Continuity Sets and Convergence of Probabilities

Why Boundaries Matter

Weak convergence compares probability measures through their integrals against bounded continuous functions. Probabilities of sets are different: the indicator of a set can jump at its boundary, so weak convergence does not automatically make every event probability converge.

The Portmanteau theorem gives a useful way to handle this. If the limiting measure assigns no probability to the boundary of a Borel set, then the probabilities of that set do converge. The boundary is precisely where the set’s indicator can change abruptly.

Continuity Sets

Definition: Let $\mu$ be a Borel probability measure on $\mathbb{R}^d$. A Borel set $A\subseteq\mathbb{R}^d$ is a continuity set for $\mu$ if its boundary has $\mu$-measure zero: $$ \mu(\partial A)=0. $$ Here $\partial A=\overline{A}\setminus A^\circ$, where $\overline{A}$ is the closure and $A^\circ$ is the interior of $A$.

This name refers to continuity of the set’s indicator in a measure-theoretic sense: any discontinuities of the indicator occur on the boundary, and a continuity set has a boundary that is negligible for $\mu$. The set itself need not be open or closed.

For a sequence of probability measures $(\mu_n)$, weak convergence $\mu_n\Rightarrow\mu$ means convergence against every bounded continuous test function. The Portmanteau theorem gives, in particular, the following bounds: for every open set $G$ and every closed set $F$,

\mu(G)\leq \liminf_{n\to\infty}\mu_n(G), \qquad \limsup_{n\to\infty}\mu_n(F)\leq \mu(F).

These inequalities are the key ingredients in the continuity-set result.

The Continuity-Set Theorem

Theorem: Suppose $\mu_n\Rightarrow\mu$ are Borel probability measures on $\mathbb{R}^d$. If $A\subseteq\mathbb{R}^d$ is Borel and $\mu(\partial A)=0$, then $$ \lim_{n\to\infty}\mu_n(A)=\mu(A). $$

Proof. The interior $A^\circ$ is open, and the closure $\overline{A}$ is closed. Since $A^\circ\subseteq A\subseteq\overline{A}$, monotonicity of each probability measure gives $$ \mu_n(A^\circ)\leq\mu_n(A)\leq\mu_n(\overline{A}). $$ The open-set and closed-set Portmanteau bounds therefore imply $$ \mu(A^\circ) \leq \liminf_{n\to\infty}\mu_n(A) \leq \limsup_{n\to\infty}\mu_n(A) \leq \mu(\overline{A}). $$ The closure decomposes into the disjoint union $A^\circ\cup\partial A$. The assumption $\mu(\partial A)=0$ consequently gives $\mu(\overline{A})=\mu(A^\circ)$. Also, $A^\circ\subseteq A\subseteq\overline{A}$, so the same assumption gives $\mu(A)=\mu(A^\circ)=\mu(\overline{A})$. The lower and upper bounds thus agree, proving the claimed limit. $\square$

The proof works by placing the probabilities of $A$ between those of an open set and a closed set. When the boundary has zero limiting probability, those two bounds meet. This is why the condition is useful: it removes any gap between the interior and closure bounds.

Worked Example: Intervals and Rays on the Real Line

Let $a<b$. For the closed interval $A=[a,b]\subseteq\mathbb{R}$, the interior is $(a,b)$ and the closure is $[a,b]$. Thus $$ \partial A=\{a,b\}. $$ For the open interval $(a,b)$, the interior is $(a,b)$ and its closure is also $[a,b]$, so its boundary is again $\{a,b\}$. Both intervals are continuity sets for $\mu$ exactly when $$ \mu(\{a\})+\mu(\{b\})=0. $$ Because the terms are nonnegative, this is equivalent to having no $\mu$-mass at either endpoint. Whenever this condition holds and $\mu_n\Rightarrow\mu$, the theorem gives convergence of the probabilities of each interval.

A ray illustrates the same endpoint issue. For $A=(-\infty,t]$, every point less than $t$ is interior, points greater than $t$ are outside the closure, and $t$ belongs to the boundary. Hence $\partial A=\{t\}$. The ray is a continuity set exactly when $\mu(\{t\})=0$, in which case $\mu_n((-\infty,t])\to\mu((-\infty,t])$.

Worked Example: A Rectangle in the Plane

Take $a<b$ and $c<d$, and let $R=[a,b]\times[c,d]\subseteq\mathbb{R}^2$. Its interior is $(a,b)\times(c,d)$, and its closure is $R$. Removing the interior leaves the four edges: $$ \partial R= ([a,b]\times\{c\}) \cup([a,b]\times\{d\}) \cup(\{a\}\times[c,d]) \cup(\{b\}\times[c,d]). $$ The corners are already included in these edges. Thus $R$ is a continuity set for $\mu$ precisely when the union of its edges has $\mu$-measure zero.

In particular, suppose $\mu$ has a density with respect to two-dimensional Lebesgue measure. Each edge is a line segment and has two-dimensional Lebesgue measure zero. A finite union of such edges still has Lebesgue measure zero, so it also has $\mu$-measure zero. Therefore, for any weakly convergent sequence with this limiting measure, the probabilities of $R$ converge.

