common dominating measure for a family of measures

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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 $mu_i$ wrt to $\lambda$.

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

-

What is $f_i$ supposed to be? – Aaron Tikuisis Aug 20 at 12:49
the densities/radon nikodym derivative – warsaga Aug 20 at 12:58

1 Answer

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 $\mathbb R$.

-

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.  – Jochen Wengenroth Aug 20 at 14:17
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.924249529838562, "perplexity_flag": "middle"}