Measurability of probability densities as maps from a parameter space

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Consider the measurable space $(X:=(0,\infty)^d, \mathcal{X})$, for some $d \in \mathbb{N}_+$, where $\mathcal{X}$ denotes the usual Borel $\sigma$-field. Let $(\Theta, \sigma(\tau))$ denote a parameter space endowed with the Borel $\sigma$-filed induced by the topology $\tau$ on $\Theta$. Let $\mathcal{P}:=\{P_\theta: \, \theta \in \Theta\}$ be a class of probability measures on $(X,\mathcal{X})$ which are absolutely continuous with respect to the Lebsegue measure $\lambda$. Then, by separability of $X$, there exists a positive $\mathcal{X}\times \sigma(\tau)$-measurable map $X \times \Theta \mapsto (0, \infty):(x,\theta) \mapsto p(x, \theta)$ such that, for every $\theta$, $ p(\cdot,\theta)$ is a Lebesgue density of $P_\theta$.

My questions are the following:

a) denote by $\sigma(w_X)$ the Borel $\sigma$-field induced by the topology of pointwise convergence on the space of probability densities on $X$. Consider the statement:

S.1: the map $\theta \mapsto p(\cdot,\theta)$ is measurable with respect to the $\sigma$-fields $\sigma(\tau)$ (on the initial space) and $\sigma(w_X)$ (on the arrival space).

Is it correct?

b) Denote by $\Vert P-Q \Vert_{TV}:=\sup_\mathcal{X}|P(A)-Q(A)|$ the total variation distance between two probability measures on $(X,\mathcal{X})$. Recall that, by Scheffé's lemma, pointwise convergence of a sequence of probability density functions $p(\cdot, \theta_n)$ to a density $p(\cdot,\theta)$ entails that $2\Vert P_{\theta_n} - P_\theta \Vert_{TV}=\Vert p(\cdot, \theta_n)-p(\cdot, \theta)\Vert_{L_1(\lambda)}=o(1)$. Bearing this fact in mind and denoting by $\sigma(L_1(\lambda))$ the Borel $\sigma$-field induced by the $L_1(\lambda)$-metric topology on the space of probability densities on $X$, could we conclude from S.1 (provided it is true) that the statement

S.2: the map $\theta \mapsto p(\cdot,\theta)$ is measurable with respect to the $\sigma$-fields $\sigma(\tau)$ (on the initial space) and $\sigma(L_1(\lambda))$ (on the arrival space)

is also correct?