Here is a modification of the proof of Proposition 1.3.3 that addresses the issues you brought up. There also appear to be some issues with how atoms are used within the proof and these are also addressed. First, we need to cover some preliminary results. These will be based on the book [1] as much as is possible.
We start off by defining atoms. Let $(S, \mathcal{A}, \mu )$. We say $A\in\mathcal{A}$ is an atom (of $(S, \mathcal{A}, \mu)$) if $\mu (A) > 0$ and for all $B\in\mathcal{A}$ with $B\subseteq A$ we have $\mu (B) = \mu (A)$ or $\mu (B) = 0$. As defined in [1], we say $(S, \mathcal{A}, \mu )$ is atomic if the $\sigma$-algebra generated by the atoms of $\mu$ is equal to $\mathcal{A}$. If $A, B\in \mathcal{A}$, we define $A\Delta B = (A\setminus B) \cup (B\setminus A)$. We say two sets $A, B\in \mathcal{A}$ are equivalent if $\mu (A\Delta B) = 0$. Note that this defines an equivalence relation on the members of $\mathcal{A}$.
Proposition 1. Let $(S, \mathcal{A}, \mu )$ be an atomic measure space. Let $A\in\mathcal{A}$ be a $\sigma$-finite set and let $\{A_{i} : i\in I\}$ be a maximal collection of pairwise disjoint atoms each contained in $A$, where the collection is maximal in the sense that there is no other atom $B\in\mathcal{A}$ such that $B\subseteq A$ and $B\cap A_{i} = \emptyset$ for all $i\in I$. Then $I$ is countable, $\bigcup_{i\in I} A_{i} \in \mathcal{A}$ and $\mu (A\setminus \bigcup_{i\in I} A_{i}) = 0$. Furthermore, for every $B\in\mathcal{A}$ with $B\subseteq A$ there is $I_{0}\subseteq I$ such that $B$ and $\bigcup_{i\in I_{0}} A_{i}$ are equivalent.
Proof. We first show $I$ is countable and $\bigcup_{i\in I} A_{i}\in \mathcal{A}$. Since $A$ is $\sigma$-finite and $\mu (A) > 0$ we can write $A = \bigcup_{j\in J} D_{j}$ where $J$ is countable, $D_{j}\in \mathcal{A}$ for each $j\in J$ and $0 < \mu (D_{j}) < \infty$ for each $j\in J$. For each $j\in J$ and $k\in\mathbb{N}$ the set $D_{j}$ has finite measure and hence the set
\begin{equation}
\{i\in I : \mu (A_{i}\cap D_{j}) \geq \tfrac{1}{k} \}
\end{equation}
is finite. It follows from
\begin{align*}
I = \{i\in I : \mu (A_{i}) > 0 \} &= \bigcup_{j\in J} \{i\in I : \mu (A_{i}\cap D_{j}) > 0 \} \\
&= \bigcup_{j\in J} \bigcup_{k\in\mathbb{N}} \{i\in I : \mu (A_{i}\cap D_{j}) \geq \tfrac{1}{k} \} \\
\end{align*}
that $I$ is countable. In particular we have $\bigcup_{i\in I} A_{i}\in \mathcal{A}$.
We next show $N := A\setminus \bigcup_{i\in I} A_{i}$ satisfies $\mu (N) = 0$. First we obtain a characterisation of $\mathcal{A}$. Define
\begin{equation}
\mathcal{B} := \Big\{\bigcup_{i\in I_{0}} A_{i} : I_{0}\subseteq I \Big\} \cup \Big\{N\cup \bigcup_{i\in I_{0}} A_{i} : I_{0}\subseteq I \Big\} .
\end{equation}
Since the collection $\{N\}\cup\{A_{i} : i\in I\}$ consists of pairwise disjoint sets whose union is $A$, we see that $\mathcal{B}$ is a $\sigma$-algebra on $A$. We also have $\mathcal{B}\subseteq \mathcal{A}$ because $I$ is countable. Now define
\begin{equation}
\mathcal{A}' := \{E\in\mathcal{A} : \text{there is } B\in \mathcal{B} \text{ such that } E\cap A \text{ and } B \text{ are equivalent} \} .
