Is this shorter attempt on the existence of the maximizer in Kantorovich duality is correct?

I'm reading section 3.4 Existence of Maximisers to the Dual Problem in this lecture notes. The proof is quite involved and requires a complicated approximation argument.

Below is my straightforward approach which is so naive that I'm not sure if it's correct. Could you have a check on my attempt?

Let $$X,Y$$ be Polish spaces and $$c:X \times Y \to \mathbb [0, +\infty)$$. We fix Borel probability measures (b.p.m.) $$\mu \in \mathcal P(X)$$ and $$\nu \in \mathcal P(Y)$$.

• Let $$\Phi_c$$ be the collection of $$(\varphi: X \to \mathbb R, \psi: Y \to \mathbb R) \in L_1(\mu) \times L_1(\nu)$$ such that $$\varphi(x)+\psi(y) \le c(x,y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y\in Y$$.

• Let $$J (\varphi, \psi) := \int_X \varphi d \mu + \int \psi d \nu \quad \forall (\varphi, \psi) \in \Phi_c.$$

• We assume a regularity condition that there are $$c_X: X \to \mathbb R$$ and $$c_Y: Y \to \mathbb R$$ such that $$c_X \in L_1(\mu), c_Y \in L_1 (\nu)$$, and $$c (x, y) \le c_X(x)+c_Y(y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y\in Y$$.

Then the maximization $$\sup_{(\varphi, \psi) \in \Phi_c} J (\varphi, \psi)$$ has a solution.

My attempt: We need the following essential lemma.

Lemma: For each $$(\varphi, \psi) \in \Phi_c$$, there is an improved pair $$(\overline \varphi, \overline\psi) \in \Phi_c$$ such that

• $$J (\varphi, \psi) = J (\overline \varphi, \overline\psi)$$.
• $$\varphi (x) \le c_X (x)$$ and $$\psi (y) \le c_Y (y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y\in Y$$.

Let $$(\varphi_n, \psi_n) \subset \Phi_c$$ be a maximizing sequence. Let $$(\overline \varphi_n, \overline\psi_n) \in \Phi_c$$ be the improved pair of $$(\varphi_n, \psi_n)$$ as in the Lemma, i.e.,

• $$J (\varphi_n, \psi_n) = J (\overline \varphi_n, \overline\psi_n)$$.
• $$\overline \varphi_n (x) \le c_X (x)$$ and $$\overline \psi_n (y) \le c_Y (y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y\in Y$$.

Let $$\varphi := \limsup_n \overline\varphi_n \quad \text{and} \quad \psi := \limsup_n \overline \psi.$$

Then $$\varphi$$ is $$\mu$$-measurable and $$\psi$$ is $$\nu$$-measurable. For every $$n \in \mathbb N$$, we have

• $$\overline \varphi_n (x) \le \varphi (x) \le c_X (x)$$ for all $$\mu$$-a.e. $$x\in X$$.
• $$\overline \psi (y) \le \psi (y) \le c_Y (y)$$ for all $$\nu$$-a.e. $$y \in Y$$.

This means $$\varphi, \psi$$ are bounded by $$\mu$$-integrable and $$\nu$$-integrable functions respectively, so $$\varphi \in L_1(\mu)$$ and $$\psi \in L_1 (\nu)$$. Clearly, $$J(\varphi_n, \psi_n) \le J(\varphi, \psi) \quad \forall n\in \mathbb N.$$

Hence $$\sup_{(\varphi, \psi) \in \Phi_c} J (\varphi, \psi) \le J(\varphi, \psi).$$

It remains to show that $$\varphi (x)+ \psi(y) \le c(x,y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y \in Y$$. For each $$n \in \mathbb N$$, $$\overline \varphi_n (x)+ \overline\psi_n(y) \le c(x,y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y \in Y$$. Notice that the countable union of null sets is again a null set. Then $$\limsup_n \overline \varphi_n (x)+ \limsup_n \overline\psi_n(y) \le c(x,y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y \in Y$$. This completes the proof.

I have realized that I made a stupid mistake in claiming

$$\limsup_n \overline \varphi_n (x)+ \limsup_n \overline\psi_n(y) \le \limsup_n [\overline \varphi_n (x)+ \overline\psi_n(y)].$$

The correct proof by the author is as follows.

