I've been following along in https://lchizat.github.io/files2020ot/lecture1.pdf to learn about optimal transport theory and ran into some confusion in Chapter 4 "The dual problem"...
Let $X,Y$ be polish spaces. For $\mu \in \mathcal{P}(X), \nu \in \mathcal{P}(Y)$ let $\Pi(\mu, \nu)$ be the set of probability measures defined on $X \times Y$ with marginals $\mu, \nu$, i.e.
$$ \Pi(\mu, \nu) = \left\{ \gamma \in \mathcal{P}(X \times Y) \mid \gamma(A \times Y) = \mu(A),\, \gamma(X \times B) = \nu(B) \right\} $$
Let $\mathcal{M}^+$ denote the set of non-negative Borel regular measures and let $\mathcal{C}_b(U)$ be the set of bounded continuous real valued functions defined on $U$.
Given $\gamma \in \mathcal{M}^+$, consider the following equation $$ \sup_{(\varphi, \psi) \in \mathcal{C}_b(X) \times \mathcal{C}_b(Y)} \int_X \varphi \,\mathrm{d} \mu + \int_Y \psi \,\mathrm{d} \nu - \int_{X \times Y} \left(\varphi(x) + \psi(y) \right) \,\mathrm{d} \gamma = \begin{cases} 0 & \gamma \in \Pi(\mu, \nu) \\ +\infty & \text{otherwise} \end{cases} $$
Is it necessary for $X,Y$ to be compact in order for the above equation to be correct?
I (naively) think the answer is no, my reasoning is as follows:
Suppose $\gamma \in \Pi(\mu, \nu)$. We can consider $\phi, \psi \ge 0$ since any non-negative pair with achieve a larger $\sup$ than a possibly-negative pair. By definition
$$ \int_{X \times Y} \varphi(x) \,\mathrm{d}\gamma = \sup \left\{ \int_{X \times Y} s \, \mathrm{d} \gamma \mid s = \sum_i \alpha_i \chi_{E_i},\, s \le \varphi,\, E_i \, \gamma-\text{measurable}, \alpha_i \in [0, \infty) \right\} $$
Also, by construction
$$ \int_{X \times Y} \sum_i \alpha_i \chi_{A_i \times Y} \,\mathrm{d} \gamma = \sum_i \alpha_i \gamma \left( A_i \times Y \right) = \sum_i \alpha_i \mu \left( A_i \right) = \int_X \sum_i \alpha_i \chi_{A_i} \,\mathrm{d} \mu \\ \implies \int_{X \times Y} \varphi(x) \,\mathrm{d} \gamma \ge \int_X \varphi \,\mathrm{d} \mu $$
However, we also have, for any measurable $E_i \subset X \times Y$, $E_{i_y} = \{ x \mid (x, y) \in E_i \}$
$$ \int_X \sum_i \alpha_i \chi_{E_{i_y}} \,\mathrm{d} \mu = \sum_i \alpha_i \mu \left( E_{i_y} \right) = \sum_i \alpha_i \gamma \left( E_{i_y} \times Y \right) \ge \sum_i \alpha_i \gamma \left( E_i \right) = \int_{X \times Y} \sum_i \alpha_i \chi_{E_i} \,\mathrm{d} \gamma \\ \implies \int_X \varphi \,\mathrm{d} \mu \ge \int_{X \times Y} \varphi(x) \,\mathrm{d} \gamma $$
Combining the above statements we get $$ \int_{X \times Y} \varphi(x) \,\mathrm{d} \gamma = \int_X \varphi \,\mathrm{d} \mu $$
By similar logic we find $$ \int_{X \times Y} \psi(y) \,\mathrm{d} \gamma = \int_Y \psi \,\mathrm{d} \nu $$
So that
$$ \int_{X \times Y} \varphi(x) + \psi(y) \,\mathrm{d} \gamma = \int_X \varphi \,\mathrm{d}\mu + \int_Y \psi \,\mathrm{d} \nu $$
And finally
$$ \int_X \varphi \,\mathrm{d} \mu + \int_Y \psi \,\mathrm{d} \nu - \int_{X \times Y} \left(\varphi(x) + \psi(y) \right) \,\mathrm{d} \gamma = 0 $$
Now suppose $\gamma \not\in \Pi(\mu, \nu)$. Without loss of generality suppose $\gamma(A \times Y) \neq \mu(A)$ for some measurable $A \subset X$. Again without loss of generality suppose $\mu(A) > \gamma(A \times Y)$ (otherwise work with $A_i^c$). Put $\psi = 0$ and $\varphi_n \in \mathcal{C}_b$ so that $\varphi_n \to \chi_E$ (e.g. by convolving $\chi_E$ with a sequence of mollifiers). Put $\varphi_n' = n \varphi_n$ then
$$ \int_X \varphi_n' \,\mathrm{d} \mu + \int_Y \psi \,\mathrm{d} \nu - \int_{X \times Y} \left(\varphi_n(x)' + \psi(y) \right) \,\mathrm{d} \gamma \to n \left( \mu(E) - \gamma(E \times Y) \right) $$
i.e. it's unbounded so that
$$ \sup_{(\varphi, \psi) \in \mathcal{C}_b(X) \times \mathcal{C}_b(Y)} \int_X \varphi \,\mathrm{d} \mu + \int_Y \psi \,\mathrm{d} \nu - \int_{X \times Y} \left(\varphi(x) + \psi(y) \right) \,\mathrm{d} \gamma = + \infty $$
Notably I never need $X,Y$ to be compact in the above proofs, but then again the proofs might be wrong-- let me know!
N.B. This question came up because the linked pdf states "we remind that $X,Y$ are compact spaces" before stating the above equation, but I couldn't figure out why!