\end{equation}
Using that $\mathcal{B}$ is a $\sigma$-algebra on $A$, it readily follows that $\mathcal{A}'$ is a $\sigma$-algebra on $S$ contained in $\mathcal{A}$ from the following two elementary observations.
$({\rm i})$ Suppose $J$ is countable, $E_{j}, F_{j}\in \mathcal{A}$ for each $j\in J$ and $E_{j}$ is equivalent to $F_{j}$ for each $j\in J$. Then $\bigcup_{j\in J} E_{j}$ is equivalent to $\bigcup_{j\in J} F_{j}$.
$({\rm ii})$ Suppose $E, F\in \mathcal{A}$ with $E\subseteq A$ and $F\subseteq A$. If $E$ and $F$ are equivalent then $A\setminus E$ and $A\setminus F$ are equivalent.
It is also straightforward to observe that each atom of $(S, \mathcal{A}, \mu )$ belongs to $\mathcal{A}'$. Hence we have $\mathcal{A}' = \mathcal{A}$.
Now suppose for a contradiction that $\mu (N) > 0$. If $N$ contains an atom of $(S, \mathcal{A}, \mu )$ then this contradicts that the collection $\{A_{i} : i\in I\}$ is maximal. Hence there is some $N'\in\mathcal{A}$ such that $N' \subseteq N$ and $0 < \mu (N') < \mu (N)$. Since $\mu (N') > 0$ and $N'$ is disjoint from $A_{i}$ for each $i\in I$, we see that $N'$ is not equivalent to a set of the form $\bigcup_{i\in I_{0}} A_{i}$ for any $I_{0}\subseteq I$. And as $\mu (N\setminus N') > 0$ we see that $N'$ is not equivalent to a set of the form $N\cup \bigcup_{i\in I_{0}} A_{i}$ for any $I_{0}\subseteq I$. Hence $N' \not\in \mathcal{A}'$. But as $\mathcal{A} = \mathcal{A}'$ this is a contradiction. Therefore $\mu (N) = 0$.
The final statement that for every $B\in\mathcal{A}$ with $B\subseteq A$ there is some $I_{0}\subseteq I$ such that $B$ and $\bigcup_{i\in I_{0}} A_{i}$ are equivalent follows from the characterisation of $\mathcal{A}'$.
Lemma 1. Let $(S, \mathcal{A}, \mu )$ be a measure space and let $p\in [1, \infty )$. The linear span of the set $\{1_{A}\otimes x : A\in \mathcal{A}, \mu (A) < \infty \text{ and } x\in X \}$ is dense in $L^{p}(S; X)$.
Proof. See Lemma 1.2.19 of [1].
Proposition 2. Let $p, q\in [1, \infty ]$ such that $\tfrac{1}{p} + \tfrac{1}{q} = 1$. For each $g\in L^{q}(S; X^{*})$ there is a linear functional $\phi_{g}\in (L^{p}(S; X))^{*}$ such that
\begin{equation}
\langle f, \phi_{g} \rangle = \int_{S} \langle f(s), g(s) \rangle \, d\mu (s)
\end{equation}
for all $f\in L^{p}(S; X)$. Furthermore, the mapping $g\mapsto \phi_{g}$ defines an isometry of $L^{q}(S;X^{*})$ onto a closed subspace of $(L^{p}(S; X))^{*}$.
Proof. The first claim follows from the Hölder inequality and the second claim follows from Proposition 1.3.1 of [1].
Lemma 2. Let $\{a_{i} : i\in I\} \subseteq [0, \infty )$ and assume there is some $M\in [0, \infty )$ such that
\begin{equation}
\sum_{i\in F} a_{i} \leq M
\end{equation}
for every finite subset $F$ of $I$. Then $\{i\in I : a_{i} \neq 0\}$ is countable.
