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Nov 21, 2022 at 16:13 comment added Nate River Let us continue this discussion in chat.
Nov 21, 2022 at 15:22 comment added Akira @NateRiver I have sent an email to IEHS where Villani is currently working. They have forwarded my email to Villani. If we are lucky, we will get a reply from him...
Nov 21, 2022 at 15:18 comment added Nate River Ha, I think that is beyond my skill level as well.. if it’s not available anywhere online you may well have to email people in the field.
Nov 21, 2022 at 14:46 comment added Akira @NateRiver Could you have a look at my closely related question here?
Nov 21, 2022 at 11:15 comment added Akira @NateRiver In the linked version, $f \in L_\infty (\mu)$ and thus $h:y \mapsto \int_X f\mathrm d\mu_y$ belongs to $L_\infty (\nu)$. On the other hand, $g \in L_1 (\nu)$. So $hg \in L_1 (\nu)$ by Hölder's inequality. So it seems we can generalize to $f \in L_1 (\mu)$ with a trade-off that $g \in L_\infty (\nu)$.
Nov 21, 2022 at 11:08 comment added Nate River Indeed, I believe so.
Nov 21, 2022 at 11:07 comment added Akira @NateRiver you meant that the function $f$ in claims 1. and 2. of this version can be generalized from being bounded measurable to being just $\mu$-integrable, right?
Nov 21, 2022 at 1:40 comment added Akira @NateRiver Thank you so much for your elaboration! I got it. Greinecker also came to the same conclusion as yours.
Nov 21, 2022 at 1:38 history edited Akira CC BY-SA 4.0
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Nov 21, 2022 at 1:35 comment added Nate River I had a look at this and I wasn’t able to resolve it either.. there are versions that hold for either nonnegative, bounded, or $L^1$ functions, but we cannot say that the integrand is any of these.
Nov 20, 2022 at 10:57 history asked Akira CC BY-SA 4.0