Application of Radon Nikodym Theorem

Question

Let $f_n,g_n \in L^1(\mathbb{R})$ be two sequences of non negative functions such that $h_n:=\frac{g_n}{f_n} \leq K$. Suppose, $f_n \buildrel\ast\over\rightharpoonup \mu_f$ and $g_n \buildrel\ast\over\rightharpoonup \mu_g$ in $\mathcal{M}(\mathbb{R}).$

I have the following doubts.

1. Can we apply Radon–Nikodym theorem to show that there exists $h\in L^{\infty}(\mathbb{R})$ such that $\mu_g=h\mu_f$ and $h_n \rightarrow h$ in the sense of distribution.
2. If not, suppose in additon assume that $h_n \rightarrow h$ in $L^1(\mathbb{R}).$ Now, do we have $\mu_g=h\mu_f?$

Any help would be greatly appreciated.

Answer 1

Yes, to the first half of (1). Since $g_n\le Kf_n$ we have $$\int\phi d\mu_g\le K\int\phi d\mu_f\quad(\phi\in C_c(\Bbb R), \phi\ge0),$$hence $$\mu_g\le K\mu_f.$$Hence $\mu_g<<\mu_f$, so $$d\mu_g=hd\mu_f,$$and hence hence $h\le K$ a.e.[$\mu_f$].

Haven't shown that $h_n\to h$ in any interesting sense (I didn't see that part of the question until after posting this.)

Edit. Thanks to N. S. for giving a simple counterexample to the second part of (1). Say $f_n\to0$, $g_n=(2+(-1)^n)f_n$. (Here $\mu_f=0$, so any $h$ whatever gives $d\mu_g=hd\mu_f$. Except we can't take $h=\lim h_n$ because $\lim h_n$ does not exist.)