Is strong convergence of measures equivalent to convergence in measure of the Radon Nikodym derivatives?

Question

Let $X$ be a measure space, and suppose $\mu_i$ are probability measures on $X$ that are absolutely continuous with respect to another probability measure $\mu$. Is strong convergence of $\mu_i$ to $\mu$ equivalent to convergence in measure (wrt $\mu$) of the Radon nikodym derivatives $\frac{d\mu_i}{d\mu}$ to $1$?

Answer 1

Let $A_n:=\{x\colon f_n(x)\le1\}$ and $B_n:=\{x\colon f_n(x)>1\}$, where $f_n:=\frac{d\mu_n}{d\mu}$. Then the total variation of $\mu_n-\mu$ is $$\|\mu_n-\mu\|=\int_{B_n}(f_n-1)d\mu+\int_{A_n}(1-f_n)d\mu=2\int_{A_n}(1-f_n)d\mu\to0$$ by dominated convergence if $f_n\to1$ in measure wrt $\mu$; the latter displayed equality is the key, even though simple, observation here.

Vice versa,
$$\|\mu_n-\mu\|=\int_X|f_n-1|d\mu.$$ So, as noted by Nate Eldredge, $\|\mu_n-\mu\|\to0$ means that $f_n\to1$ in $L^1(\mu)$, which implies, by Markov's inequality $$\mu\{x\colon|f_n(x)-1|>\epsilon\}\le\frac1\epsilon\,\int_X|f_n-1|d\mu$$ for all $\epsilon>0$, that $f_n\to1$ in measure wrt $\mu$.

Thus, $\mu_n\to\mu$ in total variation iff $\frac{d\mu_n}{d\mu}\to1$ in measure wrt $\mu$.