Suppose I am given a countable family $(\mu_n)$ of finite Borelmeasures on a compact interval $[0,T]$. Can I find a dominating measure $\mu$ (with $\mu_n \ll \mu$ for all $n$), such that all RadonNikodym densities $\frac{d\mu_n}{d\mu}$ are essentially bounded by a constant $K$ (independent of n)?

$\begingroup$ 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
2 Answers
No, take any finite measure $\mu$ and look at the family $n \mu$ for n integral.

$\begingroup$ I am guessing some assumptions were missing...? $\endgroup$ Oct 6, 2016 at 14:06

$\begingroup$ Thanks for the simple clear counterexample. Do you have a counterexample for the case that the mass of all measures is uniformly bounded (e.g. only probability measures)? $\endgroup$– MKROct 7, 2016 at 9:08
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 RadonNikodym densities cannot be uniformly bounded on any set of positive measure.