@JHM If $c$ is continuous then $\varphi^c$ is upper semi-continuous and thus measurable. From the condition $c(x,y) \le c_X (x) + c_Y(y) \quad \forall (x, y) \in X \times Y$, we can prove that $\varphi^c$ is also $\nu$-integrable. Of course, $c$ being continuous is sufficient for most applications. In this thread, I'm curious about the machinery that allows the author to obtain the result in case $c$ is just lower semi-continuous.

@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...

@MichaelGreinecker It's a typo. I meant $\int f \mathrm d \mu_y$, not $\int F \mathrm d \mu_y$.

@NateRiver Could you have a look at my closely related question here?

@JHM It seems from your comment that Exercise 2.36 has no solution... I would like to send an email to Villani but I could not find his email address.

@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 $g \in L_\infty (\nu)$.

@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?

@NateRiver Thank you so much for your elaboration! I got it. Greinecker also came to the same conclusion as yours.

Thank you so much for your response. That's exactly what I'm asking for. Now I understand why the canonical isometric isomorphism is beautiful...

Thank you so much for your help. Good night!