Preliminaries
Let $T>0$ be a positive constant, $d,n \in \mathbb{N}_{\geq 1}$ be natural numbers and $(\Omega, \mathcal{G}, \mathbb{P})$ be a probability space.
For $m \in \mathbb{N}_{\geq 0}$, we define a sequence of functions
$$X_m : \Omega \times [0,T] \times \mathbb{R}^n \rightarrow \mathbb{R}^d $$
such that, for all $m \in \mathbb{N}_{\geq 0}$,
$$ \omega \mapsto X_m(\omega, t, \theta) $$ is $\mathcal{G}/\mathcal{B}(\mathbb{R}^d)$ measurable, $$ t \mapsto X_m(\omega, t, \theta) $$ is continuous, and $$ \theta \mapsto X_m(\omega, t, \theta) $$ is continuous as well.
Now assume, that there is an $A \subseteq \Omega$ with $\mathbb{P}(A) = 1$, such that for all $\tilde{\omega} \in A$, it holds that, for all $t \in [0,T]$ and $\theta \in \mathbb{R}^n$, $$ \lim_{m \rightarrow \infty} X_m(\tilde{\omega}, t, \theta) $$ exists and is finite.
Now define a new function
$$ X : \Omega \times [0,T] \times \mathbb{R}^n \rightarrow \mathbb{R}^d $$ via
$$ X(\omega, t, \theta) := \lim_{m \rightarrow \infty} \mathbb{1}_{A}(\omega) X_m(\omega, t, \theta). $$
Moreover, we assume, that, for all $\tilde{\omega} \in A$ the sequence of functions $$ t \mapsto X_m(\tilde{\omega}, t, \theta) $$ converges uniformly on $[0,T]$ towards $$t \mapsto X(\tilde{\omega}, t, \theta).$$
Question
Can we conclude from all of this, that $$ \theta \mapsto X_m(\omega, t, \theta) $$ is $\mathcal{B}(\mathbb{R}^n) / \mathcal{B}(\mathbb{R}^d) $ -measurable? And can we even conclude from this, that $$ \theta \mapsto X_m(\omega, t, \theta) $$ is continuous?
Which conditions would be sufficient to guarantee either measurability or even continuity for $$ \theta \mapsto X_m(\omega, t, \theta) ?$$