Timeline for optimal transport, measurable selection
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 13, 2019 at 21:11 | comment | added | Taras Banakh | @Ryan So the problem reduced to the proof of the surjectivity of the maps $Ppr_X$. But I do not know how to prove this theorem without selection theorems. Maybe for your purposes it suffices to refer to some known results on the surjectivity of maps between spaces of measures? | |
| Feb 13, 2019 at 21:09 | comment | added | Taras Banakh | @Ryan Using the surjectivity of the map $Ppr_X:P(D_v)\to P(A_v)$, we can choose a $\sigma$-additive Borel measure $\mu_v$ on the Borel subset $D_v$ of $X\times Y$ such that $Ppr_X(\mu_v)=\nu{\restriction}A_v\setminus \bigcup_{u<v}A_u$. Then $\mu=\sum_{v\in V}\mu_v$ is a measure withessing the required equality. | |
| Feb 13, 2019 at 18:47 | vote | accept | Ryan | ||
| Feb 13, 2019 at 18:47 | comment | added | Ryan | Hi Taras, thanks for you prompt respond! Could you please elaborate on your previous comment, or refer me to some known results that you mentioned? Thanks again! | |
| Feb 13, 2019 at 6:38 | comment | added | Taras Banakh | @Ryan Oh sorry! I did not read your question carefully. You asked about the proof WITHOUT selections and I red it as WITH selections. Yes, I think there is a proof without these selection results, instead one can use the known results on the surjectivity of the map $Pf:PX\to PY$, induced by a surjective map $f:X\to Y$ from a Polish space $X$ to a compact space $Y$. | |
| Feb 12, 2019 at 22:40 | comment | added | Ryan | Thanks for your answer. I guess the Uniformization Theorem here is some analogy of the Von Neumann measurable selection theorem? Just wondering if there is a more elementary proof without using such results. | |
| Feb 12, 2019 at 7:58 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 21 characters in body
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| Feb 12, 2019 at 7:52 | history | answered | Taras Banakh | CC BY-SA 4.0 |