Maps Continuous Almost Surely Under the Limit Law

When Continuity Everywhere Is More Than We Need

The continuous mapping theorem says that applying a continuous function to weakly convergent random variables preserves weak convergence. In many applications, however, the map of interest has a small set of discontinuities. A threshold indicator jumps at its threshold; an integer-part map jumps at every integer. If the limiting distribution assigns probability zero to those locations, the jumps need not obstruct convergence.

The important point is that the discontinuities are judged under the limiting measure. The approximating measures may assign positive probability to the discontinuity set. The theorem below requires only that the limiting measure assigns it probability zero.

Definition: For a map \(f:\mathbb{R}^d\to\mathbb{R}^m\), its discontinuity set is \(D_f=\{x\in\mathbb{R}^d:f\text{ is not continuous at }x\}\). The map is continuous almost surely under a probability measure \(\mu\) if \(\mu(D_f)=0\).

We use weak convergence in its bounded-continuous-test form: \(\mu_n\) converges weakly to \(\mu\) if the integrals of every bounded continuous real-valued function converge. We also use the Portmanteau continuity-set conclusion: if \(\mu_n\) converges weakly to \(\mu\) and \(\mu(\partial A)=0\), then \(\mu_n(A)\to\mu(A)\).

A Portmanteau Integral Bound

The key step is to show that weak convergence controls a bounded measurable function when its discontinuities have zero limiting probability. This fact makes the theorem applicable even though composing with a merely measurable map need not produce a continuous test function.

Proposition: Suppose \(\mu_n\) converges weakly to a probability measure \(\mu\) on \(\mathbb{R}^d\). If \(g:\mathbb{R}^d\to\mathbb{R}\) is bounded and Borel measurable, and \(\mu(D_g)=0\), then \(\int g\,d\mu_n\to\int g\,d\mu\).

Proof. Choose \(M\geq 0\) such that \(|g(x)|\leq M\) for every \(x\). If \(M=0\), then \(g\) is identically zero and the claim follows immediately. Otherwise, \(g+M\) takes values in \([0,2M]\). For any probability measure \(\nu\), the layer-cake identity gives

\int (g+M)\,d\nu = \int_0^{2M}\nu(\{x:g(x)+M>t\})\,dt.

For a real number \(s\), consider \(A_s=\{x:g(x)>s\}\). If \(x\notin D_g\) and \(g(x)\neq s\), continuity of \(g\) at \(x\) gives a neighbourhood of \(x\) on which \(g\) remains on the same side of \(s\) as \(g(x)\). Thus \(x\notin\partial A_s\). Consequently,

\partial A_s\subseteq D_g\cup\{x:g(x)=s\}.

The set of real \(s\) for which \(\mu(\{g=s\})>0\) is at most countable. Indeed, for each positive integer \(r\), there can be only finitely many disjoint level sets with measure at least \(1/r\); every level set of positive measure has measure at least \(1/r\) for some \(r\). Therefore, outside a countable set of \(s\), we have \(\mu(\partial A_s)=0\), since \(\mu(D_g)=0\).

For every such \(s\), the Portmanteau continuity-set conclusion gives \(\mu_n(A_s)\to\mu(A_s)\). In the layer-cake identity, this is pointwise convergence for almost every \(t\in[0,2M]\), with \(s=t-M\). The integrands are bounded between zero and one. The dominated convergence theorem therefore yields

\int (g+M)\,d\mu_n \longrightarrow \int (g+M)\,d\mu.

Since each measure is a probability measure, subtracting \(M\) from both integrals proves the required convergence for \(g\). \(\square\)

The Almost-Sure Continuity Mapping Theorem

Theorem: Let \(\mu_n\) and \(\mu\) be probability measures on \(\mathbb{R}^d\), with \(\mu_n\) converging weakly to \(\mu\). Let \(f:\mathbb{R}^d\to\mathbb{R}^m\) be Borel measurable and suppose \(\mu(D_f)=0\). Then the image measures \(\mu_n\circ f^{-1}\) converge weakly to \(\mu\circ f^{-1}\).

Proof. To verify weak convergence of the image measures, take an arbitrary bounded continuous function \(\varphi:\mathbb{R}^m\to\mathbb{R}\). The composition \(g=\varphi\circ f\) is bounded and Borel measurable. At every point where \(f\) is continuous, the composition \(\varphi\circ f\) is continuous, so \(D_g\subseteq D_f\). Hence \(\mu(D_g)=0\). The proposition just proved gives

\int \varphi\,d(\mu_n\circ f^{-1}) = \int (\varphi\circ f)\,d\mu_n \longrightarrow \int (\varphi\circ f)\,d\mu = \int \varphi\,d(\mu\circ f^{-1}).

This holds for every bounded continuous \(\varphi\), which is precisely weak convergence of the image measures. \(\square\)

In random-variable language, if \(X_n\) converges in distribution to \(X\), \(f\) is Borel measurable, and \(\mathbb{P}(X\in D_f)=0\), then \(f(X_n)\) converges in distribution to \(f(X)\). The statement concerns the laws, so \(X_n\) and \(X\) need not be defined on the same probability space.