Proof. Let $\varepsilon > 0$. We will show that $\{i\in I : a_{i} \geq \varepsilon \}$ is finite. Let $F\subseteq I$ be finite such that $a_{i} \geq \varepsilon$ for all $i\in F$. Let $|F|$ denote the cardinality of $F$. We have
\begin{equation}
|F|\varepsilon = \sum_{i\in F} \varepsilon \leq \sum_{i\in F} a_{i} \leq M .
\end{equation}
Hence $|F| \leq \tfrac{M}{\varepsilon}$. This means the set $\{i\in I : a_{i} \geq \varepsilon \}$ does not contain a subset with cardinality $\lfloor \tfrac{M}{\varepsilon} \rfloor + 1$. Hence $\{i\in I : a_{i} \geq \varepsilon \}$ is finite. As $\varepsilon > 0$ was arbitrary, this shows that
\begin{equation}
\{i\in I : a_{i} \neq 0 \} = \bigcup_{n\in\mathbb{N}} \{i\in I : a_{i} \geq \tfrac{1}{n} \}
\end{equation}
is countable as it is a countable union of finite sets.
We now prove the result from the question.
Proposition 1.3.3 (from [1]). Let $1 < p < \infty$ and $\tfrac{1}{p} = \tfrac{1}{q} = 1$, and let $(S, \mathcal{A}, \mu )$ be an atomic measure space. Then for any Banach space $X$ we have
\begin{equation}
(L^{p}(S;X))^{*} = L^{q}(S;X^{*})
\end{equation}
isometrically. In particular, for any set $I$ we have $(\ell^{p}(I; X))^{*} = \ell^{q}(I; X^{*})$ isometrically. If $\mu$ is $\sigma$-finite the same result holds for $p = 1$.
Proof. We split the proof into two cases.
Case 1. Suppose $p = 1$ and $(S, \mathcal{A}, \mu )$ is $\sigma$-finite. Let $\phi \in (L^{1}(S; X))^{*}$. By Proposition 2 it is sufficient to find some $g\in L^{1}(S;X^{*})$ such that
\begin{equation}
\langle f, \phi \rangle = \int_{S} \langle f(s), g(s)\rangle \, d\mu (s)
\end{equation}
for all $f\in L^{1}(S;X)$. Let $\{A_{i} : i\in I\}$ be a maximal collection of pairwise disjoint atoms each having finite measure. Such a collection can be obtained using Zorn's lemma. Since $S$ is $\sigma$-finite, we see by Proposition 1 that $I$ is countable and $\mu (S\setminus \bigcup_{i\in I} A_{i} ) = 0$.
Let $i\in I$. We consider the linear map
\begin{equation}
x\mapsto \mu (A_{i})^{-1} \langle 1_{A_{i}}\otimes x, \phi \rangle \tag{1}
\end{equation}
from
$X$ into $\mathbb{K}$. Since $\|1_{A_{i}}\otimes x \|_{L^{1}(S;X)} = \mu (A_{i}) \|x\|$ and hence
\begin{equation}
|\mu (A_{i})^{-1} \langle 1_{A_{i}}\otimes x, \phi \rangle | \leq \mu (A_{i})^{-1} \|1_{A_{i}}\otimes x \|_{L^{1}(S;X)} \|\phi\|_{(L^{1}(S;X))^{*} } = \|\phi\|_{(L^{1}(S;X))^{*} } \|x\|
\end{equation}
for each $x\in X$, we see that the map $(1)$ is a bounded linear functional on $X$ with norm less than or equal to $\|\phi\|_{(L^{1}(S;X))^{*} }$. Hence there is some $x_{A_{i}}^{*} \in X^{*}$ with $\|x_{A_{i}}^{*}\| \leq \|\phi\|_{(L^{1}(S;X))^{*} }$ such that
\begin{equation}
\langle x, x_{A_{i}}^{*} \rangle = \mu (A_{i})^{-1} \langle 1_{A_{i}}\otimes x, \phi \rangle
\end{equation}
for all $x\in X$.
