Why Weak Convergence Has Several Forms
Weak convergence is defined by how probability measures integrate bounded continuous functions. Yet many applications ask about probabilities of events: whether a random vector lies in a region, for example. An event indicator is usually discontinuous, so weak convergence does not directly guarantee convergence of its probabilities.
The Portmanteau theorem resolves this tension. It shows that bounded continuous tests, inequalities for open and closed sets, and convergence on sets with negligible boundary are different formulations of the same convergence of probability measures.
For a set \(A\subseteq\mathbb{R}^d\), its boundary is \(\partial A=\overline{A}\setminus A^\circ\), where \(\overline{A}\) is its closure and \(A^\circ\) is its interior. A Borel set \(A\) is a continuity set for \(\mu\) if \(\mu(\partial A)=0\).
The Portmanteau Theorem
- \(\mu_n\Rightarrow\mu\).
- For every closed set \(F\subseteq\mathbb{R}^d\), \(\limsup_{n\to\infty}\mu_n(F)\leq\mu(F)\).
- For every open set \(G\subseteq\mathbb{R}^d\), \(\liminf_{n\to\infty}\mu_n(G)\geq\mu(G)\).
- For every Borel continuity set \(A\) for \(\mu\), \(\mu_n(A)\to\mu(A)\).
The set bounds are deliberately one-sided. A closed set cannot acquire more limiting probability than its \(\mu\)-probability, in the limsup sense. An open set cannot lose probability below its \(\mu\)-probability, in the liminf sense. When the boundary has zero \(\mu\)-probability, these two bounds meet and give convergence.
From Continuous Tests to Set Bounds
We first establish the implication from weak convergence to the open-set bound. We use the standard regularity of Borel probability measures on \(\mathbb{R}^d\): for every open \(G\), $$ \mu(G)=\sup\{\mu(K):K\subseteq G,\ K\text{ compact}\}. $$ We also need a continuous cutoff around a compact subset of an open set. Here is a direct construction.
Fix a compact set \(K\subseteq G\), with \(G\) open. If \(K\) is empty, the cutoff \(h=0\) suffices. Otherwise, for each \(x\in K\), choose \(r_x>0\) so that the closed ball with center \(x\) and radius \(2r_x\) lies in \(G\). The balls of radius \(r_x\) cover \(K\), so compactness gives a finite subcover with centers \(x_1,\ldots,x_m\) and radii \(r_1,\ldots,r_m\).
For each \(i\), let \(\phi_i\) equal \(1\) on the ball of radius \(r_i\) centered at \(x_i\), decrease linearly to \(0\) between radii \(r_i\) and \(2r_i\), and equal \(0\) outside the ball of radius \(2r_i\). Each \(\phi_i\) is continuous, takes values in \([0,1]\), and vanishes outside \(G\). The function \(h=\min(1,\sum_{i=1}^m\phi_i)\) is continuous, equals \(1\) on \(K\), takes values in \([0,1]\), and vanishes outside \(G\).
Now suppose \(\mu_n\Rightarrow\mu\). Since \(0\leq h\leq 1\) and \(h\) vanishes outside \(G\), we have \(\int h\,d\mu_n\leq\mu_n(G)\). Weak convergence and \(h=1\) on \(K\) therefore give $$ \liminf_{n\to\infty}\mu_n(G) \geq \lim_{n\to\infty}\int h\,d\mu_n =\int h\,d\mu \geq\mu(K). $$ Taking the supremum over compact \(K\subseteq G\) and using regularity proves \(\liminf_n\mu_n(G)\geq\mu(G)\). This also covers \(G=\varnothing\), for which both sides are zero.
For a closed set \(F\), its complement \(F^c\) is open. Since all measures in question are probability measures, $$ \limsup_{n\to\infty}\mu_n(F) =1-\liminf_{n\to\infty}\mu_n(F^c) \leq 1-\mu(F^c) =\mu(F). $$ Thus weak convergence gives both set bounds.
How the Set Bounds Give Continuity-Set Convergence
Suppose the closed-set bound holds. Taking complements yields the open-set bound: for an open \(G\), the set \(G^c\) is closed, and $$ \liminf_{n\to\infty}\mu_n(G) =1-\limsup_{n\to\infty}\mu_n(G^c) \geq 1-\mu(G^c) =\mu(G). $$
Now let \(A\) be a Borel continuity set for \(\mu\). Its interior \(A^\circ\) is open and its closure \(\overline{A}\) is closed, with \(A^\circ\subseteq A\subseteq\overline{A}\). The boundary condition implies $$ \mu(A^\circ)=\mu(A)=\mu(\overline{A}), $$ because both differences are contained in \(\partial A\). Applying the open-set bound to \(A^\circ\) and the closed-set bound to \(\overline{A}\) gives $$ \mu(A) \leq\liminf_{n\to\infty}\mu_n(A) \leq\limsup_{n\to\infty}\mu_n(A) \leq\mu(A). $$ The liminf and limsup are equal to \(\mu(A)\), so \(\mu_n(A)\to\mu(A)\).
From Continuity Sets Back to Continuous Tests
It remains to show that convergence on all continuity sets implies weak convergence. Let \(f:\mathbb{R}^d\to\mathbb{R}\) be bounded and continuous. If \(f\) is identically zero there is nothing to prove. Otherwise choose \(M>0\) such that \(|f(x)|\leq M\) for all \(x\), and set \(g=(f+M)/(2M)\). Then \(g\) is continuous and takes values in \([0,1]\).
The distribution of \(g\) under \(\mu\) has at most countably many atoms: for each positive integer \(k\), there can be only finitely many levels \(t\) with \(\mu(g=t)\geq 1/k\), and every level of positive mass belongs to one of these finite collections for some \(k\). Thus we can choose finite partitions $$ t_0<t_1<\cdots<t_m,\qquad t_0<0,\quad t_m>1, $$ with arbitrarily small maximum gap and with \(\mu(g=t_j)=0\) at every internal level \(t_j\). The endpoints outside \([0,1]\) also have zero level mass.
