The cartesian $\mathcal{O}$-modules form a weak Serre subcategory by Lemma \ref{lemma-Serre-subcat-cartesian-modules} (the required hypotheses hold by the discussion in Remark \ref{remark-augmentation-ringed}). Since the restriction functor $g_n^* : \textit{Mod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}_n)$ are exact, it follows that $\mathcal{A}_{total}$ is a weak Serre subcategory. \medskip\noindent Let us show that $a^* : \mathcal{A} \to \mathcal{A}_{total}$ is an equivalence of categories with inverse given by $La_*$. We already know that $La_*a^*\mathcal{F} = \mathcal{F}$ by the bounded version (Lemma \ref{lemma-hypercovering-equivalence-bounded-modules}). It is clear that $a^*\mathcal{F}$ is in $\mathcal{A}_{total}$ for $\mathcal{F}$ in $\mathcal{A}$. Conversely, assume that $\mathcal{G} \in \mathcal{A}_{total}$. Because $\mathcal{G}$ is cartesian we see that $\mathcal{G} = a^*\mathcal{F}$ for some $\mathcal{O}_\mathcal{C}$-module $\mathcal{F}$ by Lemma \ref{lemma-hypercovering-descent-modules}. We want to show that $\mathcal{F}$ is in $\mathcal{A}$. Take $U \in \Ob(\mathcal{C})$. We have to show that the restriction of $\mathcal{F}$ to $\mathcal{C}/U$ is in $\mathcal{A}_U$. As usual, write $K_0 = \{U_{0, i}\}_{i \in I_0}$. Since $K$ is a hypercovering, the map $\coprod_{i \in I_0} h_{U_{0, i}} \to *$ becomes surjective after sheafification. This implies there is a covering $\{U_j \to U\}_{j \in J}$ and a map $\tau : J \to I_0$ and for each $j \in J$ a morphism $\varphi_j : U_j \to U_{0, \tau(j)}$. Since $\mathcal{G}_0 = a_0^*\mathcal{F}$ we find that the restriction of $\mathcal{F}$ to $\mathcal{C}/U_j$ is equal to the restriction of the $\tau(j)$th component of $\mathcal{G}_0$ to $\mathcal{C}/U_j$ via the morphism $\varphi_j : U_j \to U_{0, \tau(i)}$. Hence by (\ref{item-restriction}) we find that $\mathcal{F}|_{\mathcal{C}/U_j}$ is in $\mathcal{A}_{U_j}$ and in turn by (\ref{item-local}) we find that $\mathcal{F}|_{\mathcal{C}/U}$ is in $\mathcal{A}_U$. \medskip\noindent In particular the statement of the lemma makes sense. The lemma now follows from Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-equivalence-unbounded-one}. Assumption (1) is clear (see Remark \ref{remark-augmentation-ringed}). Assumptions (2) and (3) we proved in the preceding paragraph. Assumption (4) is immediate from (\ref{item-bounded-dimension}). For assumption (5) let $\mathcal{B}_{total}$ be the set of objects $U/U_{n, i}$ of the site $(\mathcal{C}/K)_{total}$ such that $U \in \mathcal{B}$ where $\mathcal{B}$ is as in (\ref{item-bounded-dimension}). Here we use the description of the site $(\mathcal{C}/K)_{total}$ given in Section \ref{section-simplicial-semi-representable}. Moreover, we set $\text{Cov}_{U/U_{n, i}}$ equal to $\text{Cov}_U$ and $d_{U/U_{n, i}}$ equal $d_U$ where $\text{Cov}_U$ and $d_U$ are given to us by (\ref{item-bounded-dimension}). Then we claim that condition (5) holds with these choices. This follows immediately from Lemma \ref{lemma-sanity-check-simplicial-semi-representable} and the fact that $\mathcal{F} \in \mathcal{A}_{total}$ implies $\mathcal{F}_n \in \mathcal{A}_{K_n}$ and hence $\mathcal{F}_{n, i} \in \mathcal{A}_{U_{n, i}}$. (The reader who worries about the difference between cohomology of abelian sheaves versus cohomology of sheaves of modules may consult Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-cohomology-modules-abelian-agree}.)