Let $(\Omega,\mathcal F,P)$ be a probability space and $(T,\mathcal A,\mu)$ be a finite measure space. Let $X:\Omega\times T\to \mathbb R$ be measurable w.r.t the product sigma-algebra $\mathcal F \otimes \mathcal A$ and assume that $\int_{\Omega\times T} X(\omega,t)^2 d(P\otimes \mu)(\omega,t)<\infty$, i.e., $X\in L^2(P\otimes \mu)$.
Let $A=\{w\in \Omega: \int_T X(\omega,t)^2 d\mu(t)<\infty\}$, so that $P(A)=1$ and define the random function $Y$
$$\begin{align} \Omega &\to L^2(\mu) \\ \omega &\mapsto X(\omega,\cdot)1_A(\omega) \end{align} $$
I consider $L^2(\mu)$ as a Hilbert space, so the $\sigma$-algebra on $L^2(\mu)$ is the corresponding Borel $\sigma$-algebra.
Question: I want to show that $Y$ is measurable.
By Lemma 4.1 in Kallenberg's Foundations of Modern Probability Theory, it suffices to show the inclusion $$ \mathcal B(L^2(\mu)) \subset L^2(\mu)\cap \mathcal B(\mathbb R)^T,$$ where $\mathcal B(\mathbb R)^T$ is the $\sigma$-algebra on $\mathbb R^T$ generated by the evaluation maps, $ L^2(\mu)\cap \mathcal B(\mathbb R)^T$ denotes its trace on $L^2(\mu)$.
Is this inclusion even true ? Is there another way to prove measurability of $Y$ ?