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Given a family $\{\mu \}_{i\in I}$ on a Polish space (complete, separable metric space) $X$. When does there exist a measure $\lambda$ such that $$\mu_i=f_i \lambda$$

where the $f_i$ are densities (Radon-Nikodym) of $\mu_i$ with respect to $\lambda$.

EDIT: What is a verifiable condition in the case $I$ is uncountable.

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  • $\begingroup$ What is $f_i$ supposed to be? $\endgroup$ – Aaron Tikuisis Aug 20 '12 at 12:49
  • $\begingroup$ the densities/radon nikodym derivative $\endgroup$ – warsaga Aug 20 '12 at 12:58
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At least for finite measures (the tags seem to suggest this question concerns mainly probability measures), this is equivalent to $\{\mu_i:i\in I\}$ being separable under the variation norm.

The variation norm on finite measures dominated by some finite measure $\nu$ agrees with the $L_1(\nu)$-norm of the corresponding Radon-Nikodym derivatives, so separability of $\{\mu_i:i\in I\}$ follows in this case from separability of $L_1(\nu)$.

Conversely, if the sequence of nonzero finite measures $\langle\lambda_n\rangle$ is dense in $\{\mu_i:i\in I\}$ under the variation norm, then whenever $\mu_i(A)>0$ for some $i\in I$, then there is some $n$ such that $\lambda_n(A)>0$. It follows that $\{\mu_i:i\in I\}$ is dominated by the probability measure $\kappa$ given by $$\kappa(A)=\sum_{n=1}^\infty \frac{1}{2^n \lambda_n(X)}\lambda_n(A).$$

Relating this to the comment by Jochen Wengenroth: If $I$ is compact and the function $i\mapsto\nu_i$ is continuous when the range is endowed with the variation distance, the range will be compact and be separable as a compact metrizable space, so in this case, a finite dominating measure will exist.

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A countable family of sigma-finite measures, yes.

Can we drop a condition?

Drop sigma-finite: consider two measures on $\mathbb R$: Lebesgue measure and counting measure.

Drop countable: on $\mathbb R$ consider the family of measures, one of them is Lebesgue measure, and the rest are the unit point masses at all the points of $\mathbb R$.

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  • $\begingroup$ I think that the OP has something about the dependence $i\mapsto \mu_i$ in mind. I could imagine positive answers for an interval $I$ and continuous dependence if the topology considered on all (finite) measures on $X$ is strong enough. $\endgroup$ – Jochen Wengenroth Aug 20 '12 at 14:17

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