Tightness prevents mass from escaping to infinity. Prokhorov's theorem identifies tightness as the compactness criterion for probability measures on Polish spaces.
Tight families
Tightness says most mass lies in one compact set, uniformly over the family.
Definition. A family \(\mathcal P\) of probability measures on a metric space \(S\) is tight if for every \(\varepsilon>0\) there is compact \(K\subset S\) such that \(P(K)\ge1-\varepsilon\) for all \(P\in\mathcal P\).
Proposition. On \(\mathbb R^d\), a family is tight if and only if \(\lim_{M\to\infty}\sup_{P\in\mathcal P}P(\|x\|>M)=0\).
Proof. Closed balls are compact in \(\mathbb R^d\), so the tail condition gives tightness. Conversely, if \(K\) is compact, it is contained in some closed ball \(\{\|x\|\le M\}\), so mass outside that ball is at most mass outside \(K\).
Relative compactness
Tightness is designed to produce weakly convergent subsequences.
Theorem. Every tight sequence of probability measures on \(\mathbb R^d\) has a weakly convergent subsequence.
Proof. Choose a countable dense subset \(\{f_j\}\) of the unit ball of bounded Lipschitz functions under uniform convergence on compact sets. Since \(|\int f_j\,\mathrm{d}P_n|\le1\), diagonal extraction gives a subsequence along which every one of these integrals converges. Tightness makes this subsequence Cauchy against every bounded Lipschitz \(f\): choose a compact \(K\) carrying mass at least \(1-\varepsilon\), approximate \(f|_K\) uniformly by some \(f_j|_K\), and bound both tails by \(2\varepsilon\).
The limiting map \(L(f)=\lim_k\int f\,\mathrm{d}P_{n_k}\) is positive, linear, and satisfies \(L(1)=1\). Its restriction to \(C_c(\mathbb R^d)\) is represented by a Radon measure \(P\) through the Riesz-Markov theorem. Tightness gives \[ P(\mathbb R^d)=1: \] for compact \(K\) carrying uniformly at least \(1-\varepsilon\), continuous compactly supported majorants of \(\mathbf1_K\) show \(P(\mathbb R^d)\ge1-\varepsilon\). The bounded-Lipschitz determination theorem then gives \(P_{n_k}\Rightarrow P\).
Prokhorov theorem
Polish spaces admit the same compactness criterion.
Theorem. On a Polish space, a family of probability measures is relatively compact for weak convergence if and only if it is tight.
Proof outline. A Polish space has a countable convergence-determining family of bounded uniformly continuous functions. Tightness reduces each such function, uniformly over the family of measures, to its values on compact sets; diagonal extraction therefore constructs a positive limiting functional. The Daniell-Stone representation theorem turns it into a probability measure, with tightness preventing loss of total mass.
For the converse, let \(\mathcal C=\overline{\mathcal P}\) be weakly compact and fix \(\varepsilon>0\). Choose \(\varepsilon_m>0\) with \(\sum_m\varepsilon_m<\varepsilon\) and radii \(r_m\downarrow0\). Every \(P\in\mathcal C\) is tight, so choose compact \(K_{P,m}\) with \(P(K_{P,m})>1-\varepsilon_m\). Portmanteau makes \[ \{Q:Q(K_{P,m}^{\,r_m})>1-\varepsilon_m\} \] a weak neighborhood of \(P\). Finitely many of these neighborhoods cover compact \(\mathcal C\). Let \(C_m\) be the finite union of the corresponding closed \(r_m\)-neighborhoods. Then \(Q(C_m)>1-\varepsilon_m\) for every \(Q\in\mathcal C\).
Set \(L_m=\bigcap_{j\le m}C_j\) and \(K=\bigcap_mL_m\). Each \(C_m\) lies within distance \(r_m\) of a finite union of compact sets. Since \(r_m\to0\), \(K\) is totally bounded; it is closed in the complete space, hence compact. Moreover, \[ Q(L_m)\ge1-\sum_{j\le m}\varepsilon_j, \] so continuity from above gives \(Q(K)\ge1-\varepsilon\) uniformly over \(Q\in\mathcal C\). Thus \(\mathcal P\) is tight.
Definition (Prokhorov metric). For probability measures on a separable metric space, \[ d_{\mathrm P}(P,Q) =\inf\left\{\varepsilon>0: \begin{array}{l} P(A)\le Q(A^\varepsilon)+\varepsilon,\\ Q(A)\le P(A^\varepsilon)+\varepsilon \end{array} \text{ for every Borel }A \right\}, \] where \(A^\varepsilon=\{x:d(x,A)<\varepsilon\}\). On Polish spaces, \(d_{\mathrm P}\) metrizes weak convergence.
Moment criteria
Moment bounds imply tightness in Euclidean spaces.
Proposition. If \(\sup_{P\in\mathcal P}\int \|x\|^p\,\mathrm{d}P(x)\lt \infty\) for some \(p>0\), then \(\mathcal P\) is tight on \(\mathbb R^d\).
Proof. Markov's inequality gives \(P(\|x\|>M)\le M^{-p}\int\|x\|^p\,\mathrm{d}P\). Taking the supremum over \(P\) and sending \(M\to\infty\) gives the tail criterion for tightness.
Tightness and uniqueness of subsequential limits
Compactness produces candidates; a determining class identifies the only possible candidate.
Remark. If \((P_n)\) is tight and every weakly convergent subsequence has the same limit \(P\), then \(P_n\Rightarrow P\). Otherwise some bounded-continuous test would stay separated from its \(P\)-integral along a subsequence; Prokhorov would give that subsequence a further weak limit, contradicting uniqueness.