Suppose I am given a countable family $(\mu_n)$ of finite Borel-measures on a compact interval $[0,T]$. Can I find a dominating measure $\mu$ (with $\mu_n \ll \mu$ for all $n$), such that all Radon-Nikodym densities $\frac{d\mu_n}{d\mu}$ are essentially bounded by a constant $K$ (independent of n)?
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$\begingroup$ Note that the collection of measures domiated by $K\mu$ is uniformly integrable, hence (their RN derivatives are) weakly compact in $L^1(\mu)$ which is a separable infinite dimensional space. Take any countable collection which is not weakly precompact. If you want a norm uniformly bounded example, a dense sequence in the unit ball will do. $\endgroup$– Uri BaderOct 8, 2016 at 16:29
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2 Answers
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No, take any finite measure $\mu$ and look at the family $n \mu$ for n integral.
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$\begingroup$ I am guessing some assumptions were missing...? $\endgroup$ Oct 6, 2016 at 14:06
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$\begingroup$ Thanks for the simple clear counter-example. Do you have a counter-example for the case that the mass of all measures is uniformly bounded (e.g. only probability measures)? $\endgroup$– MKROct 7, 2016 at 9:08
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If you are looking for probability measures, just modify @michael's answer as follows: let $\mu_n$ be the uniform measure on $[0,1/n)$. In fact, by taking 'sliding bumps' ($\mu_{m,n}$ is uniform measure on $[m/n,(m+1)/n)$, for $0\leq m <n$), the Radon-Nikodym densities cannot be uniformly bounded on any set of positive measure.
