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 $1$?
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2$\begingroup$ 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." $\endgroup$– Fedor PetrovJan 29, 2020 at 10:04
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$\begingroup$ @ 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. $\endgroup$– R WJan 29, 2020 at 10:34
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$\begingroup$ I did indeed mean convergence in the sense stated by wiki.. $\endgroup$– James BaxterJan 29, 2020 at 10:37
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$\begingroup$ If it’s meant in the other sense is it true? $\endgroup$– James BaxterJan 29, 2020 at 10:38
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$\begingroup$ It seems like it is indeed true, if I didn’t make any mistakes. $\endgroup$– James BaxterJan 29, 2020 at 10:41
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1 Answer
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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 $\mu$.
answered Jan 29, 2020 at 14:58