[proofplan] We use the direct method. First choose a minimizing sequence of transport plans. The fixed marginal constraints imply tightness of this sequence on the product space, so [Prokhorov's theorem](/theorems/1172) gives a weakly convergent subsequence. We then show that the marginal constraints are closed under [weak convergence](/page/Weak%20Convergence) and use the Portmanteau lower semicontinuity theorem for the bounded-below lower semicontinuous cost to pass the minimizing property to the limit. [/proofplan]

[step:Choose a minimizing sequence with finite costs] Let $V \in \mathbb{R}$ denote the Kantorovich value: \begin{align*} V := \inf_{\gamma \in \Pi(\mu,\nu)} \int_{X \times Y} c(x,y)\, d\gamma(x,y). \end{align*} By hypothesis, $V < \infty$. Since $c$ is bounded below, there exists $a \in \mathbb{R}$ such that $c(x,y) \geq a$ for every $(x,y) \in X \times Y$, so the integral functional cannot take the value $-\infty$. Thus $V \in \mathbb{R}$. For each $k \in \mathbb{N}$, choose $\pi_k \in \Pi(\mu,\nu)$ such that \begin{align*} \int_{X \times Y} c(x,y)\, d\pi_k(x,y) \leq V + \frac{1}{k}. \end{align*} This is possible by the definition of the infimum. The sequence $(\pi_k)_{k \in \mathbb{N}}$ is a minimizing sequence in $\Pi(\mu,\nu)$. [/step]

[step:Use the fixed marginals to prove tightness on $X \times Y$]We prove that $(\pi_k)_{k \in \mathbb{N}}$ is tight in $\mathcal{P}(X \times Y)$. Let $\varepsilon > 0$. Since $X$ and $Y$ are Polish spaces, the probability measures $\mu$ and $\nu$ are tight. Hence there exist compact sets $K_X \subset X$ and $K_Y \subset Y$ such that \begin{align*} \mu(X \setminus K_X) < \frac{\varepsilon}{2} \end{align*} and \begin{align*} \nu(Y \setminus K_Y) < \frac{\varepsilon}{2}. \end{align*} The set $K_X \times K_Y$ is compact in $X \times Y$. Since each $\pi_k$ has marginals $\mu$ and $\nu$, we have \begin{align*} \pi_k((X \times Y) \setminus (K_X \times K_Y)) \leq \pi_k((X \setminus K_X) \times Y) + \pi_k(X \times (Y \setminus K_Y)). \end{align*} Using the marginal identities gives \begin{align*} \pi_k((X \setminus K_X) \times Y) = \mu(X \setminus K_X) \end{align*} and \begin{align*} \pi_k(X \times (Y \setminus K_Y)) = \nu(Y \setminus K_Y). \end{align*} Therefore \begin{align*} \pi_k((X \times Y) \setminus (K_X \times K_Y)) < \varepsilon \end{align*} for every $k \in \mathbb{N}$. This proves tightness.[/step]

[guided]The point of this step is that although the plans $\pi_k$ may vary, their two marginals do not vary. Tightness of the marginals therefore forces tightness of the whole sequence of joint measures. Let $\varepsilon > 0$. Because $X$ and $Y$ are Polish spaces, every Borel probability measure on each space is tight. Applying this to $\mu \in \mathcal{P}(X)$ and $\nu \in \mathcal{P}(Y)$, choose compact sets $K_X \subset X$ and $K_Y \subset Y$ satisfying \begin{align*} \mu(X \setminus K_X) < \frac{\varepsilon}{2} \end{align*} and \begin{align*} \nu(Y \setminus K_Y) < \frac{\varepsilon}{2}. \end{align*} The product $K_X \times K_Y$ is compact in $X \times Y$. We now estimate how much mass $\pi_k$ can put outside this product compact set. The inclusion \begin{align*} (X \times Y) \setminus (K_X \times K_Y) \subset ((X \setminus K_X) \times Y) \cup (X \times (Y \setminus K_Y)) \end{align*} gives, by subadditivity of the probability measure $\pi_k$, \begin{align*} \pi_k((X \times Y) \setminus (K_X \times K_Y)) \leq \pi_k((X \setminus K_X) \times Y) + \pi_k(X \times (Y \setminus K_Y)). \end{align*} Now we use the defining property $\pi_k \in \Pi(\mu,\nu)$. The first marginal of $\pi_k$ is $\mu$, so \begin{align*} \pi_k((X \setminus K_X) \times Y) = \mu(X \setminus K_X). \end{align*} The second marginal of $\pi_k$ is $\nu$, so \begin{align*} \pi_k(X \times (Y \setminus K_Y)) = \nu(Y \setminus K_Y). \end{align*} Combining the three estimates yields \begin{align*} \pi_k((X \times Y) \setminus (K_X \times K_Y)) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} The compact set $K_X \times K_Y$ works uniformly for all $k$, which is exactly tightness of the sequence $(\pi_k)_{k \in \mathbb{N}}$ in $\mathcal{P}(X \times Y)$.[/guided]

