The Continuity-Point Criterion for CDF Convergence

Why Continuity Points Are the Right Test

A cumulative distribution function is monotone, but it need not be continuous: a jump records positive probability at a point. Consequently, convergence of CDF values at every real number is stronger than is needed for convergence in distribution. The natural test points are the continuity points of the limiting CDF. At those points, small changes in the threshold cannot expose a persistent discrepancy in probability.

We will prove both directions of this criterion. The first uses continuous functions that bound the indicator of a half-line. The converse uses monotonicity to control probabilities on short intervals, then approximates a bounded continuous function over a large, finite range.

Definition: Let \(X_n\) and \(X\) be real-valued random variables, with CDFs \(F_n\) and \(F\). We use the standard weak-convergence formulation of convergence in distribution: \(X_n\xrightarrow{d}X\) means that \(E[g(X_n)]\to E[g(X)]\) for every bounded continuous function \(g:\mathbb{R}\to\mathbb{R}\). A continuity point of \(F\) is a real number \(x\) at which \(F\) is continuous.

The Continuity-Point Criterion

Theorem: For real-valued random variables \(X_n\) and \(X\), with CDFs \(F_n\) and \(F\),

\(X_n\xrightarrow{d}X\) if and only if \(F_n(x)\to F(x)\) for every continuity point \(x\) of \(F\).

A CDF is nondecreasing, is right-continuous, and has limits \(0\) and \(1\) at negative and positive infinity. We use these standard properties in the proof. We also need that the continuity points of a CDF are dense in \(\mathbb{R}\). To see why, every discontinuity of a CDF is a positive jump. For each positive integer \(k\), there can be only finitely many jumps of size at least \(1/k\), since the sum of disjoint jumps is at most \(1\). Every positive jump is at least \(1/k\) for some \(k\), so there are at most countably many discontinuities. A countable set cannot contain any open interval; therefore continuity points are dense.

Proof: Distributional Convergence Gives CDF Convergence

Suppose \(X_n\xrightarrow{d}X\), and let \(x\) be a continuity point of \(F\). For a fixed \(\delta>0\), define two continuous functions. Let \(g^-_\delta\) equal \(1\) on \((-\infty,x-\delta]\), decrease linearly from \(1\) to \(0\) on \([x-\delta,x]\), and equal \(0\) on \([x,\infty)\). Let \(g^+_\delta\) equal \(1\) on \((-\infty,x]\), decrease linearly from \(1\) to \(0\) on \([x,x+\delta]\), and equal \(0\) on \([x+\delta,\infty)\). Both functions are bounded and continuous, and for every real \(t\),

g^-_\delta(t)\leq \mathbf{1}\{t\leq x\}\leq g^+_\delta(t).

Taking expectations under \(X_n\), and using convergence of expectations for bounded continuous functions, gives

E[g^-_\delta(X)]\leq \liminf_{n\to\infty}F_n(x) \leq \limsup_{n\to\infty}F_n(x)\leq E[g^+_\delta(X)].

As \(\delta\) decreases to zero, \(g^-_\delta(t)\) converges to \(\mathbf{1}\{t<x\}\), while \(g^+_\delta(t)\) converges to \(\mathbf{1}\{t\leq x\}\). Both sequences are bounded by \(1\), so the dominated convergence theorem applies. Continuity of \(F\) at \(x\) means \(P(X=x)=0\), and hence \(P(X<x)=P(X\leq x)=F(x)\). The lower and upper bounds therefore both tend to \(F(x)\), proving \(F_n(x)\to F(x)\). \(\square\)

Worked Example: Shifting Point Masses

Let \(X_n=1+1/n\) with probability one, and let \(X=1\) with probability one. Their CDFs are \(F_n(x)=\mathbf{1}\{x\geq 1+1/n\}\) and \(F(x)=\mathbf{1}\{x\geq 1\}\). The only discontinuity point of \(F\) is \(1\).

If \(x<1\), then \(F_n(x)=F(x)=0\) for every \(n\). If \(x>1\), then \(1+1/n\leq x\) for all sufficiently large \(n\), so \(F_n(x)=F(x)=1\) eventually. Thus the CDFs converge at every continuity point of \(F\).

At \(x=1\), however, \(F_n(1)=0\) for every \(n\), while \(F(1)=1\). This does not contradict the criterion: \(1\) is not a continuity point of \(F\). Directly, for every bounded continuous \(g\), \(E[g(X_n)]=g(1+1/n)\to g(1)=E[g(X)]\), confirming \(X_n\xrightarrow{d}X\).

Proof: CDF Convergence Gives Distributional Convergence

Now suppose \(F_n(x)\to F(x)\) at every continuity point \(x\) of \(F\). We prove convergence of expectations for an arbitrary bounded continuous \(g\). Write \(M=\sup_{t\in\mathbb{R}}|g(t)|\). If \(M=0\), the claim is immediate, so assume \(M>0\).

Fix \(\varepsilon>0\). Since \(F\) tends to \(0\) and \(1\) in the two tails, and its continuity points are dense, choose continuity points \(a<b\) such that

P(X\leq a)<\eta,\qquad P(X>b)<\eta,

where \(\eta>0\) will be chosen small enough that \(6M\eta<\varepsilon/2\). By the assumed CDF convergence at \(a\) and \(b\), for all sufficiently large \(n\),

P(X_n\leq a)<2\eta,\qquad P(X_n>b)<2\eta.

