It seems that there is no reference where the notion of total integral closure is discussed in detail. But a good place to look at is Bhatt's notes on perfectoid spaces, especially at Proposition 5.2.5. It explains the main usage of total closures in theory of perfectoid spaces.

**Definition:** Let $A \subset B$ be an extension of rings then the total integral closure $A_{tic}$ of $A$ in $B$ is the set of all $f\in B$ such that $f^{\mathbf N}$ is contained in a finitely generated $A$-submodule of $B$.

**Remark:** It can be easily checked that $A_{tic}$ is an $A$-subalgebra of $B$.

**Remark:** The total integral closure coincides with the usual integral closure of $A$ in $B$ when $A$ is noetherian. So this definition is interesting only for non-noetherian rings.

**Warning:** The name may be a bit misleading as it need not be the case that $A_{tic}$ is itself totally integrally closed in $B$.

Even though this definition makes perfect sense for any arbitrary extension of abstract rings, we will use it only for a Tate ring (and a ring of definitions therein). So suppose that we have a Tate ring $A$ and a couple of definition $(A_0, \varpi)$ ($\varpi$ is assumed to be a pseudo-uniformizer). Then one of the main applications of our definition is that the subring $A^{\circ}$ can be recovered from this data as the total integral closure of $A_0$ in $A$.

**Proposition 1:** Let $A, A_0, \varpi$ be as above, then $A^{\circ}$ is equal to the total integral closure of $A_0$ in $A$.

Let us postpone the proof of this proposition and discuss its applications. We start with the following corollary:

**Corollary 2:** A ring of definition $A_0$ in a Tate ring $A$ is equal to $A^{\circ}$ if and only if it is totally integrally closed in $A$.

Note that Corollary $2$ is false if we assume that $A_0$ is only integrally closed in $A$ as there are examples of perfectoid pairs $(R, R^+)$ with $R^+ \neq R^{\circ}$.

**Corollary 3:** Let $A$ be a uniform Tate ring then $A^{\circ}$ is totally integrally closed in $A$.

**Proof:** We note that $A^{\circ}$ is a ring of definition in $A$ since $A$ is uniform. Then we can just apply Corollary $2$ to finish the proof.

Corollary $2$ is a pretty useful criterion to check that various rings of definitions are actually equal to the subrings of power-bounded elements. These two corollaries lie at the core of the proof of Proposition 5.2.5 in Bhatt's notes that describes the category of uniform Banach $K$-algebras in more algebraic terms. This, in turn, is used (together with a similar Proposition 5.2.6) in the proof of tilting equivalence for perfectoid $K$-algebras. Look at the proof of Theorem 6.2.5 for more details.

**Proof of Proposition 1:** The first thing to note is that $f\in A$ is power-bounded if and only if there is an integer $M$ such that
$$
f^{\mathbf N} \subset \varpi^{-M}A_0.
$$
This follows from the definition of a power-bounded element taking into account that $A$ is Tate. We claim that this implies that

$$
A^{\circ} \subset A_{0, tic}.
$$
Indeed, for any element $f\in A^{\circ}$ we have an inclusion $f^{\mathbf N} \subset \varpi^{-M}A_0$ for some $M$. It is a finitely generated $A_0$-module (really it is just isomorphic to $A_0$) as there is a surjection
$$
A_0 \xrightarrow{\varpi^{-M}} \varpi^{-M}A_0,
$$
so $f\in A_{0, tic}$.

Now we want to show that
$$
A_{0, tic} \subset A^{\circ}.
$$
Pick any element $f\in A_{0, tic}$ then the set $f^{\mathbf N}$ is contained in some finitely generated $A_0$-submodule of $A$. Denote this submodule by $M$ and choose some set of generators $x_1, \dots, x_n$. Recall that $A\cong A_0[\frac{1}{\varpi}]$ as $A$ is Tate, so we conclude that each $x_i$ can be written as
$$
x_i=a_i/\varpi^{M_i}
$$
for some $a_i\in A_0$ and $n_i\in \mathbf N$. Set $M:=\max_{i=1}^n{M_i}$ then it is easy to see that $M\subset \varpi^{-M}A_0$. Thus we see that
$$
f^{\mathbf N} \subset \varpi^{-M}A_0.
$$
So $f$ is power-bounded by the observation at the beginning of the proof. We conclude that $A^0\subset A_{0, tic}$.

Finally, I need to say that in the context of rigid geometry many of those subtleties with total integral closure disappear.

**Theorem:** Let $K$ be a complete non-archimedean field with a pseudo-uniformizer $\varpi$, and let $A$ be a topologically finitely generated $K$-algebra. Suppose that $A_0$ is an open topologically finitely generated $K^{\circ}$-subalgebra of $A$ such that it is integrally closed in $A$ and $A_0[\frac{1}{\varpi}]=A$ (i.e. $A_0$ is a ring of definition in $A$). Then $A_0$ is equal to $A^{\circ}$.

**Proof:** Pick any surjection
$$
K^{\circ}\langle X_1, \dots, X_n \rangle \to A_0
$$
that exists by the assumption on $A_0$. We use Theorem 6.3.5/1 from the book "Non-Archimedean analysis" by Bosch, Guntzer, and Remmert to conclude that $A^{\circ}$ is integral over $K^{\circ}\langle X_1, \dots, X_n \rangle$, in particular, it is integral over $A_0$. Finally, we recall that $A_0$ is integrally closed in $A$, so $A^{\circ}$ must be equal to $A_0$.