Let $(\Omega,\mathcal{F})$ be a measurable space. Given a subset $B\subset\mathbb{R}$, which is not a Borel set, i.e. $B\notin \mathcal{B}(\mathbb{R})$.
How can I conclude now that $B\times \Omega\notin \mathcal{B}(\mathbb{R})\otimes \mathcal{F}$?
If it does not hold generally, what do we need to assume additionally in order for the statement to hold?
Pick a probability measure $\mathbb{P}$ on $(\Omega,\mathcal{F})$ (e.g. the Dirac measure centered at some point $\omega_0 \in \Omega$.) Let $B \subseteq \mathbb{R}$ be a set such that $B \times \Omega \in \mathcal{B}(\mathbb{R}) \otimes \mathcal{F}$. Then it follows from Tonelli's theorem that the mapping
$$x \mapsto \int_{\Omega} 1_{B \times \Omega}(x,\omega) \, d\mathbb{P}(\omega) = 1_B(x)$$
is measurable, i.e. $B \in \mathcal{B}(\mathbb{R})$. This shows the implication
$$B \times \Omega \in \mathcal{B}(\mathbb{R}) \otimes \mathcal{F} \implies B \in \mathcal{B}(\mathbb{R}), \tag{1}$$
and so
$$B \notin \mathcal{B}(\mathbb{R}) \implies B \times \Omega \notin \mathcal{B}(\mathbb{R}) \otimes \mathcal{F}.$$
Remark: The implication $(1)$ can be also shown by considering the measurability of suitable mappings (meaning that there is no need for integration theory/Tonelli's theorem to prove $(1)$). The reasoning goes as follows: For fixed $\omega_0 \in \Omega$ it is not difficult to see that the mapping
$$(\mathbb{R},\mathcal{B}(\mathbb{R})) \ni x \mapsto \iota(x):=(x,\omega_0) \in (\mathbb{R} \times \Omega,\mathcal{B}(\mathbb{R})) \otimes \mathcal{F})$$
is measurable. Now if $B \subseteq \mathbb{R}$ is such that $B \times \Omega \in \mathcal{B}(\mathbb{R}) \otimes \mathcal{F}$, then the measurability of $\iota$ implies
$$\iota^{-1}(B \times \Omega) = B \in \mathcal{B}(\mathbb{R}).$$