Why Discontinuities Need a Probability Condition
The continuous mapping theorem says that a continuous transformation preserves convergence in distribution. Many useful transformations are not continuous everywhere: threshold indicators jump at a cutoff, rounding jumps at integers, and a direction map is undefined or unstable at the zero vector. In these cases, the location of the discontinuities matters. A discontinuity causes no asymptotic problem if the limiting random variable lands there with probability zero.
This condition is about the limit distribution, not about whether the approximating variables avoid the discontinuities exactly. The approximating variables may land on a troublesome point with positive probability for every \(n\); what matters is whether that probability persists in the limit.
The Discontinuous Mapping Theorem
Let \(X_n\) and \(X\) be random vectors in \(\mathbb{R}^k\), and let \(g:\mathbb{R}^k\to\mathbb{R}^m\) be Borel measurable. The discontinuity set \(D_g\) consists of all points at which \(g\) is not continuous. Measurability ensures that \(g(X_n)\) and \(g(X)\) are random vectors.
Here is a proof that makes the role of the null set precise. We use two consequences of convergence in distribution. First, for any closed set \(C\), the Portmanteau inequality gives \( \limsup_{n\to\infty}P(X_n\in C)\leq P(X\in C) \). For an open set \(G\), the corresponding inequality is \( P(X\in G)\leq\liminf_{n\to\infty}P(X_n\in G) \). Consequently, if a Borel set \(A\) has \(P(X\in\partial A)=0\), where \(\partial A\) is its boundary, then \(P(X_n\in A)\to P(X\in A)\): apply the open-set inequality to the interior of \(A\), the closed-set inequality to its closure, and use \(P(X\in\operatorname{int} A)=P(X\in\overline A)=P(X\in A)\).
For completeness, the closed-set inequality follows directly from bounded-continuous-function convergence. For closed \(C\), define \(u_r(x)=\max\{0,1-r\,d(x,C)\}\), where \(d(x,C)\) is the distance from \(x\) to \(C\). Each \(u_r\) is bounded and continuous, and \(1_C\leq u_r\). Thus \( \limsup_n P(X_n\in C)\leq E[u_r(X)] \). As \(r\to\infty\), \(u_r(X)\to 1_C(X)\); bounded convergence gives \(E[u_r(X)]\to P(X\in C)\). The open-set inequality follows by taking complements. This establishes the set-probability facts used below.
To prove the lemma, first suppose \(0\leq q\leq M\). At any point where \(q\) is continuous and \(q(x)\neq t\), membership in \(\{q>t\}\) is locally constant. Therefore \( \partial\{q>t\}\subseteq D_q\cup\{q=t\} \). There are at most countably many \(t\) for which \(P(q(X)=t)>0\): for each positive integer \(j\), there can be only finitely many disjoint level sets with probability at least \(1/j\). Hence, for almost every \(t\in[0,M]\), the boundary of \(\{q>t\}\) has probability zero under \(X\). The set-probability result above gives \(P(q(X_n)>t)\to P(q(X)>t)\) for almost every such \(t\). The tail-integral identity, obtained by integrating \(1_{\{q>t\}}\) over \(t\), gives \( E[q(X_n)]=\int_0^M P(q(X_n)>t)\,dt \). The integrands are bounded by 1, so dominated convergence yields \(E[q(X_n)]\to E[q(X)]\). For a general bounded real-valued \(q\), add a constant to make it nonnegative; subtracting the same constant from both expectations proves the claim. \(\square\)
Now let \(h:\mathbb{R}^m\to\mathbb{R}\) be any bounded continuous function and set \(q=h\circ g\). Whenever \(g\) is continuous at \(x\), \(h\circ g\) is continuous at \(x\); thus \(D_q\subseteq D_g\), and \(P(X\in D_q)=0\). The lemma gives \(E[h(g(X_n))]\to E[h(g(X))]\). This holds for every bounded continuous \(h\), which is the test-function definition of \(g(X_n)\xrightarrow{d}g(X)\). \(\square\)
Null Boundaries and Event Probabilities
A particularly useful case is an indicator map. For a Borel set \(A\subseteq\mathbb{R}^k\), the map \(g(x)=1_A(x)\) is continuous at every point outside \(\partial A\). It follows that a null boundary is enough to make the indicator transformation preserve convergence.
The last conclusion also follows by applying the continuity-set result directly. The mapping formulation adds a useful interpretation: the event indicator is a discontinuous transformation, and its only possible discontinuities lie at the boundary. This is why convergence of probabilities is often stated for sets whose boundaries have zero limiting probability.
Worked Example: A Threshold Indicator
Let \(X_n\sim N(0,1+1/n)\), so \(X_n\xrightarrow{d}X\) for \(X\sim N(0,1)\). Consider \(g(x)=1_{\{x>1\}}\). Its discontinuity set is \(D_g=\{1\}\), and \(P(X=1)=0\) because the normal distribution has a density. The theorem therefore gives \(g(X_n)\xrightarrow{d}g(X)\).