For each partition define $$ q(y)=t_0+\sum_{j=1}^{m}(t_j-t_{j-1})\mathbf{1}_{\{y\geq t_j\}}, \qquad 0\leq y\leq 1. $$ If \(y\in[t_k,t_{k+1})\), then \(q(y)=t_k\); hence \(|q(y)-y|\) is at most the largest partition gap. The events \(\{g\geq t_j\}\) are Borel continuity sets: continuity of \(g\) implies their boundaries are contained in \(\{g=t_j\}\), which has \(\mu\)-measure zero. By the assumed continuity-set convergence, their probabilities under \(\mu_n\) converge to their probabilities under \(\mu\). Since \(q(g)\) is a finite linear combination of these event indicators and a constant, it follows that $$ \int q(g)\,d\mu_n\longrightarrow\int q(g)\,d\mu. $$ If the maximum partition gap is at most \(\eta\), then \(|q(g)-g|\leq\eta\) everywhere. Since \(\mu_n\) and \(\mu\) are probability measures, $$ \left|\int g\,d\mu_n-\int g\,d\mu\right| \leq 2\eta+ \left|\int q(g)\,d\mu_n-\int q(g)\,d\mu\right|. $$ Taking the limit superior and then choosing partitions with \(\eta\) arbitrarily small proves convergence of the integrals of \(g\). The identity \(f=2Mg-M\) gives convergence of the integrals of \(f\). This is weak convergence and completes the equivalence proof. \(\square\)
Worked Example: Point Masses Moving Toward the Origin
Worked Example: Point Masses Moving Toward the Origin
Let \(e_1=(1,0,\ldots,0)\in\mathbb{R}^d\), and take \(\mu_n=\delta_{e_1/n}\) and \(\mu=\delta_0\), where \(\delta_x\) denotes unit mass at \(x\). For every bounded continuous \(f\), $$ \int f\,d\mu_n=f(e_1/n)\longrightarrow f(0)=\int f\,d\mu, $$ so \(\mu_n\Rightarrow\mu\).
Consider the open ball \(A=\{x:|x|<2\}\). Its boundary is the sphere of radius \(2\), which does not contain \(0\), so \(\mu(\partial A)=0\). For every \(n\), \(|e_1/n|=1/n\leq1<2\); also \(0\in A\). Thus \(\mu_n(A)=1=\mu(A)\) for every \(n\), including \(n=1\). The continuity-set conclusion applies exactly as stated.
Worked Example: A Small Amount of Mass Escapes
Worked Example: A Small Amount of Mass Escapes
Define \(\nu_n=(1-1/n)\delta_0+(1/n)\delta_{ne_1}\). For any bounded continuous \(f\), $$ \left|\int f\,d\nu_n-f(0)\right| =\frac{1}{n}|f(ne_1)-f(0)| \leq\frac{2\|f\|_\infty}{n}\longrightarrow0. $$ Hence \(\nu_n\Rightarrow\delta_0\), even though the locations \(ne_1\) move arbitrarily far from the origin.
Take the closed set \(F=\{x:|x|\geq1\}\). The origin is not in \(F\), and \(ne_1\in F\) for every \(n\in\mathbb{N}\), so \(\delta_0(F)=0\) and \(\nu_n(F)=1/n\). In particular, \(\limsup_n\nu_n(F)=0\leq\delta_0(F)\), as required by the closed-set bound.
Worked Example: An Event Probability Can Fail to Converge
Worked Example: An Event Probability Can Fail to Converge
Keep \(\mu_n=\delta_{e_1/n}\Rightarrow\delta_0\), and let \(A=\{x\in\mathbb{R}^d:x_1\leq0\}\). For every \(n\), the first coordinate of \(e_1/n\) is \(1/n>0\), so \(\mu_n(A)=0\). But the first coordinate of \(0\) is \(0\), so \(\delta_0(A)=1\). Therefore \(\mu_n(A)\) does not converge to \(\delta_0(A)\).
There is no contradiction with the theorem: \(\partial A=\{x:x_1=0\}\), and \(\delta_0(\partial A)=1\), not zero. The theorem guarantees convergence for continuity sets, not for every Borel event.
Why the Boundary Condition Matters
An event indicator \(\mathbf{1}_A\) is continuous away from the boundary of \(A\), but it can jump there. If the limiting measure assigns positive probability to that boundary, small movements of mass near it may change whether points count as inside or outside the event. The half-space example makes this effect explicit: the measures approach a point on the boundary from the positive side, while the limiting point itself belongs to the event.
The four formulations are useful in different situations. Bounded continuous tests are convenient for defining weak convergence and deriving distributional limits. Open- and closed-set bounds provide robust inequalities even when event probabilities do not converge. Continuity sets are the appropriate setting when an application requires an actual limit for a probability.
Check Your Understanding
Use the equivalent formulations to answer the following questions.
- State the Portmanteau bound for a closed set and the corresponding bound for an open set.
- Why does a Borel set \(A\) with \(\mu(\partial A)=0\) satisfy \(\mu(A^\circ)=\mu(A)=\mu(\overline{A})\)?
- For \(\mu_n=\delta_{e_1/n}\) and \(\mu=\delta_0\), why is \(\{x:x_1\leq0\}\) not a continuity set for \(\mu\)?
- In the escaping-mass example, calculate \(\nu_n(\{x:|x|\geq1\})\) and its limit.
- Which part of the continuity-set-to-weak-convergence argument allows bounded continuous functions to be approximated by finite combinations of event indicators?