$ \newcommand{\bR}{\mathbb{R}} \newcommand{\bT}{\mathbb{T}} \newcommand{\bN}{\mathbb{N}} \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}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cG}{\mathcal{G}} \newcommand{\diff}{\mathop{}\!\mathrm{d}} $ We need the following result:
Lemma Let $B \in \cA \otimes \cE$. For $\omega \in \cA$$\omega \in \Omega$, we define $B_\omega := \{ x \in E : (\omega, x) \in B\}$. First, $B_\omega \in \cE$. We define $f_B : \Omega \to \bR_+$ by $f_B (\omega) := \mu(B_\omega)$. Second, $f_B$ is measurable.
WLOG, we assume $f$ is non-negative. There is a sequence $(f^n)$ of simple functions such that $f_n \uparrow f$ everywhere. We assume $$ 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.