[step:Extract a weakly convergent subsequence by Prokhorov compactness] The product $X \times Y$ is Polish. Since $(\pi_k)_{k \in \mathbb{N}}$ is tight in $\mathcal{P}(X \times Y)$, Prokhorov's theorem applies and gives a subsequence $(\pi_{k_j})_{j \in \mathbb{N}}$ and a probability measure $\pi \in \mathcal{P}(X \times Y)$ such that \begin{align*} \pi_{k_j} \rightharpoonup \pi \end{align*} weakly in $\mathcal{P}(X \times Y)$ as $j \to \infty$. (citing a result not yet in the wiki: Prokhorov's theorem) [/step]

[step:Pass the marginal constraints to the weak limit] Let \begin{align*} \operatorname{pr}_X: X \times Y &\to X \end{align*} denote the first coordinate projection, and let \begin{align*} \operatorname{pr}_Y: X \times Y &\to Y \end{align*} denote the second coordinate projection. Both projections are continuous. Let $\varphi: X \to \mathbb{R}$ be a bounded [continuous function](/page/Continuous%20Function). Then $\varphi \circ \operatorname{pr}_X: X \times Y \to \mathbb{R}$ is bounded and continuous. By weak convergence, \begin{align*} \int_{X \times Y} \varphi(\operatorname{pr}_X(x,y))\, d\pi(x,y) = \lim_{j \to \infty} \int_{X \times Y} \varphi(\operatorname{pr}_X(x,y))\, d\pi_{k_j}(x,y). \end{align*} Since $(\operatorname{pr}_X)_{\#}\pi_{k_j} = \mu$, the right-hand side equals \begin{align*} \lim_{j \to \infty} \int_X \varphi(x)\, d\mu(x) = \int_X \varphi(x)\, d\mu(x). \end{align*} Therefore $(\operatorname{pr}_X)_{\#}\pi = \mu$. The same argument with an arbitrary bounded continuous function $\psi: Y \to \mathbb{R}$ gives $(\operatorname{pr}_Y)_{\#}\pi = \nu$. Hence $\pi \in \Pi(\mu,\nu)$. [/step]

[step:Use lower semicontinuity of the cost to attain the infimum] Define the shifted cost \begin{align*} \tilde c: X \times Y &\to [0,\infty] \end{align*} \begin{align*} (x,y) &\mapsto c(x,y) - a. \end{align*} Since $c$ is lower semicontinuous and $a \in \mathbb{R}$ is constant, $\tilde c$ is lower semicontinuous and nonnegative. By the Portmanteau lower semicontinuity theorem applied to the weak convergence $\pi_{k_j} \rightharpoonup \pi$, \begin{align*} \int_{X \times Y} \tilde c(x,y)\, d\pi(x,y) \leq \liminf_{j \to \infty} \int_{X \times Y} \tilde c(x,y)\, d\pi_{k_j}(x,y). \end{align*} Because $\pi$ and each $\pi_{k_j}$ are probability measures, subtracting the constant shift gives \begin{align*} \int_{X \times Y} c(x,y)\, d\pi(x,y) \leq \liminf_{j \to \infty} \int_{X \times Y} c(x,y)\, d\pi_{k_j}(x,y). \end{align*} Using the minimizing property of $(\pi_k)_{k \in \mathbb{N}}$, \begin{align*} \liminf_{j \to \infty} \int_{X \times Y} c(x,y)\, d\pi_{k_j}(x,y) \leq \lim_{j \to \infty} \left(V + \frac{1}{k_j}\right) = V. \end{align*} Thus \begin{align*} \int_{X \times Y} c(x,y)\, d\pi(x,y) \leq V. \end{align*} Since $\pi \in \Pi(\mu,\nu)$ and $V$ is the infimum over $\Pi(\mu,\nu)$, we also have \begin{align*} V \leq \int_{X \times Y} c(x,y)\, d\pi(x,y). \end{align*} Therefore \begin{align*} \int_{X \times Y} c(x,y)\, d\pi(x,y) = V, \end{align*} so $\pi$ attains the Kantorovich infimum. (citing a result not yet in the wiki: Portmanteau lower semicontinuity theorem) [/step]