Let $(X,\Sigma)$ be a measure space, let $\mathcal P(X)=\{ p : (X,\Sigma,p) \text{ is a probability space} \}$. We use the total variation distance, i.e. for any $p, q\in\mathcal P(X)$, $\| p-q \| = \sup_{A\in \Sigma} p(A)-q(A)$.
Let $\alpha(p,q)=\inf\{ \alpha\in [1,\infty[ : \forall A\in\Sigma, \frac{1}{\alpha} p(A)\leq q(A)\leq \alpha p(A) \}$, notice that $\{ (p,q) \in\mathcal P(X)^2 : \alpha(p,q)<\infty\}$ is an equivalence relation and $\alpha(p,q)<\infty$ implies that both $p\ll q$ and $q\ll p$. Also $\alpha(p,q)=1$ iff $p=q$. Let $F(p)=\{ q\in\mathcal P(X) : \alpha(p,q)< \infty\}$.
Let $p\in\mathcal P(X)$ and $\{p_n\}$ in $F(p)$ such that $p_n$ converges to $p$ in total variation, is it true that $\alpha(p,p_n)\to 1$ ?
What I know is that this cannot hold in general if $p_n$ is not in $F(p)$, so my question is about if adding this constraint would be enough.
This question is related to an attempt of solving this other question for a specific underlying space. Also if $\alpha$ is a studied object I would be happy to get sources on that.