Define $g\colon S\to X^{*}$ by
\begin{equation}
g(s) := \sum_{i\in I} (1_{A_{i}} \otimes x_{A_{i}}^{*})(s)
\end{equation}
Since $I$ is countable and $\{A_{i} : i\in I\}$ is pairwise disjoint we have that $g$ is well-defined and Bochner measurable. Furthermore, since the collection $\{A_{i} : i\in I\}$ is pairwise disjoint, we have
\begin{equation}
\|g(s)\| \leq \sup\{ \|(1_{A_{i}} \otimes x_{A_{i}}^{*})(s)\| : i\in I\} = \sup\{ \|x_{A_{i}}^{*}\| : i\in I\} \leq \|\phi\|_{(L^{1}(S;X))^{*} }
\end{equation}
for each $s\in S$. It follows that $g\in L^{\infty}(S; X^{*})$ with $\|g\|_{L^{\infty}(S; X^{*})} \leq \|\phi\|_{(L^{1}(S; X))^{*}}$.
Define $\phi_{g} \colon L^{1}(S; X) \to \mathbb{K}$ by
\begin{equation}
\langle f, \phi_{g} \rangle := \int_{S} \langle f(s), g(s) \rangle \, d\mu (s) .
\end{equation}
By Proposition 2 we have $\phi_{g}\in (L^{1}(S; X))^{*}$. We claim $\phi_{g} = \phi$. Fix $x\in X$ and define
\begin{equation}
\mathcal{C} := \{C\in\mathcal{A} : \mu (C) < \infty \text{ and } \langle 1_{C}\otimes x, \phi \rangle = \langle 1_{C}\otimes x, \phi_{g} \rangle \} .
\end{equation}
We wish to show $\mathcal{C}$ contains every member of $\mathcal{A}$ with finite measure. After that we will use density to show that $\phi_{g} = \phi$.
We first show $A_{i}\in \mathcal{C}$ for every $i\in I$. Let $i\in I$. We have
\begin{align*}
\langle 1_{A_{i}}\otimes x, \phi \rangle = \mu (A_{i}) \langle x, x_{A_{i}}^{*} \rangle &= \int_{S} \langle (1_{A_{i}}\otimes x)(s), (1_{A_{i}} \otimes x_{A_{i}}^{*})(s) \rangle \, d\mu (s) \\
&= \int_{S} \langle (1_{A_{i}}\otimes x)(s), g(s) \rangle \, d\mu (s) = \langle 1_{A_{i}}\otimes x, \phi_{g} \rangle .
\end{align*}
Hence $A_{i}\in\mathcal{C}$. Now let $C\in\mathcal{A}$ with $\mu (C) < \infty$. Because $\mu (S\setminus \bigcup_{i\in I} A_{i}) = 0$, using Proposition 1 there is $I_{0}\subseteq I$ such that $C$ is equivalent to $\bigcup_{i\in I_{0}} A_{i}$. We consequently find a sequence $(C_{n})_{n\in\mathbb{N}}$ of finite unions of members of $\{A_{i} : i\in I_{0}\}$ such that $\lim_{n\to\infty} \mu (C\Delta C_{n}) = 0$. As $\mathcal{C}$ is closed under finite unions we have that $C_{n}\in\mathcal{C}$ for each $n\in\mathbb{N}$. Combining this with the continuity of both $\phi$ and $\phi_{g}$ as well as the dominated convergence theorem, we see
\begin{equation}
\langle 1_{C} \otimes x, \phi \rangle = \lim_{n\to\infty} \langle 1_{C_{n}}\otimes x, \phi \rangle = \lim_{n\to\infty} \langle 1_{C_{n}} \otimes x, \phi_{g} \rangle = \langle 1_{C}\otimes x, \phi \rangle .
