[proofplan] The proof reduces the compactness criterion to two strictly more primitive compactness facts for probability measures on Polish spaces. First, the Prokhorov selection theorem says that every tight sequence of Borel probability measures on a Polish space has a weakly convergent subsequence. Second, every weakly compact subset of $\mathcal{P}(Z)$ is uniformly tight. Since the [weak topology](/page/Weak%20Topology) on $\mathcal{P}(Z)$ is metrizable when $Z$ is Polish, [sequential compactness](/page/Sequential%20Compactness) and compactness agree, so these two facts give the two directions of the equivalence. [/proofplan] [step:Identify the weak topology and the tightness condition] Let $\mathbb{R}$ denote the [real numbers](/page/Real%20Numbers) and $\mathbb{N}$ the natural numbers. Let $\mathcal{B}(Z)$ denote the Borel $\sigma$-algebra of the Polish space $Z$. The space $\mathcal{P}(Z)$ consists of all probability measures $\mu: \mathcal{B}(Z) \to [0,1]$. The topology of [weak convergence](/page/Weak%20Convergence) on $\mathcal{P}(Z)$ is the topology for which a sequence $(\mu_n)_{n=1}^\infty$ in $\mathcal{P}(Z)$ converges weakly to $\mu \in \mathcal{P}(Z)$ precisely when \begin{align*} \lim_{n \to \infty} \int_Z f(z)\,d\mu_n(z) = \int_Z f(z)\,d\mu(z) \end{align*} for every bounded [continuous function](/page/Continuous%20Function) $f: Z \to \mathbb{R}$. A family $\mathcal{C} \subset \mathcal{P}(Z)$ is tight precisely when for every $\varepsilon > 0$ there exists a compact set $K_\varepsilon \subset Z$ such that \begin{align*} \inf_{\mu \in \mathcal{C}} \mu(K_\varepsilon) \geq 1 - \varepsilon. \end{align*} These are exactly the weak topology and the uniform tightness condition appearing in [Prokhorov's theorem](/theorems/1172) for Polish spaces. [guided] We first make sure that the words in the statement match the hypotheses of the [compactness theorem](/theorems/2748) being used. The measurable structure on $Z$ is its Borel $\sigma$-algebra $\mathcal{B}(Z)$, so every measure under discussion is a map \begin{align*} \mu: \mathcal{B}(Z) \to [0,1] \end{align*} with $\mu(Z)=1$ and countable additivity. Weak convergence on $\mathcal{P}(Z)$ is convergence against bounded continuous test functions. Thus a sequence $(\mu_n)_{n=1}^\infty$ converges weakly to $\mu$ when, for every bounded continuous map $f: Z \to \mathbb{R}$, one has \begin{align*} \lim_{n \to \infty} \int_Z f(z)\,d\mu_n(z) = \int_Z f(z)\,d\mu(z). \end{align*} The tightness condition is uniform over the whole family $\mathcal{C}$: once $\varepsilon > 0$ is fixed, a single compact set $K_\varepsilon \subset Z$ must work for every measure $\mu \in \mathcal{C}$. This is stronger than saying each individual measure is tight. Written quantitatively, it says \begin{align*} \inf_{\mu \in \mathcal{C}} \mu(K_\varepsilon) \geq 1 - \varepsilon. \end{align*} These are precisely the objects in the Polish-space version of Prokhorov's theorem. [/guided] [/step] [step:Derive relative compactness from tightness using the selection theorem] Assume that $\mathcal{C}$ is tight. We use the following external prerequisite, the Prokhorov selection theorem: if $Z$ is a Polish space and $(\mu_n)_{n=1}^{\infty}$ is a tight sequence in $\mathcal{P}(Z)$, then there exist a subsequence $(\mu_{n_k})_{k=1}^{\infty}$ and a measure $\mu \in \mathcal{P}(Z)$ such that $\mu_{n_k}$ converges weakly to $\mu$. The hypotheses of the selection theorem are satisfied. The space $Z$ is Polish by assumption. If $(\mu_n)_{n=1}^{\infty}$ is any sequence in $\mathcal{C}$, then the tightness of $\mathcal{C}$ gives, for every $\varepsilon > 0$, a compact set $K_\varepsilon \subset Z$ such that \begin{align*} \mu_n(K_\varepsilon) \geq 1 - \varepsilon \end{align*} for