Why Discontinuities Matter
Weak convergence behaves well under continuous transformations: if random vectors converge weakly and a map is continuous at the relevant values, transforming the vectors preserves the expected weak limit. The continuity qualification matters. When the limiting law places positive probability on a discontinuity, even a simple deterministic sequence can reveal a failure.
We will examine three familiar maps: sign, a threshold indicator, and the floor function. In each case, the input laws converge weakly to a point mass at a discontinuity, while the transformed laws do not converge to the point mass obtained by transforming the limit. The examples show a possibility, not a rule: a charged discontinuity can break convergence, but it does not have to.
For a constant random vector equal to \(x\), its law is the point mass \(\delta_x\). If \(x_n\to x\), then \(\delta_{x_n}\Rightarrow\delta_x\): for every bounded continuous \(f\), continuity gives \(f(x_n)\to f(x)\). Thus deterministic inputs already provide valid tests of whether a transformation preserves weak convergence.
A Criterion for Point-Mass Laws
Proof. If \(y_n\to y\), then for every bounded continuous \(f\), \(f(y_n)\to f(y)\). Since \(\int f\,d\delta_{y_n}=f(y_n)\) and \(\int f\,d\delta_y=f(y)\), this is exactly \(\delta_{y_n}\Rightarrow\delta_y\).
Conversely, suppose \(\delta_{y_n}\Rightarrow\delta_y\). The function \(f(z)=\min\{1,\|z-y\|\}\) is bounded and continuous, so weak convergence implies \(f(y_n)\to f(y)=0\). Given any \(\varepsilon>0\), set \(r=\min\{\varepsilon,1\}/2\). Eventually \(f(y_n)<r<1\), which forces \(\|y_n-y\|=f(y_n)<r<\varepsilon\). Therefore \(y_n\to y\).
\(\square\)
This criterion turns a discontinuity into a direct counterexample. If a measurable map \(g:\mathbb{R}^d\to\mathbb{R}^k\) is discontinuous at \(x\), the sequential definition of continuity gives a sequence \(x_n\to x\) for which \(g(x_n)\) does not converge to \(g(x)\). The input point masses converge, \(\delta_{x_n}\Rightarrow\delta_x\), but the theorem shows that \(\delta_{g(x_n)}\) cannot converge weakly to \(\delta_{g(x)}\). The examples below make this mechanism explicit.
Worked Example: The Sign Map at Zero
Worked Example: The Sign Map at Zero
Define \(\operatorname{sgn}:\mathbb{R}\to\mathbb{R}\) by \(\operatorname{sgn}(x)=-1\) for \(x<0\), \(\operatorname{sgn}(0)=0\), and \(\operatorname{sgn}(x)=1\) for \(x>0\). Consider the constant random variables \(X_n=1/n\) and \(X=0\). Since \(1/n\to0\), their laws satisfy \(\mathcal{L}(X_n)=\delta_{1/n}\Rightarrow\delta_0=\mathcal{L}(X)\).
But \(1/n>0\) for every \(n\), so \(\operatorname{sgn}(X_n)=1\), whereas \(\operatorname{sgn}(X)=0\). Hence the transformed laws are \(\delta_1\) for every \(n\), not a sequence converging weakly to \(\delta_0\). For instance, the bounded continuous test function \(h(t)=t/(1+|t|)\) has integral \(1/2\) under \(\delta_1\) and \(0\) under \(\delta_0\). The limiting law of \(X\) assigns probability one to the discontinuity at zero.
The point is not that the input is random in any complicated way. It is that inputs arbitrarily close to zero on the positive side all receive the same transformed value, \(1\), while the value assigned at zero is \(0\).
Worked Example: A Threshold Map
Worked Example: A Threshold Map
Fix \(c\in\mathbb{R}\), and define \(H_c(x)=\mathbf{1}_{(c,\infty)}(x)\). Thus \(H_c(x)=1\) when \(x>c\), and \(H_c(x)=0\) when \(x\le c\). This map is discontinuous at \(c\). Take \(X_n=c+1/n\) and \(X=c\). For every bounded continuous \(\varphi:\mathbb{R}\to\mathbb{R}\), \(\varphi(c+1/n)\to\varphi(c)\); hence \(\delta_{c+1/n}\Rightarrow\delta_c\).
Nevertheless, \(H_c(X_n)=1\) for every \(n\), and \(H_c(X)=0\). The transformed laws are \(\delta_1\) and \(\delta_0\), respectively. The test function \(h(t)=t/(1+|t|)\) again gives integrals \(1/2\) and \(0\), proving that the transformed laws do not converge to the claimed limit. Here too the limiting input law charges the discontinuity: \(\delta_c(\{c\})=1\).
A threshold indicator is often used to represent an event probability, such as the probability that a variable exceeds a cutoff. This example warns that weak convergence alone does not guarantee convergence of such probabilities when the cutoff is an atom of the limiting distribution. The continuity-set criterion from Portmanteau gives the corresponding positive condition when the boundary has limiting probability zero.