\end{equation}
Hence $C\in\mathcal{C}$. This means that $\phi$ and $\phi_{g}$ agree on all functions of the form $1_{C}\otimes x$ where $C\in\mathcal{A}$ with $\mu (C) < \infty$ and $x\in X$. Since the linear span of such functions is dense in $L^{1}(S;X)$ by Lemma 1, it follows from the continuity of $\phi$ and $\phi_{g}$ that $\phi = \phi_{g}$.
Case 2. Suppose $p \in (1,\infty )$. We do not assume $(S, \mathcal{A}, \mu )$ is $\sigma$-finite. Let $\phi \in (L^{p}(S; X))^{*}$. As in the first case, by Proposition 2 it is sufficient to find some $g\in L^{q}(S;X^{*})$ such that
\begin{equation}
\langle f, \phi \rangle = \int_{S} \langle f(s), g(s)\rangle \, d\mu (s)
\end{equation}
for all $f\in L^{p}(S;X)$. Let $\{A_{i} : i\in I\}$ be a maximal collection of pairwise disjoint atoms each having finite measure such that
\begin{equation}
\{i\in I : \langle 1_{A_{i}}\otimes x, \phi \rangle = 0 \text{ for all } x\in X \} = \emptyset . \tag{2}
\end{equation}
Such a collection can be obtained using Zorn's lemma.
Let $i\in I$. We consider the linear map
\begin{equation}
x\mapsto \mu (A_{i})^{-\tfrac{1}{p}} \langle 1_{A_{i}}\otimes x, \phi \rangle \tag{3}
\end{equation}
from $X$ into $\mathbb{K}$. Since $\|1_{A_{i}}\otimes x \|_{L^{p}(S;X)} = \mu (A_{i})^{\tfrac{1}{p}} \|x\|$ and hence
\begin{equation}
|\mu (A_{i})^{-\tfrac{1}{p}} \langle 1_{A_{i}}\otimes x, \phi \rangle | \leq \mu (A_{i})^{-\tfrac{1}{p}} \|1_{A_{i}}\otimes x \|_{L^{p}(S;X)} \|\phi\|_{(L^{p}(S;X))^{*} } = \|\phi\|_{(L^{p}(S;X))^{*} } \|x\|
\end{equation}
for each $x\in X$, we see that the map $(3)$ is a bounded linear functional on $X$ with norm less than or equal to $\|\phi\|_{(L^{p}(S;X))^{*} }$. Hence there is some $x_{A_{i}}^{*} \in X^{*}$ with $\|x_{A_{i}}^{*}\| \leq \|\phi\|_{(L^{p}(S;X))^{*} }$ such that
\begin{equation}
\langle x, x_{A_{i}}^{*} \rangle = \mu (A_{i})^{-\tfrac{1}{p}} \langle 1_{A_{i}}\otimes x, \phi \rangle
\end{equation}
for all $x\in X$. Note that by $(2)$ we have $\| x_{A_{i}}^{*} \| \neq 0$ for every $i\in I$, that is we have
\begin{equation}
I = \{i\in I : \| x_{A_{i}}^{*} \| \neq 0 \} . \tag{4}
\end{equation}
We now show
\begin{equation}
\sup \Big\{ \sum_{i\in F} \|x_{A_{i}}^{*}\|^{q} : F\subseteq I \text{ finite} \Big\} \leq \|\phi\|_{(L^{p}(S; X))^{*}}^{q} . \tag{5}
\end{equation}
Once this is done, we deduce that $I$ is countable and use this to define our candidate function $g\in L^{q}(S; X^{*})$.