Worked Examples

Worked Example: A Threshold Indicator

Let \(\mu\) be the standard normal law on \(\mathbb{R}\), and fix \(c\in\mathbb{R}\). Define \(f(x)=1\) when \(x>c\), and \(f(x)=0\) when \(x\leq c\). This Borel map is discontinuous only at \(c\). The standard normal distribution has a density, so \(\mu(\{c\})=0\). Thus, for any probability measures \(\mu_n\) converging weakly to \(\mu\), the theorem gives \(\mu_n\circ f^{-1}\) converging weakly to \(\mu\circ f^{-1}\).

More explicitly, the output is supported on \(\{0,1\}\). Its probability of taking the value \(1\) is \(\mu_n((c,\infty))\), and the boundary of \((c,\infty)\) is \(\{c\}\), which has \(\mu\)-measure zero. Portmanteau therefore gives

\mu_n((c,\infty))\longrightarrow \mu((c,\infty))=1-\Phi(c),

where \(\Phi\) is the standard normal distribution function. The probability of output \(0\) converges as well, since it is one minus the probability of output \(1\). Hence the entire two-point output law converges.

Worked Example: The Integer-Part Map

Let \(\mu\) again be standard normal, and let \(f(x)=\lfloor x\rfloor\), the greatest integer less than or equal to \(x\). The map is continuous at every noninteger and discontinuous at every integer, so \(D_f=\mathbb{Z}\). Since \(\mu\) has a density, \(\mu(\mathbb{Z})=0\), and the theorem applies to any \(\mu_n\) converging weakly to \(\mu\).

For each \(k\in\mathbb{Z}\), the event that the output equals \(k\) is \([k,k+1)\). Its boundary is \(\{k,k+1\}\), a set of standard normal probability zero. Therefore Portmanteau gives

\mu_n([k,k+1)) \longrightarrow \mu([k,k+1)) = \Phi(k+1)-\Phi(k).

Thus each individual output probability converges to the corresponding integer-part probability under the standard normal law. The theorem also gives weak convergence of the complete output distributions, not merely convergence of one selected probability.

Worked Example: Classifying a Multivariate Observation

Let \(\mu\) be the law of a pair \((Z_1,Z_2)\) of independent standard normal random variables on \(\mathbb{R}^2\). Define \(f(x,y)=1\) if both \(x>0\) and \(y>0\), and \(f(x,y)=0\) otherwise. The map can be discontinuous only on the coordinate axes: away from the axes, the signs of both coordinates are locally unchanged. Each axis has \(\mu\)-probability zero, because each marginal has a density. Hence \(\mu(D_f)=0\).

For any \(\mu_n\) converging weakly to \(\mu\), the image laws converge. To calculate the limiting probability of output \(1\), independence and symmetry give

\mu(\{(x,y):x>0,\ y>0\}) = \mathbb{P}(Z_1>0)\mathbb{P}(Z_2>0) = \frac12\cdot\frac12 = \frac14.

Equivalently, the boundary of the positive quadrant lies in the axes and has \(\mu\)-measure zero, so Portmanteau gives convergence of its probability under \(\mu_n\) to \(1/4\). The probability of output \(0\) consequently converges to \(3/4\).

What the Null-Discontinuity Condition Does—and Does Not—Say

The condition \(\mu(D_f)=0\) does not say that \(f\) is continuous everywhere, nor that the approximating measures avoid the discontinuities. It says that the limiting law places no probability on the points where continuity fails. The Portmanteau argument works because, for almost every threshold of a bounded test function composed with \(f\), the corresponding level set has a boundary of limiting probability zero.

A common mistake is to demand \(\mu_n(D_f)=0\) for every \(n\). That is not part of the theorem. Weak convergence is tested against bounded continuous functions, and the proof handles the composition by controlling its discontinuities relative to \(\mu\), the limiting measure. Conversely, measurability alone is not the condition that makes the conclusion work: the null-discontinuity hypothesis is what allows the Portmanteau bounds to control the composed test functions.

This extension is useful whenever a statistical or probabilistic quantity is defined by a rule with boundary cases, thresholds, or partitions. If those boundaries have zero probability under the limiting law, the rule can still transmit weak convergence even though it is not globally continuous.

Check Your Understanding

Use the theorem and the Portmanteau reasoning above to answer the following questions.

Which measure must assign probability zero to \(D_f\) in the almost-sure continuity mapping theorem?
Why is \(\varphi\circ f\) continuous at every continuity point of \(f\) when \(\varphi\) is continuous?
For the integer-part map, what is its discontinuity set, and why does a standard normal law assign it probability zero?
Why does the level set \(\{g>s\}\) have a boundary of \(\mu\)-measure zero for all but at most countably many \(s\), when \(\mu(D_g)=0\)?
In the multivariate classification example, why does the probability of output \(1\) converge to \(1/4\)?