Define $ \mathcal{X}:=([0,1], \mathscr{B}([0,1]))$ - i.e. the unit interval endowed with its Borel $\sigma$-algebra - and let $\mathcal{M}$ be the class of probability measures on $\mathcal{X}$ which are absolutely continuous with respect to the Lebesgue measure.
Two typical topologies which are used to define Borel $\sigma$-fields on $\mathcal{M}$ are the topology of weak convergence of measures, defined via
$\int_0^1f(w)P_n(dw) \to \int_0^1f(w)P(dw), \quad \forall f\in C([0,1])\cap\ell^\infty([0,1]),$
and the metric topology induced by the total variation distance, defined via
$d_{TV}(P,Q):=\sup_{B \in \mathscr{B}([0,1])}|P(B)-Q(B)|\\ \hspace{4.7em}=\sup\{\int_0^1\psi(w)P(dw)-\int_0^1\psi(w)Q(dw): \psi:[0,1]\mapsto[0,1], \text{measurable}\}$
for all $P,Q \in \mathcal{M}$. Define by $\mathscr{B}_W$ and $\mathscr{B}_{TV}$ the associated Borel $\sigma$-fields. Note that since $d_{TV}$ is separable on $\mathcal{M}$, then $\mathscr{B}_{TV}$ coincide with the ball $\sigma$-algebra.
Question 1 Is it true that $\mathcal{B}_W=\mathcal{B}_{TV}$?
Since total variation defines a stronger topology, we should have that $\mathcal{B}_W \subset \mathcal{B}_{TV}$. On the other hand, $(\mathcal{M}, \mathcal{B}_W )$ is Borel isomorphic to the class $(\mathcal{F},\mathscr{B}(L_1))$, where $\mathcal{F}:=\{f:[0,1]\mapsto \mathbb{R}_+, \Vert f \Vert_1=1\}$, and $\mathscr{B}(L_1)$ is the Borel-$\sigma$-algebra induced by the $L_1$-topology on $\mathcal{F}$. Loosely speaking, $\mathcal{F}$ is the class of density functions pertaining to the measures in $\mathcal{M}$. With this in mind, and knowing that also $2d_{TV}(P,Q)=\Vert p-q \Vert_1$, where $p$ and $q$ are the densities of $P$ and $Q$, I would conclude that $$ \{Q \in \mathcal{M}:d_{TV}(P,Q)<\epsilon\} = \{Q \in \mathcal{M}:2 \Vert p-q \Vert_1<\epsilon\} \in \mathcal{B}_W, \quad \forall \epsilon>0, \, \forall P \in \mathcal{M} $$ and hence deduce $\mathcal{B}_{TV}\subset \mathcal{B}_W$. Is this correct?
Question 2 Can we obtain a similar result by showing that $$ \{Q \in \mathcal{M}: \, \sup\{\int_0^1\psi(w)p(dw)-\int_0^1\psi(w)q(dw): \psi:[0,1]\mapsto[0,1], \text{measurable}\}<\epsilon\} \in \mathcal{B}_W, $$ for all $\epsilon>0$ and $ P \in \mathcal{M}$? If yes, how?