The function \(g\) is uniformly continuous on the compact interval \([a,b]\). Choose a partition \(a=x_0<x_1<\cdots<x_m=b\), with every \(x_j\) a continuity point of \(F\), and with the largest interval width small enough that the variation of \(g\) within any interval is less than a number \(\rho>0\), chosen so \(2\rho<\varepsilon/2\). Such a partition exists because continuity points are dense: a sufficiently fine finite grid can have each interior grid point perturbed slightly to a continuity point while keeping all widths small.

Approximate \(g\) on each interval \((x_{j-1},x_j]\) by the constant \(g(x_j)\). The probability of that interval under \(X_n\) is \(F_n(x_j)-F_n(x_{j-1})\), which converges to \(F(x_j)-F(x_{j-1})\). Thus the finite sums

S_n=\sum_{j=1}^{m}g(x_j)\bigl[F_n(x_j)-F_n(x_{j-1})\bigr] \quad\text{converge to}\quad S=\sum_{j=1}^{m}g(x_j)\bigl[F(x_j)-F(x_{j-1})\bigr].

The sum \(S_n\) approximates \(E[g(X_n)\mathbf{1}\{a<X_n\leq b\}]\) with error at most \(\rho\), because the pointwise approximation error on each interval is less than \(\rho\). The analogous error for \(X\) is also at most \(\rho\). Outside \((a,b]\), the absolute contribution to each expectation is bounded by \(M\) times the probability of being outside. Consequently, for all sufficiently large \(n\),

|E[g(X_n)]-E[g(X)]| \leq 2M\eta+4M\eta+2\rho+|S_n-S|.

The first term bounds the two tails for \(X\); the second bounds the two tails for \(X_n\). By choice, \(6M\eta+2\rho<\varepsilon\), and \(S_n\to S\), so eventually the right-hand side is less than \(\varepsilon\). Since \(\varepsilon\) was arbitrary, \(E[g(X_n)]\to E[g(X)]\). This holds for every bounded continuous \(g\), proving \(X_n\xrightarrow{d}X\). \(\square\)

Worked Example: Expanding Uniform Supports

Let \(X\sim\operatorname{Uniform}(0,1)\), and define \(X_n=(1+1/n)X\). Thus \(X_n\) is uniform on \((0,1+1/n)\). For \(c_n=1+1/n\), its CDF is

F_n(x)= \begin{cases} 0,&x\leq 0,\\ x/c_n,&0<x<c_n,\\ 1,&x\geq c_n. \end{cases}

The limiting CDF is \(F(x)=0\) for \(x\leq0\), \(F(x)=x\) for \(0<x<1\), and \(F(x)=1\) for \(x\geq1\). This CDF is continuous everywhere. For \(0<x<1\), eventually \(x<c_n\), and \(F_n(x)=x/c_n\to x=F(x)\). At \(x=0\), both CDFs equal zero. At \(x=1\), \(F_n(1)=1/c_n=n/(n+1)\to1=F(1)\); for \(x>1\), \(F_n(x)=1\) for all sufficiently large \(n\). For \(x<0\), both are zero. Therefore the CDFs converge at every continuity point, and the theorem gives \(X_n\xrightarrow{d}X\).

Worked Example: Normal Distributions With Changing Variance

Let \(Z\sim N(0,1)\), set \(X_n=\sqrt{1+1/n}\,Z\), and let \(X=Z\). Then \(X_n\sim N(0,1+1/n)\). If \(\Phi\) denotes the CDF of \(N(0,1)\), scaling gives

F_n(x)=P\!\left(Z\leq \frac{x}{\sqrt{1+1/n}}\right) =\Phi\!\left(\frac{x}{\sqrt{1+1/n}}\right).

For each real \(x\), \(x/\sqrt{1+1/n}\to x\). The function \(\Phi\) is continuous, so \(F_n(x)\to\Phi(x)=F(x)\) at every real \(x\). In particular, convergence holds at all continuity points of the limiting CDF, and the criterion yields \(X_n\xrightarrow{d}X\). Here the CDF calculation establishes convergence without requiring a separate calculation for each bounded continuous test function.

Common Mistake and Why the Criterion Matters

A frequent mistake is to require \(F_n(x)\to F(x)\) at every real \(x\). The point-mass example shows why this is incorrect: distributional convergence can hold even when convergence fails at a jump of \(F\). At a jump, the event \(\{X_n\leq x\}\) can change abruptly as the threshold is crossed, so the limiting probability at the jump itself need not be approximated. The theorem requires convergence at every continuity point, not at every point.

The converse proof explains why continuity points suffice. Convergence at the partition endpoints controls probabilities of the short intervals between them through differences of CDF values. Monotonicity is what makes those interval probabilities available; uniform continuity makes a bounded continuous function nearly constant on each short interval. The tails can be made negligible because CDFs approach \(0\) and \(1\). Together, these facts turn pointwise CDF control at suitable thresholds into convergence of expectations.

Key takeaway: For real-valued random variables, \(X_n\xrightarrow{d}X\) exactly when \(F_n(x)\to F(x)\) at every continuity point of \(F\). The continuity points are essential: convergence may fail at a jump without violating convergence in distribution.

Check Your Understanding

Use the criterion and the proof strategy to answer the following questions.

Why does continuity of \(F\) at \(x\) imply \(P(X=x)=0\), and where is that fact used in the first direction of the proof?
For \(X_n=1+1/n\) and \(X=1\), identify every continuity point of \(F\) and explain why failure at \(x=1\) does not disprove convergence in distribution.
In the converse proof, why must the partition endpoints be continuity points of \(F\)?
What role does uniform continuity of \(g\) on \([a,b]\) play in approximating its expectation by a finite sum?
Explain why the tails outside \((a,b]\) can be made small for both \(X\) and, eventually, \(X_n\).