Let $F\subseteq I$ be finite. Denote by $\ell^{p}(F)$ the Banach space of functions $c\colon F\to \mathbb{K}$ with the norm $\|c\|_{\ell^{p}(F)} = (\sum_{i\in F} |c_{i}|^{p})^{\tfrac{1}{p}}$. By the Hölder inequality and a suitable choice of $c\in \ell^{p}(F)$ we have
\begin{equation}
\Big( \sum_{i\in F} \|x_{A_{i}}^{*}\|^{q} \Big)^{\tfrac{1}{q}} = \sup\Big\{\sum_{i\in F} |c_{i}| \, \|x_{A_{i}}^{*}\| : c\in \ell^{p}(F) \text{ and } \|c\|_{\ell^{p}(F)} \leq 1 \Big\} . \tag{6}
\end{equation}
Let $\varepsilon > 0$. For each $i\in F$ choose $x_{A_{i}}\in X$ with $\|x_{A_{i}}\| \leq 1$ such that $\langle x_{A_{i}}, x_{A_{i}}^{*}\rangle \geq \|x_{A_{i}}^{*}\| - \varepsilon$. Let $c\in \ell^{p}(F)$ with $\|c\|_{\ell^{p}(F)} \leq 1$ and define $\psi := \sum_{i\in F} \mu (A_{i})^{-\tfrac{1}{p}} |c_{i}| 1_{A_{i}}\otimes x_{A_{i}}$. By the Hölder inequality we see $\psi \in L^{p}(S; X)$ with $\|\psi\|_{L^{p}(S; X)} \leq 1$,
\begin{equation}
\langle \psi , \phi \rangle = \sum_{i\in F} \mu (A_{i})^{-\tfrac{1}{p}} |c_{i}| \langle 1_{A_{i}}\otimes x_{A_{i}}, \phi \rangle = \sum_{i\in F} |c_{i}| \langle x_{A_{i}}, x_{A_{i}}^{*} \rangle
\end{equation}
and
\begin{align*}
\sum_{i\in F} |c_{i}| \, \|x_{A_{i}}^{*}\| &\leq \varepsilon \sum_{i\in F} |c_{i}| + \sum_{i\in F} |c_{i}| \langle x_{A_{i}}, x_{A_{i}}^{*}\rangle \\
&= \varepsilon \sum_{i\in F} |c_{i}| + \langle \psi, \phi \rangle \\
&\leq \varepsilon \sum_{i\in F} |c_{i}| + \|\phi\|_{(L^{p}(S; X))^{*}} . \tag{7}
\end{align*}
Since $\varepsilon > 0$ was arbitrary, we combine $(6)$ and $(7)$ to obtain
\begin{equation}
\sum_{i\in F} \|x_{A_{i}}^{*}\|^{q} \leq \|\phi\|_{(L^{p}(S; X))^{*}}^{q} .
\end{equation}
This shows $(5)$. We next claim $I$ is countable. By Lemma 2 we see from $(5)$ that the set
\begin{equation}
\{i\in I : \| x_{A_{i}}^{*} \| \neq 0 \} = \{i\in I : \| x_{A_{i}}^{*} \|^{q} \neq 0 \} \tag{8}
\end{equation}
is countable. But as the set $(8)$ is equal to $I$ by $(4)$, we deduce that $I$ is countable. Furthermore, by modifying the collection on a set of measure zero if necessary, we assume the collection $\{A_{i} : i\in I\}$ is pairwise disjoint.
Define $g\colon S\to X^{*}$ by
\begin{equation}
g(s) := \sum_{i\in I} (1_{A_{i}} \otimes x_{A_{i}}^{*})(s) .
\end{equation}
Since $I$ is countable and $\{A_{i} : i\in I\}$ is pairwise disjoint we have that $g$ is well-defined and Bochner measurable. Furthermore, by $(5)$ we have $g\in L^{q}(S; X^{*})$ with $\|g\|_{L^{q}(S; X^{*})} \leq \|\phi_{g}\|_{(L^{p}(S; X))^{*}}$.
Define $\phi_{g} \colon L^{p}(S; X) \to \mathbb{K}$ by
\begin{equation}
\langle f, \phi_{g} \rangle := \int_{S} \langle f(s), g(s) \rangle \, d\mu (s) .
