Let $\{\mu_n\}$ be a sequence of Borel probability measures on $\mathcal{X} \subset \mathbb{R}^d$, $d \in \mathbb{N}\setminus\{0\}$, which are absolutely continuous with respect to the Lebesgue measure, with densities $\{f_n\}$. Similarly, let $\mu$ be an absolutely continuous probability measure, with density $f$.
A well known fact in probability theory is that convergence in distribution (weak convergence) of $\mu_n$ to $\mu$ does not entail that $f_n$ converges to $f$ pointwise almost everywhere. Another well known result (Scheffé's lemma) is that if $f_n$ converges pointwise everywhere to a probability density, then $\mu_n$ converges in total variation to the pertaining probability measure. In particular: $$ f_n(x)\to f(x), \, a.e. \, on \, \mathcal{X} \implies \sup_{B \in \mathcal{B}(\mathcal{X})}|\mu_n(B)-\mu(B)|=0 $$ where $\mathcal{B}(\mathcal{X})$ is the Borel $\sigma$-algebra of $\mathcal{X}$.
QUESTION 1: Is it possible that $f_n$ converges almost everywhere to some nondegenerate function (i.e. nonnull on a set with positive Lebesgue measure) which is not a probability density function but still $\mu_n$ weakly converges to some probability measure?
I was starting from a simple example: let's say that $\mathcal{X}=[0,1]$. We might have that $\mu_n$ assigns no mass to the boundary $\{0,1\}$ but converges weakly to a probabity measure $\mu_*$ whose restriction to $(0,1)$ is absolutely continuous, with density $f_*$ (not a probability density) and such that $\mu_*(\{0\})>0$, $\mu_*(\{1\})>0$. Let's firstly assume that $\{f_n\}$ is dominated by an integrable function $g$ and converges to some function $f_{**}$ a.e. on $(0,1)$: in this case, for every Borel $B \subset (0,1)$, we must have that
$$
\mu_n(B)=\int_B f_n(x)dx \to \int_{B} f_{**}(x)dx =\mu_*(B)
$$
But then (by the properties of Radon-Nikodym derivatives) we must have $f_{*}=f_{**}$ almost everywhere and $f_n \to f_*$ almost everywhere.
QUESTION 2 Removing the assumption of dominance by an integrable function, can $f_n$ converge to $f_{**} \neq f_*$ on a set of non-null Lebeasgue measure?
Let $$ f_n(x)=\cases{n^2x&if $\ 0\le x\le\frac{1}{2n}$\\ n^2\left(\frac{1}{n}-x\right)&if $\ \frac{1}{2n}\le x\le\frac{1}{n}$\\ \frac{n}{2(n-2)}&if $\ \frac{1}{n}\le x\le1-\frac{1}{n}$\\ n^2\left(x-1+\frac{1}{n}\right)&if $\ 1-\frac{1}{n}\le x\le1-\frac{1}{2n}$\\ n^2\left(1-x\right)&if $\ 1-\frac{1}{2n}\le x\le1$\\ 0&otherwise.} $$ Then $\ \lim_\limits{n\rightarrow\infty}f_n(x)=f_*(x)\ $ for all $\ x\ $, where $$ f_*(x)=\cases{\frac{1}{2}&if $\ 0<x<1$\\ 0&otherwise.} $$ If $\ \mu_n(B)=\displaystyle\int_B f_n(x)dx\ $, then $\ \mu_n\ $ are probability measures which converge weakly to the probability measure $\ \mu_*\ $ whose distribution function $\ F_{\mu_*}\ $ is given by $$ F_{\mu_*}(x)=\cases{0&if $\ x<0$\\ \frac{2x+1}{4}&for $\ 0\le x<1$\\ 1&for $\ 1\le x\ $.} $$ That is, $\ \mu_*\ $ has atoms of weight $\ \frac{1}{4}\ $ at $\ x=0\ $ and $\ x=1\ $, and the remaining half of its weight is uniformly distributed over the interval $\ (0,1)\ $.