[proofplan] We reduce the bounded-below function $f$ to a nonnegative lower semicontinuous function by adding a finite constant. The [Portmanteau theorem](/theorems/1171) in its lower-semicontinuous test-function form then gives the desired liminf inequality for the shifted function. Since every $\alpha_n$ and $\alpha$ is a probability measure, the added constant contributes the same finite amount to every integral and can be subtracted in the extended-real order. [/proofplan]

[step:Shift the cost to a nonnegative lower semicontinuous function]Choose $m \in \mathbb{R}$ such that $f(z) \geq m$ for every $z \in Z$, and define the finite constant $c := \max\{-m,0\}$. Define \begin{align*} g: Z \to [0,\infty], \qquad g(z) := f(z)+c. \end{align*} Then $g(z) \geq 0$ for every $z \in Z$. Since addition by a finite constant preserves lower semicontinuity, $g$ is lower semicontinuous.[/step]

[guided]The point of the shift is to place the cost inside the nonnegative test-function version of the Portmanteau theorem. Because $f$ is bounded from below, there is a finite number $m \in \mathbb{R}$ with $f(z) \geq m$ for all $z \in Z$. We choose \begin{align*} c := \max\{-m,0\}. \end{align*} This choice guarantees $c \geq 0$ and $m+c \geq 0$. Now define \begin{align*} g: Z \to [0,\infty], \qquad g(z) := f(z)+c. \end{align*} For every $z \in Z$ we have \begin{align*} g(z) = f(z)+c \geq m+c \geq 0. \end{align*} Thus $g$ is nonnegative. Moreover, if $a \in \mathbb{R}$, then \begin{align*} \{z \in Z : g(z) > a\} = \{z \in Z : f(z) > a-c\}. \end{align*} Since $f$ is lower semicontinuous, the set on the right is open in $Z$, so $g$ is lower semicontinuous. This verifies exactly the admissibility condition needed for the lower-semicontinuous Portmanteau inequality.[/guided]

[step:Apply the lower semicontinuous form of Portmanteau to the shifted cost] By the Portmanteau theorem in its nonnegative lower-semicontinuous test-function form (citing a result not yet in the wiki: Portmanteau Theorem), [weak convergence](/page/Weak%20Convergence) $\alpha_n \rightharpoonup \alpha$ implies \begin{align*} \int_Z g(z)\,d\alpha(z) \leq \liminf_{n \to \infty} \int_Z g(z)\,d\alpha_n(z). \end{align*} The hypotheses are satisfied because $Z$ is Polish, $\alpha_n$ and $\alpha$ are Borel probability measures on $Z$, $\alpha_n \rightharpoonup \alpha$ by assumption, and $g$ is nonnegative and lower semicontinuous by the previous step. [/step]

[step:Subtract the probability mass contribution of the shift] Since $\alpha$ and every $\alpha_n$ are probability measures, $\alpha(Z)=1$ and $\alpha_n(Z)=1$ for every $n \in \mathbb{N}$. Therefore \begin{align*} \int_Z g(z)\,d\alpha(z) = \int_Z f(z)\,d\alpha(z) + c \end{align*} and, for every $n \in \mathbb{N}$, \begin{align*} \int_Z g(z)\,d\alpha_n(z) = \int_Z f(z)\,d\alpha_n(z) + c. \end{align*} Substituting these identities into the Portmanteau inequality gives \begin{align*} \int_Z f(z)\,d\alpha(z) + c \leq \liminf_{n \to \infty} \left(\int_Z f(z)\,d\alpha_n(z) + c\right). \end{align*} Because $c \in \mathbb{R}$ is finite, translation commutes with the extended-real liminf: \begin{align*} \liminf_{n \to \infty} \left(\int_Z f(z)\,d\alpha_n(z) + c\right) = \left(\liminf_{n \to \infty} \int_Z f(z)\,d\alpha_n(z)\right)+c. \end{align*} Subtracting the same finite constant $c$ from both sides yields \begin{align*} \int_Z f(z)\,d\alpha(z) \leq \liminf_{n \to \infty} \int_Z f(z)\,d\alpha_n(z). \end{align*} This is the desired lower bound. [/step]