Each transformed variable takes only the values 0 and 1. Its probability of equalling 1 is \( P(g(X_n)=1)=P(X_n>1)=1-\Phi\bigl(1/\sqrt{1+1/n}\bigr) \), where \(\Phi\) is the standard normal distribution function. Since \(1/\sqrt{1+1/n}\to1\) and \(\Phi\) is continuous, \( P(g(X_n)=1)\to1-\Phi(1)=P(X>1) \). Thus the Bernoulli probabilities converge, confirming the distributional conclusion by direct calculation.
Worked Example: Rounding Down a Normal Limit
Use the same sequence \(X_n\sim N(0,1+1/n)\), with limit \(X\sim N(0,1)\), and define \(g(x)=\lfloor x\rfloor\), the greatest integer less than or equal to \(x\). The floor function is discontinuous at every integer and continuous elsewhere, so \(D_g=\mathbb{Z}\). A continuous normal distribution assigns probability zero to this countable set: \(P(X\in\mathbb{Z})=0\). Hence \(\lfloor X_n\rfloor\xrightarrow{d}\lfloor X\rfloor\).
For a direct check of one output probability, \(\lfloor X_n\rfloor=0\) exactly when \(0\leq X_n<1\). Writing \(\sigma_n=\sqrt{1+1/n}\), symmetry and the normal CDF give \( P(\lfloor X_n\rfloor=0)=P(0\leq X_n<1)=\Phi(1/\sigma_n)-\Phi(0) \). As \(n\to\infty\), this converges to \(\Phi(1)-1/2\), which equals \(P(0\leq X<1)=P(\lfloor X\rfloor=0)\). The same null-discontinuity argument applies to the other integer thresholds as well; checking one point probability illustrates the conclusion without replacing the theorem's general justification.
Worked Example: The Direction of a Random Vector
Let \(X_n\sim N_2((1+1/n,0),I_2)\), where \(I_2\) is the \(2\times2\) identity matrix. These vectors converge in distribution to \(X\sim N_2((1,0),I_2)\). Define \(g(x)=x/\lVert x\rVert\) for \(x\neq0\), and set \(g(0)=(1,0)\). The map is continuous away from the origin, but is discontinuous there: approaching zero along different directions gives different unit vectors.
The limit \(X\) has a density on \(\mathbb{R}^2\), so a singleton has probability zero and \(P(X=0)=0\). Therefore \(P(X\in D_g)=0\), and the theorem yields \( X_n/\lVert X_n\rVert\xrightarrow{d}X/\lVert X\rVert \). The arbitrary definition of \(g(0)\) has no effect on the limiting distribution because the limit vector equals zero with probability zero. The same reasoning would not justify the conclusion if the limit distribution placed positive probability at the origin.
When the Condition Fails
A positive probability at a discontinuity does not automatically mean that convergence fails, but the theorem no longer guarantees it. In fact, failure can occur even for a deterministic sequence. Let \(X_n=-1/n\), so \(X_n\xrightarrow{d}0\), and take \(g(x)=1_{\{x\geq0\}}\). This map is discontinuous at 0, and the limiting variable equals 0 with probability 1. Yet \(g(X_n)=0\) for every \(n\), whereas \(g(0)=1\). Thus \(g(X_n)\) does not converge in distribution to \(g(0)\).
The strict or non-strict inequality in a threshold can therefore matter when the limit has an atom at the cutoff. For example, \(1_{\{x>c\}}\) and \(1_{\{x\geq c\}}\) have different values at \(c\). If \(P(X=c)>0\), the limiting transformed value depends on that convention, and small changes in \(X_n\) near \(c\) may produce a different limit.
Common Mistake and Practical Use
- Mistake: “\(g\) is discontinuous somewhere, so the mapping theorem cannot be used.” Correction: Identify \(D_g\) and check \(P(X\in D_g)\). A map may be discontinuous on a large set and still satisfy the theorem if that set is null under the limit.
- Mistake: Checking whether \(P(X_n\in D_g)=0\) for each \(n\). Correction: The condition concerns \(X\), the limiting random variable. The approximating distributions may charge \(D_g\).
- Mistake: Treating every event probability as convergent under convergence in distribution. Correction: For \(P(X_n\in A)\to P(X\in A)\), check that \(P(X\in\partial A)=0\). A boundary atom can prevent convergence.
In practice, the check is short: write down the discontinuity set, determine whether the limit assigns it probability zero, and only then apply the transformation. For event probabilities, the relevant set is usually the boundary of the event. This step is especially important for thresholds, rounding rules, and transformations that normalize by a quantity that can be zero.
Check Your Understanding
Use the discontinuous mapping theorem and the boundary criterion where appropriate.
- For \(g(x)=1_{\{x>c\}}\), identify the discontinuity set and state the condition on \(X\) that ensures \(g(X_n)\xrightarrow{d}g(X)\).
- Suppose \(X_n\xrightarrow{d}X\) and \(P(X\in\mathbb{Z})=0\). What can be concluded about \(\lfloor X_n\rfloor\)?
- Why does \(P(X\in\partial A)=0\) imply \(P(X_n\in A)\to P(X\in A)\)?
- In the deterministic counterexample, identify the limiting mass at the discontinuity and explain why the transformed sequence has the wrong limit.
- A map is discontinuous on a set \(D\) of positive size in the ordinary geometric sense. Is that enough to rule out the theorem? Explain which probability must actually be checked.