\end{equation}
By Proposition 2 we have $\phi_{g}\in (L^{p}(S; X))^{*}$. We claim $\phi_{g} = \phi$. Fix $x\in X$ and define
\begin{equation}
\mathcal{C} := \{C\in\mathcal{A} : \mu (C) < \infty \text{ and } \langle 1_{C}\otimes x, \phi \rangle = \langle 1_{C}\otimes x, \phi_{g} \rangle \} .
\end{equation}
As in the first case, we wish to show $\mathcal{C}$ contains every member of $\mathcal{A}$ with finite measure. After that we will use density to show that $\phi_{g} = \phi$.
We first show $A_{i}\in \mathcal{C}$ for every $i\in I$. Let $i\in I$. We have
\begin{align*}
\langle 1_{A_{i}}\otimes x, \phi \rangle = \mu (A_{i}) \langle x, x_{A_{i}}^{*} \rangle &= \int_{S} \langle (1_{A_{i}}\otimes x)(s), (1_{A_{i}} \otimes x_{A_{i}}^{*})(s) \rangle \, d\mu (s) \\
&= \int_{S} \langle (1_{A_{i}}\otimes x)(s), g(s) \rangle \, d\mu (s) = \langle 1_{A_{i}}\otimes x, \phi_{g} \rangle .
\end{align*}
Hence $A_{i}\in\mathcal{C}$. Next, let $B\in \mathcal{A}$ be an atom with $\mu (B) < \infty$ and $B\cap \bigcup_{i\in I} A_{i} = \emptyset$. By $(2)$ and recalling the conditions under which $\{A_{i} : i\in I\}$ is maximal, we see $\langle 1_{B}\otimes x, \phi \rangle = 0$. As we see $\langle 1_{B}\otimes x, \phi_{g}\rangle = 0$ from $B\cap \bigcup_{i\in I} A_{i} = \emptyset$, it follows that $B\in \mathcal{C}$.
Now let $C\in\mathcal{A}$ with $\mu (C) < \infty$. By Proposition 1 there is $I_{0}\subseteq I$ such that $C\cap \bigcup_{i\in I} A_{i}$ is equivalent to $\bigcup_{i\in I_{0}} A_{i}$. Let $\{B_{j} : j\in J\}$ be a maximal collection of pairwise disjoint atoms each contained in $C\setminus \bigcup_{i\in I} A_{i}$. Then by Proposition 1 we have that $J$ is countable and that $C\setminus \bigcup_{i\in I} A_{i}$ is equivalent to $\bigcup_{j\in J} B_{j}$. We consequently find a sequence $(C_{n})_{n\in\mathbb{N}}$ of finite unions of members of $\{A_{i} : i\in I_{0}\} \cup \{B_{j} : j\in J\}$ such that $\lim_{n\to\infty} \mu (C\Delta C_{n}) = 0$. As $\mathcal{C}$ is closed under finite unions we have that $C_{n}\in\mathcal{C}$ for each $n\in\mathbb{N}$. Combining this with the continuity of both $\phi$ and $\phi_{g}$ as well as the dominated convergence theorem, we see
\begin{equation}
\langle 1_{C} \otimes x, \phi \rangle = \lim_{n\to\infty} \langle 1_{C_{n}}\otimes x, \phi \rangle = \lim_{n\to\infty} \langle 1_{C_{n}} \otimes x, \phi_{g} \rangle = \langle 1_{C}\otimes x, \phi \rangle .
\end{equation}
Hence $C\in\mathcal{C}$. This means that $\phi$ and $\phi_{g}$ agree on all functions of the form $1_{C}\otimes x$ where $C\in\mathcal{A}$ with $\mu (C) < \infty$ and $x\in X$. Since the linear span of such functions is dense in $L^{p}(S;X)$ by Lemma 1, it follows from the continuity of $\phi$ and $\phi_{g}$ that $\phi = \phi_{g}$.
[1] Hytönen, T., Neerven, J. van, Veraar, M. and Weis, L., Analysis in Banach spaces. Vol. I. Martingales and Littlewood-Paley theory, vol. 63 of Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern
Surveys in Mathematics. Springer, Cham, 2016.
