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

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 $$1$$?

• From wiki: "For example, as a consequence of the Riemann–Lebesgue lemma, the sequence μ_n of measures on the interval [−1, 1] given by μ_n(dx) = (1+ sin(nx))dx converges strongly to Lebesgue measure." Jan 29, 2020 at 10:04
• @ Fedor Petrov - it depends on what the OP meant by strong convergence of measures. I was surprised to learn that wiki (apparently this is the article you quote) introduces a rather artificial notion of "strong convergence" of measures (I have never come across it in real life) and distinguishes it from the convergence in total variation ($\equiv$ convergence in the strong topology on the space of measures). This is not the first time I come across highly dubious claims in wiki.
– R W
Jan 29, 2020 at 10:34
• I did indeed mean convergence in the sense stated by wiki.. Jan 29, 2020 at 10:37
• If it’s meant in the other sense is it true? Jan 29, 2020 at 10:38
• It seems like it is indeed true, if I didn’t make any mistakes. Jan 29, 2020 at 10:41

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.
$$\|\mu_n-\mu\|=\int_X|f_n-1|d\mu.$$ 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 $$\mu$$.