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Nov 5 at 23:13 comment added Anthony Quas I guess I am interested in how you arrived at the earlier question also. These kind of conditions seem as though they could be of interest in ergodic theory. There one builds abstract joinings (invariant couplings) of a pair of measures, and it can be of interest to know when the joining is supported on the graph of a function.
Nov 5 at 15:34 comment added thibault jeannin $$ D = \{ y : \pi_{Y=y} \text{ is a Dirac measure} \} $$ Using your approach, I demonstrated the following weaker result: If for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, $B_A = \{ y \in \mathbb{R} \mid \pi_{Y=y}(A) = 1 \}$ satisfies $\nu(B_A) > 0$, then for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, $$ \operatorname{essinf}_{\mathbb{1}_{B_A}(y) \, \nu(dy)} V_X(y) = 0. $$ I asked this question there: [mathoverflow.net/questions/480583/…
Nov 5 at 15:15 comment added thibault jeannin @AnthonyQuas Thanks you again for your interest! The question comes from another question: Does the assertion: for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, we have: $$ \nu\left( \left\{ y \in \mathbb{R} : \pi_{Y=y}(A) = 1 \right\} \right) > 0 $$ imply the assertion that for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, $$ \nu\left( \left\{ y \in D : \pi_{Y=y}(A) = 1 \right\} \right) > 0? $$
Oct 27 at 14:10 comment added Anthony Quas This is quite an interesting condition. Can you say anything about where the question came from?
Oct 24 at 9:03 vote accept thibault jeannin
Oct 23 at 21:12 history edited thibault jeannin CC BY-SA 4.0
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Oct 23 at 20:12 vote accept thibault jeannin
Oct 23 at 20:50
Oct 23 at 19:22 answer added Anthony Quas timeline score: 1
Oct 23 at 5:25 history edited YCor CC BY-SA 4.0
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Oct 22 at 21:02 history asked thibault jeannin CC BY-SA 4.0