every $n \in \mathbb{N}$. Hence the sequence $(\mu_n)_{n=1}^{\infty}$ is tight. By the selection theorem, it has a weakly convergent subsequence in $\mathcal{P}(Z)$. Since $Z$ is Polish, the topology of weak convergence on $\mathcal{P}(Z)$ is metrizable. In a [metrizable space](/page/Metrizable%20Space), a subset is relatively compact if and only if every sequence in the subset has a convergent subsequence with limit in the ambient space. Therefore $\mathcal{C}$ is relatively compact in $\mathcal{P}(Z)$ for weak convergence. [guided] Assume that $\mathcal{C}$ is tight. To prove relative compactness in a metrizable space, it is enough to prove the sequential form: every sequence in $\mathcal{C}$ has a subsequence that converges in the ambient space. Here the ambient space is $\mathcal{P}(Z)$ with the weak topology. Let $(\mu_n)_{n=1}^{\infty}$ be an arbitrary sequence in $\mathcal{C}$. Because tightness is uniform over the family $\mathcal{C}$, for every $\varepsilon > 0$ there is one compact set $K_\varepsilon \subset Z$ such that \begin{align*} \mu(K_\varepsilon) \geq 1 - \varepsilon \end{align*} for every $\mu \in \mathcal{C}$. Applying this to the particular measures $\mu_n \in \mathcal{C}$ gives \begin{align*} \mu_n(K_\varepsilon) \geq 1 - \varepsilon \end{align*} for every $n \in \mathbb{N}$. Thus the sequence $(\mu_n)_{n=1}^{\infty}$ is tight. Now apply the Prokhorov selection theorem. Its hypotheses are exactly that $Z$ is Polish and that the sequence of Borel probability measures is tight. The first hypothesis is part of the theorem statement, and the second was just verified. Therefore there exist a subsequence $(\mu_{n_k})_{k=1}^{\infty}$ and a Borel probability measure $\mu \in \mathcal{P}(Z)$ such that $\mu_{n_k}$ converges weakly to $\mu$. Finally, because $Z$ is Polish, weak convergence on $\mathcal{P}(Z)$ is metrizable. In metric spaces, relative compactness is equivalent to sequential relative compactness. Since every sequence in $\mathcal{C}$ has a weakly convergent subsequence with limit in $\mathcal{P}(Z)$, the family $\mathcal{C}$ is relatively compact for weak convergence. [/guided] [/step] [step:Derive tightness from relative compactness using uniform tightness of weakly compact sets] Assume that $\mathcal{C}$ is relatively compact in $\mathcal{P}(Z)$ for weak convergence, and let $\overline{\mathcal{C}}$ denote its closure in that topology. By relative compactness, $\overline{\mathcal{C}}$ is weakly compact. We use the following external prerequisite: every weakly compact subset of $\mathcal{P}(Z)$ is uniformly tight when $Z$ is Polish. The hypotheses are satisfied because $Z$ is Polish and $\overline{\mathcal{C}} \subset \mathcal{P}(Z)$ is weakly compact. Hence, for every $\varepsilon > 0$, there exists a compact set $K_\varepsilon \subset Z$ such that \begin{align*} \nu(K_\varepsilon) \geq 1 - \varepsilon \end{align*} for every $\nu \in \overline{\mathcal{C}}$. Since $\mathcal{C} \subset \overline{\mathcal{C}}$, the same compact set satisfies \begin{align*} \mu(K_\varepsilon) \geq 1 - \varepsilon \end{align*} for every $\mu \in \mathcal{C}$. Therefore $\mathcal{C}$ is tight. [/step] [step:Combine the two directions] The preceding steps prove both implications. If $\mathcal{C}$ is tight, then the Prokhorov selection theorem and metrizability of weak convergence on $\mathcal{P}(Z)$ imply that $\mathcal{C}$ is relatively compact. If $\mathcal{C}$ is relatively compact, then its weak closure is compact, and uniform tightness of weakly compact subsets of $\mathcal{P}(Z)$ implies that $\mathcal{C}$ is tight. Hence, for every family $\mathcal{C} \subset \mathcal{P}(Z)$ on a Polish space $Z$, relative compactness in the weak topology is equivalent to tightness. [/step]