Let $$(\varphi_n, \psi_n) \subset \Phi_c$$ be a maximizing sequence. WLOG, we assume $$J (\varphi_n, \psi_n) \ge 0 \quad \forall n \in \mathbb N.$$

Let $$(\overline \varphi_n, \overline\psi_n) \in \Phi_c$$ be the improved pair of $$(\varphi_n, \psi_n)$$ as in the Lemma, i.e.,

• $$J (\varphi_n, \psi_n) = J (\overline \varphi_n, \overline\psi_n)$$.
• $$\overline \varphi_n (x) \le c_X (x)$$ and $$\overline \psi_n (y) \le c_Y (y)$$ for $$\mu$$-a.e. $$x\in X$$ and $$\nu$$-a.e. $$y\in Y$$.

For $$(m, n)\in \mathbb N^2$$, let $$\alpha_{m,n} : =\min \{c_X-\overline \varphi_n, m\} \quad \text{and} \quad \beta_{m,n} : =\min \{c_Y-\overline \psi_n, m\} .$$

• Then $$0 \le \alpha_{m,n} \le m$$ and $$0 \le \beta_{m,n} \le m$$ for all $$n \in \mathbb N$$.
• This implies the sequences $$(\alpha_{m,n})_{n \in \mathbb N}$$ and $$(\beta_{m,n})_{n \in \mathbb N}$$ are bounded in $$L_2(\mu)$$ and $$L_2(\nu)$$ respectively.
• Notice that $$L_2(\mu)$$ and $$L_2(\nu)$$ are reflexive. Hence $$\overline{(\alpha_{m,n})_{n \in \mathbb N}}^{L_2(\mu)}$$ and $$\overline{(\beta_{m,n})_{n \in \mathbb N}}^{L_2(\nu)}$$ are weakly compact. By Tychonoff theorem, $$\Pi_{m \in \mathbb N} \overline{(\alpha_{m,n})_{n \in \mathbb N}}^{L_2(\mu)} \times \Pi_{m \in \mathbb N} \overline{(\beta_{m,n})_{n \in \mathbb N}}^{L_2(\nu)}$$ is compact in the product of weak topologies.
• Hence there are $$\alpha_1, \alpha_2, \ldots \in L_2(\mu)$$ and $$\beta_1, \beta_2, \ldots \in L_2(\nu)$$ and a subsequence $$\lambda \in \mathbb N^\mathbb N$$ such that for all $$m$$, we have $$\alpha_{m,\lambda(n)} \rightharpoonup \alpha_m$$ and $$\beta_{m,\lambda(n)} \rightharpoonup \beta_m$$ weakly as $$n \to \infty$$.
• WLOG, we assume $$\lambda$$ is the identity map, i.e., $$\lambda(n)=n$$ for all $$n \in \mathbb N$$.
• Also, $$0 \le \alpha_{m,n} \le \alpha_{m+1,n}$$ and $$0 \le \beta_{m,n} \le \beta_{m+1,n}$$ for all $$n \in \mathbb N$$. Weak convergence preserves pointwise inequalities, so $$0 \le \alpha_m \le \alpha_{m+1} \le m+1 \quad \text{and} \quad 0 \le \beta_m \le \beta_{m+1} \le m+1 \quad \forall m \in \mathbb N.$$

Let $$\alpha := \lim_m \alpha_m \quad \text{and} \quad \beta := \lim_m \beta_m.$$

Above limits are well-defined by the monotonicity of $$(\alpha_m)$$ and $$(\beta_m)$$. Clearly, $$\alpha$$ is $$\mu$$-measurable and $$\beta$$ is $$\nu$$-measurable. Also, $$0 \le \alpha$$ and $$0 \le \beta$$. Let's prove that $$\alpha \in L_1(\mu)$$ and $$\beta \in L_1 (\nu)$$. Clearly, $$0 \le \int \alpha d \mu \le +\infty \quad \text{and} \quad 0 \le \int \beta d \nu \le +\infty.$$

We have \begin{align*} & \int \alpha d \mu + \int \beta d \nu \\ =&\lim_m \int \alpha_m d \mu + \lim_m \int \beta_m d \nu \quad \text{by monotone convergence} \\ =&\lim_m \left (\int \alpha_m d \mu + \int \beta_m d \nu \right ) \end{align*}

Notice that $$\alpha_{m,n} \rightharpoonup \alpha_m$$ in the weak topology of $$L_2(\mu)$$ and that the map $$L_2(\mu) \to \mathbb R, f \mapsto \int f d \mu$$ is linear continuous in the norm topology $$\| \cdot \|_{L_2(\mu)}$$. So $$\int \alpha_m d\mu = \lim_n \int \alpha_{m,n} d\mu.$$

