In this situation we apply Lemma 38.18.2 to deduce that $\text{Ass}_{X/S}(\mathcal{F}) \subset g(X')$ from our assumption that $\mathcal{F}$ is pure along $X_ s$ and flat over $S$ along $X_ s$. By Divisors, Lemma 31.7.3 and Remark 31.7.4 we see that for any morphism of pointed schemes $(T, t) \to (S, s)$ we have

\[ \text{Ass}_{X_ T/T}(\mathcal{F}_ T) \subset (X_ T \to X)^{-1}(\text{Ass}_{X/S}(\mathcal{F})) \subset g(X') \times _ S T = g_ T(X'_ T). \]

Hence by Lemma 38.18.2 applied to the base change of our displayed diagram to $(T, t)$ we conclude that $\mathcal{F}_ T$ is pure along $X_ t$ as desired.
$\square$

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