Worked Example: Balls and Spheres

Let $x\in\mathbb{R}^d$ and $r>0$, and consider the closed ball $$ B=\{y\in\mathbb{R}^d:\|y-x\|\leq r\}. $$ Its interior is given by the strict inequality $\|y-x\|<r$, and its boundary is the sphere $$ \partial B=\{y\in\mathbb{R}^d:\|y-x\|=r\}. $$ The open ball has the same boundary. Consequently, either ball is a continuity set when $\mu$ assigns zero mass to this sphere.

A sphere has $d$-dimensional Lebesgue measure zero. Therefore, if $\mu$ has a density with respect to that measure, the sphere has $\mu$-measure zero and the ball probabilities converge under weak convergence to $\mu$. The condition $r>0$ matters for this boundary description: when $r=0$, the closed ball is the singleton $\{x\}$, whose boundary is $\{x\}$, and it is a continuity set only if $\mu(\{x\})=0$.

Worked Example: A Finite Set

Let $A$ be a nonempty finite subset of $\mathbb{R}^d$. It is closed, and its interior is empty: every open ball around any point contains points outside $A$. Hence $$ \partial A=A. $$ It follows that $A$ is a continuity set for $\mu$ exactly when $\mu(A)=0$. If $\mu$ assigns positive mass to even one point of $A$, the continuity-set theorem does not supply convergence of $\mu_n(A)$. This example shows why knowing the limiting measure’s atoms can be essential when choosing a set.

For the empty set, the boundary is empty, so it is a continuity set for every $\mu$; its probabilities are identically zero. This also covers the degenerate case omitted by the nonempty-set calculation.

Stability of Continuity Sets

Proposition: If $A$ is a continuity set for $\mu$, then $A^c$ is a continuity set for $\mu$. If $A$ and $B$ are continuity sets for $\mu$, then $A\cup B$ and $A\cap B$ are continuity sets for $\mu$.

Proof. A point belongs to the boundary of a set exactly when every neighborhood of that point meets both the set and its complement. Interchanging a set and its complement does not change that condition, so $\partial(A^c)=\partial A$. For unions, a point outside $\partial A\cup\partial B$ is interior to either $A$ or $A^c$, and also interior to either $B$ or $B^c$. If it is interior to $A$ or to $B$, it is interior to $A\cup B$. If it is interior to both complements, it is interior to $(A\cup B)^c$. In either case it is not a boundary point of $A\cup B$. Thus $$ \partial(A\cup B)\subseteq\partial A\cup\partial B. $$ Since $\partial(A\cap B)=\partial(A^c\cup B^c)$, the complement identity and the union inclusion also give $$ \partial(A\cap B)\subseteq\partial A\cup\partial B. $$ If both $A$ and $B$ are continuity sets, the right-hand side has $\mu$-measure zero. Each of these boundaries therefore has $\mu$-measure zero as well, proving the proposition. $\square$

Using the Criterion Carefully

The theorem gives a sufficient condition for convergence of probabilities, not a claim that the boundary condition is necessary in every particular example. Its practical use is to check the boundary under the limiting measure before applying the result. For sets with simple geometry, this often reduces to checking whether the limit puts mass on endpoints, edges, or a sphere.

A common mistake is to assume that weak convergence directly gives convergence of $\mu_n(A)$ for every Borel set $A$. The theorem does not say that. The useful guarantee applies when the limiting measure gives the boundary no mass; that hypothesis is what makes the open and closed Portmanteau bounds coincide.

Continuity sets turn weak convergence into a concrete statement about events. Once a set has been checked, probabilities of that set can be passed to the limit without approximating its indicator by continuous functions. The geometric boundary calculations are therefore not just formal details: they identify exactly when this direct conclusion is available.

Check Your Understanding

For each question, identify the boundary first, then relate its measure to the continuity-set criterion.

What condition on $\mu$ makes the interval $[a,b]$ a continuity set?
Why do the open and closed balls of positive radius have the same boundary?
If a probability measure has a density on $\mathbb{R}^2$, why does it assign zero mass to the edges of a rectangle?
What is the boundary of a nonempty finite subset of $\mathbb{R}^d$, and when is it a continuity set?
Which two Portmanteau bounds squeeze $\mu_n(A)$ in the proof of the continuity-set theorem?

Continuity Sets and Convergence of Probabilities

What You'll Learn

Why Boundaries Matter

Continuity Sets

The Continuity-Set Theorem

Worked Example: Intervals and Rays on the Real Line

Worked Example: Intervals and Rays on the Real Line

Worked Example: A Rectangle in the Plane

Worked Example: A Rectangle in the Plane

Worked Example: Balls and Spheres

Worked Example: Balls and Spheres

Worked Example: A Finite Set

Worked Example: A Finite Set

Stability of Continuity Sets

Using the Criterion Carefully

Check Your Understanding

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