[proofplan] We reduce the bounded-below cost to a nonnegative lower semicontinuous function by subtracting a finite lower bound. The product of two Polish spaces is Polish, so the Portmanteau lower semicontinuity statement applies on $X \times Y$ to the shifted cost. Since all measures are probability measures, the shift contributes the same finite constant to every integral, and subtracting it preserves the desired liminf inequality, including the case where the right-hand side is $+\infty$. [/proofplan] [step:Shift the cost to a nonnegative lower semicontinuous function] Let $Z := X \times Y$, equipped with the [product topology](/page/Product%20Topology) and its Borel $\sigma$-algebra. Since $X$ and $Y$ are Polish spaces, $Z$ is a Polish space. By hypothesis, there exists $m \in \mathbb{R}$ such that $c(z) \geq m$ for every $z \in Z$, where $z=(x,y)$ denotes a point of $Z$. Define the shifted cost \begin{align*} \tilde c: Z &\to [0,\infty] \end{align*} \begin{align*} z &\mapsto c(z)-m. \end{align*} Because $c$ is lower semicontinuous and subtraction of the finite constant $m$ preserves lower semicontinuity, $\tilde c$ is lower semicontinuous. Since $c(z) \geq m$ for every $z \in Z$, the function $\tilde c$ is nonnegative. [/step] [step:Apply Portmanteau to the shifted cost] The sequence $(\pi_k)_{k \in \mathbb{N}}$ converges narrowly to $\pi$ in $\mathcal{P}(Z)$ by hypothesis. The space $Z$ is Polish and $\tilde c:Z\to[0,\infty]$ is lower semicontinuous. Therefore the lower semicontinuity part of the [Portmanteau theorem](/theorems/1171) applies to $\tilde c$ and gives \begin{align*} \int_Z \tilde c(z)\,d\pi(z) \leq \liminf_{k \to \infty} \int_Z \tilde c(z)\,d\pi_k(z). \end{align*} Here we are invoking the standard Portmanteau theorem for nonnegative lower semicontinuous test functions under narrow convergence (citing a result not yet in the wiki: Portmanteau theorem). [guided] The purpose of the shift is to put the cost into exactly the form required by the lower semicontinuity part of Portmanteau. We have already defined $Z := X \times Y$, and because products of Polish spaces are Polish, $Z$ is again a Polish space. The measures $\pi_k$ and $\pi$ are Borel probability measures on $Z$, and the assumption $\pi_k \to \pi$ narrowly means convergence against every bounded [continuous function](/page/Continuous%20Function) on $Z$. The Portmanteau lower semicontinuity statement says that if $\mu_k \to \mu$ narrowly on a Polish space and if \begin{align*} f: Z \to [0,\infty] \end{align*} is lower semicontinuous, then \begin{align*} \int_Z f(z)\,d\mu(z) \leq \liminf_{k \to \infty} \int_Z f(z)\,d\mu_k(z). \end{align*} We apply this with $\mu_k := \pi_k$, $\mu := \pi$, and $f := \tilde c$. The hypotheses are all verified: $Z$ is Polish, $\pi_k \to \pi$ narrowly, and $\tilde c$ is nonnegative and lower semicontinuous. Hence \begin{align*} \int_Z \tilde c(z)\,d\pi(z) \leq \liminf_{k \to \infty} \int_Z \tilde c(z)\,d\pi_k(z). \end{align*} This is the desired lower semicontinuity inequality, but for the shifted cost $\tilde c$ rather than for the original cost $c$. [/guided] [/step] [step:Remove the shift using that the measures are probabilities] For every $k \in \mathbb{N}$, since $\pi_k(Z)=1$ and $m$ is finite, the extended-real integral satisfies \begin{align*} \int_Z \tilde c(z)\,d\pi_k(z)=\int_Z c(z)\,d\pi_k(z)-m. \end{align*} Likewise, since $\pi(Z)=1$, \begin{align*} \int_Z \tilde c(z)\,d\pi(z)=\int_Z c(z)\,d\pi(z)-m. \end{align*} Substituting these identities into the Portmanteau inequality gives \begin{align*} \int_Z c(z)\,d\pi(z)-m \leq \liminf_{k \to \infty}\left(\int_Z c(z)\,d\pi_k(z)-m\right). \end{align*} Because $m \in \mathbb{R}$ is finite, the liminf commutes with subtraction of $m$: \begin{align*} \liminf_{k \to \infty}\left(\int_Z c(z)\,d\pi_k(z)-m\right)=\left(\liminf_{k \to \infty}\int_Z c(z)\,d\pi_k(z)\right)-m. \end{align*} Adding $m$ to both sides yields \begin{align*} \int_Z c(z)\,d\pi(z) \leq \liminf_{k \to \infty}\int_Z c(z)\,d\pi_k(z). \end{align*} Finally, rewriting $z=(x,y)$ and $Z=X\times Y$ gives \begin{align*} \int_{X \times Y} c(x,y)\,d\pi(x,y) \leq \liminf_{k \to \infty}\int_{X \times Y} c(x,y)\,d\pi_k(x,y). \end{align*} This is the claimed lower semicontinuity of the transport cost. [/step]