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\newcommand{\bR}{\mathbb{R}}
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\newcommand{\bP}{\mathbb{P}}
\newcommand{\bE}{\mathbb{E}}
\newcommand{\bF}{\mathbb{F}}
\newcommand{\bD}{\mathbb{D}}
\newcommand{\cA}{\mathcal{A}}
\newcommand{\cB}{\mathcal{B}}
\newcommand{\cD}{\mathcal{D}}
\newcommand{\cE}{\mathcal{E}}
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\newcommand{\diff}{\mathop{}\!\mathrm{d}}
$
We need the following result:

>[Lemma](https://math.stackexchange.com/questions/4285297/a-detailed-and-self-contained-proof-of-tonellis-theorem) Let $B \in \cA \otimes \cE$. For $\omega \in \cA$, we define $B_\omega := \{ x \in E : (\omega, x) \in B\}$. First, $B_\omega \in \cE$. We define $f_B : \Omega \to \bR_+$ by $f_B (\omega) := \mu(B_\omega)$. Second, $f_B$ is measurable.


WLOG, we assume $f$ is non-negative. There is a sequence $(f^n)$ of simple functions such that $f_n \uparrow f$ *everywhere*. We assume
$$
f^n = \sum_{k=1}^{n_k} r^n_k 1_{B^n_k},
$$
where $r^n_k \in \bR_+$ and $B^n_k \in \cA \otimes \cE$. Let
$$
B^n_{k, \omega} := \{ x \in E : (\omega, x) \in B^n_k\}.
$$

By **Lemma**, $B^n_{k, \omega} \in \cE$. We define $f^n_\omega : E \to \bR_+$ by
$$
f^n_\omega (x) := \sum_{k=1}^{n_k} r^n_k 1_{B^n_{k, \omega}} (x).
$$

Clearly, $f^n_\omega$ is measurable. We define $g^n : \Omega \to \bR_+$ by
$$
g^n (\omega) := \int_E f^n_\omega \diff \mu = \sum_{k=1}^{n_k} r^n_k \mu(B^n_{k, \omega}). 
$$

By **Lemma**, $g^n$ is measurable. We have for each $\omega \in \Omega$ that $f^n_\omega \uparrow f(\omega, \cdot)$ *everywhere*. By MCT,
$$
g^n (\omega) \uparrow \int_E f(\omega, \cdot) \diff \mu
\quad \text{as} \quad n \to \infty.
$$

The *everywhere* limit of a sequence of measurable functions is measurable, so the map $\omega \mapsto \int_E f(\omega, \cdot) \diff \mu$ is measurable. The claim then follows.