Similarly, $$\int \beta_m d\nu = \lim_n \int \beta_{m,n} d\nu.$$

Hence \begin{align*} & \int \alpha d \mu + \int \beta d \nu \\ =& \lim_m \left ( \lim_n \int \alpha_{m, n} d \mu + \lim_n \int \beta_{m, n} d \nu \right ) \\ =& \lim_m \lim_n \left ( \int \alpha_{m, n} d \mu + \int \beta_{m, n} d \nu \right ) \\ \le& \lim_m \lim_n \left ( \int (c_X-\overline \varphi_n) d \mu + \int (c_Y-\overline \psi_n) d \nu \right ) \\ =& \lim_n \left ( \int (c_X-\overline \varphi_n) d \mu + \int (c_Y-\overline \psi_n) d \nu \right ) \end{align*}

Notice that the integrals $$\int c_X d \mu, \int \overline \varphi_n d \mu, \int c_Y d \nu, \int \overline \psi_n d \nu$$ are finite, so \begin{align*} & \int \alpha d \mu + \int \beta d \nu \\ \le& \lim_n \left ( \int c_X d \mu - \int \overline \varphi_n d \mu + \int c_Y d \nu - \int \overline \psi_n d \nu \right ) \\ =& \lim_n (J(c_X, c_Y)- J(\overline \varphi_n, \overline \psi_n)) \\ =& J(c_X, c_Y) - \lim_n J(\overline \varphi_n, \overline \psi_n) \\ \le& J(c_X, c_Y) < +\infty. \end{align*}

It follows that $$\int \alpha d \mu < +\infty, \quad \int \beta d \nu < +\infty, \quad \lim_n J(\overline \varphi_n, \overline \psi_n) < +\infty.$$

So $$\alpha \in L_1 (\mu)$$ and $$\beta \in L_1 (\nu)$$. Let $$\varphi := c_X - \alpha \quad \text{and} \quad \psi := c_Y-\beta.$$

Clearly, $$\varphi \in L_1 (\mu)$$ and $$\psi \in L_1 (\nu)$$. It follows from above inequality that $$J(\varphi, \psi) = J(c_X, c_Y) - J(\alpha, \beta) \ge \lim_n J(\overline \varphi_n, \overline \psi_n).$$

We have \begin{align*} & \varphi (x)+ \psi(y) \\ =& c_X(x)-\alpha(x) +c_Y(y)-\beta(y) \\ =& c_X(x) - \lim_m \alpha_m (x) + c_Y(y) - \lim_m \beta_m (y) \\ =& c_X(x) - \lim_m \lim_n \alpha_{m, n} (x) + c_Y(y) - \lim_m \lim_n \beta_{m,n}(y) \\ =& \lim_m \lim_n (c_{X} (x)-\alpha_{m, n} (x)) + \lim_m \lim_n (c_Y(y) - \beta_{m,n}(y)) \\ =& \lim_m \lim_n \left [ \max \{ \overline\varphi_n(x), c_X(x)-m\} + \max \{ \overline\psi_n(y), c_Y(y)-m\} \right ]. \end{align*}

We have \begin{align*} & \max \{ \overline\varphi_n(x), c_X(x)-m\} + \max \{ \overline\psi_n(y), c_Y (y)-m\} \\ =& c_X(x) + c_Y(y)+\max \{ \overline\varphi_n(x) -c_X(x), -m\} + \max \{ \overline\psi_n(y) - c_Y (y), -m\} \\ \le & c_X(x) + c_Y(y)+\max \{ \overline\varphi_n(x) -c_X(x)+\overline\psi_n(y) - c_Y (y), -m\} \quad (\star)\\ \le & c_X(x) + c_Y(y)+\max \{ c(x,y)-c_X(x) - c_Y (y), -m\} \\ \end{align*}

Here $$(\star)$$ follows from this result. It follows that \begin{align*} & \varphi (x)+ \psi(y) \\ \le & \lim_m \left [c_X(x) + c_Y(y)+\max \{ c(x,y) -c_X(x)- c_Y (y), -m\} \right ] \\ = & c_X(x) + c_Y(y) + [c(x,y)-c_X(x) - c_Y (y)] \\ = & c(x,y). \end{align*}