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 $1$?



Is strong convergence of measures equivalent to convergence in measure of the Radon Nikodym derivatives?
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 $\mu$.


