When Event Probabilities Fail to Converge

Why Event Probabilities Need Special Care

Weak convergence controls averages of bounded continuous functions. An event probability, however, is the expectation of an indicator function, and an indicator can be discontinuous at the event’s boundary. If the limiting distribution assigns positive probability to that boundary, the weak limit alone may not determine how probability approaches the event from one side or the other.

This is not a contradiction with weak convergence. It is a reminder that weak convergence is not, in general, convergence of probabilities for every measurable set. The boundary of the set identifies where the indicator can fail to be continuous; the limiting mass there is the key diagnostic.

Boundaries and Continuity Sets

Definition: For a set $A\subseteq\mathbb{R}^d$, its interior $A^\circ$ consists of points having an open neighborhood contained in $A$, and its closure $\overline{A}$ consists of points every open neighborhood of which intersects $A$. Its boundary is $\partial A=\overline{A}\setminus A^\circ$. For a probability measure $\mu$, a Borel set $A$ is a continuity set for $\mu$ if $\mu(\partial A)=0$.

Let $\mu_n$ and $\mu$ be probability measures on $\mathbb{R}^d$, and suppose $\mu_n$ converges weakly to $\mu$. The Portmanteau theorem gives, 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 two inequalities provide bounds for a general Borel event, even when its probability does not converge.

Theorem: Portmanteau bounds for an event. If $\mu_n$ converges weakly to $\mu$ and $A\subseteq\mathbb{R}^d$ is Borel, then

\mu(A^\circ) \leq \liminf_{n\to\infty}\mu_n(A) \leq \limsup_{n\to\infty}\mu_n(A) \leq \mu(\overline{A}).

Proof. The inclusions $A^\circ\subseteq A\subseteq\overline{A}$ imply $\mu_n(A^\circ)\leq\mu_n(A)\leq\mu_n(\overline{A})$ for every $n$. Since $A^\circ$ is open, the open-set Portmanteau inequality gives $\mu(A^\circ)\leq\liminf_n\mu_n(A^\circ)\leq\liminf_n\mu_n(A)$. Since $\overline{A}$ is closed, the closed-set inequality gives $\limsup_n\mu_n(A)\leq\limsup_n\mu_n(\overline{A})\leq\mu(\overline{A})$. The middle inequality between the lower and upper limits holds for every real sequence. Thus all the stated inequalities follow. $\square$

The gap between the outer bounds is precisely controlled by the boundary, because $\overline{A}\setminus A^\circ=\partial A$. In particular, if $\mu(\partial A)=0$, then $\mu(A^\circ)=\mu(\overline{A})=\mu(A)$, and the bounds squeeze the event probabilities to one value.

Corollary: Convergence on continuity sets. If $\mu_n$ converges weakly to $\mu$ and $A$ is a Borel continuity set for $\mu$, then $$ \lim_{n\to\infty}\mu_n(A)=\mu(A). $$ Proof. The boundary is null, so $\mu(\overline{A})=\mu(A^\circ)=\mu(A)$. The event bounds therefore give $\mu(A)\leq\liminf_n\mu_n(A)\leq\limsup_n\mu_n(A)\leq\mu(A)$. The lower and upper limits coincide with $\mu(A)$, which proves convergence. $\square$

Worked Examples: Failure and Convergence

Worked Example: Approaching a Boundary from Inside

On $\mathbb{R}$, let $\mu_n=\delta_{1/n}$ and $\mu=\delta_0$, where $\delta_x$ is the probability measure placing all its mass at $x$. For any bounded continuous $f:\mathbb{R}\to\mathbb{R}$,

\int f\,d\mu_n=f(1/n)\longrightarrow f(0)=\int f\,d\mu,

so $\mu_n$ converges weakly to $\mu$. Choose the event $A=(0,\infty)$. Its boundary is $\{0\}$, and $\mu(\partial A)=1$. For every $n$, $1/n>0$, so $\mu_n(A)=1$; but $0\notin A$, so $\mu(A)=0$. Thus the event probabilities converge to $1$, not to the limiting event probability $0$. The positive boundary mass allows this discrepancy.

Worked Example: Alternation Prevents Any Probability Limit

Define $x_n=(-1)^n/n$, $\mu_n=\delta_{x_n}$, and $\mu=\delta_0$. For every bounded continuous $f$, $x_n\to0$ implies $f(x_n)\to f(0)$. Since $\int f\,d\mu_n=f(x_n)$ and $\int f\,d\mu=f(0)$, this verifies $\mu_n\Rightarrow\mu$.

