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Apr 8 at 20:19 comment added Martin Väth Yes, the measurability of $f$ implies the existence of a product measurable $g$ whenever $X$ is an ideal space ($X=L_\infty(\Bbb R^d)$ is such a space). The converse may fail for non-regular ideal spaces (like $X=L_\infty(\Bbb R^d$): For product-measurable $g$ the function $f$ is not (strongly Bochner) measurable, in general.
Aug 10, 2023 at 13:36 comment added Akira May I ask if such measurable representation exists if $X=L_\infty (\mathbb{R}^d)$?
Nov 22, 2020 at 15:45 history edited Martin Väth CC BY-SA 4.0
Fix name and monograph reference.
Nov 22, 2020 at 11:16 history edited Martin Väth CC BY-SA 4.0
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Nov 22, 2020 at 11:09 history answered Martin Väth CC BY-SA 4.0