Worked Example: The Floor Function at an Integer
Worked Example: The Floor Function at an Integer
The floor map \(\lfloor x\rfloor\) is the greatest integer not exceeding \(x\). It is discontinuous at every integer. Let \(X_n=1-1/n\) for \(n\ge2\), and let \(X=1\). Since \(X_n\to1\), bounded continuous tests give \(\delta_{X_n}\Rightarrow\delta_1\).
For every \(n\ge2\), \(0\le 1-1/n<1\), so \(\lfloor X_n\rfloor=0\). At the limit, \(\lfloor X\rfloor=\lfloor1\rfloor=1\). Therefore \(\mathcal{L}(\lfloor X_n\rfloor)=\delta_0\) for every \(n\), while \(\mathcal{L}(\lfloor X\rfloor)=\delta_1\). The bounded continuous test \(h(t)=t/(1+|t|)\) has integral \(0\) under \(\delta_0\) and \(1/2\) under \(\delta_1\), so the transformed laws fail to converge to the law of the transformed limit.
For the floor map, the direction of approach is particularly visible. Approaching an integer \(m\) from below gives floor value \(m-1\), while the value at \(m\) is \(m\). Approaching from above gives floor value \(m\). A jump at the limiting input can therefore change the transformed limit according to which side the sequence approaches from.
Discontinuity Allows Failure, but Does Not Force It
Worked Example: A Threshold Approached from Below
Keep the threshold map \(H_c(x)=\mathbf{1}_{(c,\infty)}(x)\), but now let \(X_n=c-1/n\) and \(X=c\). Again \(X_n\to c\), so \(\delta_{X_n}\Rightarrow\delta_c\). This time \(H_c(X_n)=0\) for every \(n\), and \(H_c(X)=0\) as well. The transformed laws are all \(\delta_0\), so they do converge to the law of \(H_c(X)\), even though the limiting law assigns probability one to the discontinuity.
This contrast is important. A condition requiring continuity almost surely under the limiting law is a convenient sufficient condition for a continuous mapping conclusion. If the condition fails, one cannot infer that the conclusion fails; one must examine the transformation and the way the input laws approach the discontinuity. In the examples above, the chosen approaches produce a mismatch. In the threshold example approached from below, they do not.
The General Failure Mechanism
The point-mass criterion gives a useful general statement. Let \(X_n\) and \(X\) be random vectors in \(\mathbb{R}^d\) with laws \(\mu_n\) and \(\mu\), and suppose \(\mu_n\Rightarrow\mu\). For a measurable map \(g:\mathbb{R}^d\to\mathbb{R}^k\), the transformed laws are \(g_{\#}\mu_n\) and \(g_{\#}\mu\). A discontinuity at a point \(x\) supplies a counterexample whenever the input laws are chosen as point masses approaching \(x\) along a sequence whose images fail to approach \(g(x)\).
More precisely, discontinuity at \(x\) means that there are \(\varepsilon>0\) and points \(x_n\) with \(\|x_n-x\|<1/n\) but \(\|g(x_n)-g(x)\|\ge\varepsilon\). Set \(X_n=x_n\) and \(X=x\) deterministically. Then \(\delta_{x_n}\Rightarrow\delta_x\), but \(g(x_n)\) does not converge to \(g(x)\). By the point-mass theorem, \(\delta_{g(x_n)}\) does not converge weakly to \(\delta_{g(x)}\).
This construction explains why discontinuities matter without claiming that every sequence approaching one must fail. A given sequence may approach along a side on which the map behaves compatibly with its value at the limit, or the input laws may already agree with the limit law. The failure examples establish that weak convergence does not, by itself, control transformation by a map that is discontinuous at a point charged by the limiting law.
A Common Pitfall
A common mistake is to invoke the continuous mapping principle after checking only that the map is measurable. Measurability ensures that the transformed random variable and its law are defined; it does not ensure that weak convergence is preserved. The sign, threshold, and floor maps above are all measurable, yet each can break the expected conclusion.
It is also a mistake to reverse the implication and claim that positive limiting probability of the discontinuity set guarantees failure. The threshold approached from below disproves that claim. The useful distinction is:
- If the discontinuity set has limiting probability zero, the usual almost-sure continuity condition applies.
- If the limiting law charges a discontinuity, that condition is unavailable, and convergence must be checked by another argument.
- In that second case, failure is possible, as the explicit point-mass examples show, but is not automatic.
These examples make the continuity hypothesis substantive rather than cosmetic. In applications, a sign change near zero, a threshold at an atom, or rounding near an integer can alter a limiting distribution even when the original variables converge weakly.
Check Your Understanding
Use the definitions and examples above to answer the following questions.
- Why does \(X_n=1/n\) converge weakly in law to \(X=0\), even though the sign-transformed laws do not converge to the law of \(\operatorname{sgn}(X)\)?
- For the threshold map \(H_c(x)=\mathbf{1}_{(c,\infty)}(x)\), what happens to the transformed laws when the inputs approach \(c\) from below?
- State the point-mass criterion for weak convergence and identify the bounded continuous test used to prove its forward implication from weak convergence to ordinary convergence.
- At which points is the floor map discontinuous, and what are the floor values immediately below and at an integer \(m\)?
- Does positive limiting probability of a map's discontinuity set imply that transformed weak convergence must fail? Explain using an example here.