For $A=(0,\infty)$, the probability $\mu_n(A)$ equals $1$ when $n$ is even, because then $x_n=1/n>0$. It equals $0$ when $n$ is odd, because then $x_n=-1/n<0$. The sequence of event probabilities alternates between $0$ and $1$, and therefore has no limit. Meanwhile $\partial A=\{0\}$ and $\mu(\partial A)=1$. This example shows that failure need not merely mean convergence to the wrong value: the probabilities may fail to converge at all.

Worked Example: A Charged Boundary Does Not Force Failure

Keep $\mu_n=\delta_{1/n}$ and $\mu=\delta_0$, but take $A=[0,\infty)$. The boundary is again $\{0\}$, with $\mu(\partial A)=1$. Nevertheless, every $1/n$ belongs to $A$, and $0$ also belongs to $A$. Consequently, for every $n$,

\mu_n(A)=1=\mu(A).

The event probabilities converge exactly to the limiting event probability. Positive mass on the boundary removes the general guarantee of convergence to $\mu(A)$; it does not assert that convergence must fail for every such event or sequence.

Worked Example: A Continuity Set Has Convergent Probabilities

Again let $\mu_n=\delta_{1/n}$ and $\mu=\delta_0$, but now take $C=(-1,1)$. Its boundary is $\{-1,1\}$, which has $\mu$-probability zero, so $C$ is a continuity set. At $n=1$, the point $1/n=1$ is not in $C$, and hence $\mu_1(C)=0$. For every $n\geq2$, $0<1/n<1$, so $1/n\in C$ and $\mu_n(C)=1$. Thus the single initial value is followed by probabilities all equal to $1$, giving

\lim_{n\to\infty}\mu_n(C)=1=\mu(C).

This also illustrates why a finite number of exceptional initial terms does not affect convergence. The continuity-set conclusion concerns the limit as $n$ grows, not equality of probabilities for every $n$.

What the Boundary Bounds Do—and Do Not—Say

The event bounds localize the possible lower and upper limits between $\mu(A^\circ)$ and $\mu(\overline{A})$. The difference between those two quantities is $\mu(\partial A)$. If the boundary has zero limiting probability, the interval collapses, forcing convergence to $\mu(A)$. If the boundary has positive limiting probability, the interval may have positive width, so the bounds alone need not determine a single limit.

There are two distinct questions to keep separate:

Does $\mu_n(A)$ converge? The bounds may leave room for oscillation, as in the alternating example.
If it converges, must its limit equal $\mu(A)$? The inside-approach example converges to $1$, although $\mu(A)=0$.

In the examples, weak convergence is verified by testing bounded continuous functions, while the event is examined through its indicator. The indicator is discontinuous at the boundary, so weak convergence does not directly require its integrals to converge. By contrast, a continuity set has a boundary that the limiting law almost surely avoids, and the Portmanteau bounds then force convergence.

A Practical Diagnostic

Identify the event and its boundary.
Compute $A^\circ$, $\overline{A}$, and $\partial A$ rather than relying only on a visual impression of the event.

Evaluate limiting boundary mass.
If $\mu(\partial A)=0$, the continuity-set result guarantees $\mu_n(A)\to\mu(A)$.

If boundary mass is positive, check the sequence itself.
Use the interior and closure bounds, then calculate event membership or probabilities directly. Boundary mass signals that convergence is not guaranteed, not that it is impossible.

A common mistake is to claim that weak convergence always implies convergence of event probabilities. Another is to reverse the continuity-set criterion and claim that any positive boundary mass forces failure. Neither is correct. The criterion gives a robust sufficient condition; examples are needed to diagnose particular events when the boundary is charged.

Check Your Understanding

Use the boundary bounds and the examples above to answer the following questions.

For $\mu_n=\delta_{1/n}$, $\mu=\delta_0$, and $A=(0,\infty)$, what are $\mu_n(A)$, $\mu(A)$, and $\mu(\partial A)$?
Why does the event $A=(0,\infty)$ in the alternating example have probabilities with no limit, despite weak convergence of the measures?
State the Portmanteau bounds for a Borel event $A$.
Why does $\mu(\partial A)=0$ force $\mu_n(A)\to\mu(A)$?
Does positive limiting boundary mass imply that the event probabilities must fail to converge? Give the example above that answers this question.