Distribution Functions and Weak Convergence on the Line

Why Distribution Functions Help

Weak convergence is defined by testing probability laws against bounded continuous functions. That definition is flexible, but checking an entire class of functions can be inconvenient. On the real line, a probability law can also be described by a single nondecreasing function: its distribution function.

The key is that pointwise convergence of these functions is not required everywhere. The correct test points are the continuity points of the limiting distribution function. At a jump, an arbitrarily small shift of mass can change the value of a distribution function at that exact point, even when the probability laws converge weakly.

Central criterion: For probability measures on the real line, weak convergence is equivalent to convergence of distribution functions at every continuity point of the limiting distribution function.

Definitions

Definition: Let $\mu$ be a probability measure on the Borel subsets of $\mathbb{R}$. Its distribution function is the function $F:\mathbb{R}\to[0,1]$ defined, for every real number $x$, by $$ F(x)=\mu((-\infty,x]). $$ For probability measures $\mu_n$ with distribution functions $F_n$, the measures $\mu_n$ converge weakly to $\mu$, written $\mu_n\Rightarrow\mu$, if, for every bounded continuous function $f:\mathbb{R}\to\mathbb{R}$, $$ \int_{\mathbb{R}} f\,d\mu_n \longrightarrow \int_{\mathbb{R}} f\,d\mu. $$

The interval in the definition of $F(x)$ includes its right endpoint. Thus, if $\mu$ assigns positive probability to the single point $x$, then $F$ has a jump there. In particular, the jump size is $\mu(\{x\})$.

The Distribution-Function Criterion

Theorem: Let $\mu_n$ and $\mu$ be probability measures on $\mathbb{R}$, with distribution functions $F_n$ and $F$, respectively. Then $$ \mu_n\Rightarrow\mu \quad\Longleftrightarrow\quad F_n(x)\longrightarrow F(x) \text{ for every continuity point }x\text{ of }F. $$

Both directions matter. Weak convergence controls the probability of a half-line when its endpoint carries no mass under the limiting law. Conversely, enough half-line probabilities determine the probabilities of bounded intervals, and these can be used to approximate the integrals defining weak convergence.

Proof: Weak Convergence Gives Convergence at Continuity Points

Proof. Assume $\mu_n\Rightarrow\mu$, and fix a continuity point $x$ of $F$. For $\delta>0$, define continuous functions $g^-_\delta,g^+_\delta:\mathbb{R}\to[0,1]$ as follows. 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 $[x+\delta,\infty)$. Then $$ g^-_\delta(t)\leq \mathbf{1}_{(-\infty,x]}(t)\leq g^+_\delta(t) \qquad\text{for every }t\in\mathbb{R}. $$ Integrating with respect to $\mu_n$ gives $$ \int g^-_\delta\,d\mu_n\leq F_n(x)\leq\int g^+_\delta\,d\mu_n. $$ Both ramp functions are bounded and continuous, so weak convergence implies convergence of their integrals. Consequently, $$ \int g^-_\delta\,d\mu \leq \liminf_{n\to\infty}F_n(x) \leq \limsup_{n\to\infty}F_n(x) \leq \int g^+_\delta\,d\mu. $$ The lower integral is at least $F(x-\delta)$, since $g^-_\delta=1$ on $(-\infty,x-\delta]$. The upper integral is at most $F(x+\delta)$, since $g^+_\delta=0$ on $[x+\delta,\infty)$. Therefore $$ F(x-\delta)\leq \liminf_{n\to\infty}F_n(x) \leq \limsup_{n\to\infty}F_n(x)\leq F(x+\delta). $$ As $\delta$ decreases to zero, continuity of $F$ at $x$ makes both outer terms tend to $F(x)$. Thus $F_n(x)\to F(x)$, as required.

$\square$

Proof: Convergence at Continuity Points Gives Weak Convergence

Proof. Assume $F_n(x)\to F(x)$ at every continuity point $x$ of $F$. We prove convergence of integrals against an arbitrary bounded continuous function $f:\mathbb{R}\to\mathbb{R}$. Write $M=\sup_{t\in\mathbb{R}}|f(t)|$. If $M=0$, all the integrals are zero, so suppose $M>0$. Fix $\eta>0$. Since $F(x)\to0$ as $x\to-\infty$, $F(x)\to1$ as $x\to\infty$, and the continuity points of a distribution function are dense, we can choose continuity points $a<b$ such that $$ F(a)<\eta \qquad\text{and}\qquad 1-F(b)<\eta. $$ The density fact follows because a distribution function is nondecreasing and has at most countably many jumps: for each positive integer $k$, there are only finitely many jumps of size at least $1/k$. In particular, continuity points can be chosen arbitrarily far into either tail. On the compact interval $[a,b]$, $f$ is uniformly continuous. Choose a finite partition $$ a=x_0<x_1<\cdots<x_m=b $$ whose endpoints are continuity points of $F$ and whose interval lengths are small enough that the oscillation of $f$ on each $[x_{j-1},x_j]$ is at most $\varepsilon$. Such a partition exists because continuity points are dense. For each $j$, choose $t_j\in[x_{j-1},x_j]$. The mass of each partition interval converges, since $$ \mu_n((x_{j-1},x_j]) =F_n(x_j)-F_n(x_{j-1}) \longrightarrow F(x_j)-F(x_{j-1}) =\mu((x_{j-1},x_j]). $$ Therefore the finite sums $$ \sum_{j=1}^{m} f(t_j)\mu_n((x_{j-1},x_j]) \longrightarrow \sum_{j=1}^{m} f(t_j)\mu((x_{j-1},x_j]). $$ For either measure, replacing $f$ on each partition interval by $f(t_j)$ changes the integral over $(a,b]$ by at most $\varepsilon$, because the oscillation is at most $\varepsilon$ and the total mass is at most $1$. It remains to control the tails. For $\mu$, their total mass is $$ \mu((-\infty,a]\cup(b,\infty))=F(a)+1-F(b)<2\eta. $$ For all sufficiently large $n$, convergence at $a$ and $b$ gives $F_n(a)<2\eta$ and $1-F_n(b)<2\eta$, so the corresponding tail mass under $\mu_n$ is less than $4\eta$. The absolute value of the integral over any set is at most $M$ times its probability. Combining the tail and partition estimates yields $$ \limsup_{n\to\infty} \left|\int f\,d\mu_n-\int f\,d\mu\right| \leq 2\varepsilon+6M\eta. $$ Both $\varepsilon$ and $\eta$ can be chosen arbitrarily small. Hence the limsup is zero, and $\int f\,d\mu_n\to\int f\,d\mu$. This holds for every bounded continuous $f$, so $\mu_n\Rightarrow\mu$.

$\square$

Worked Examples

Worked Example: Gaussian Laws with Shifting Means

Let $\mu_n$ be the normal law with mean $1/n$ and variance $1$, and let $\mu$ be the normal law with mean $0$ and variance $1$. Write $\Phi$ for the standard normal distribution function. For every real $x$, $$ F_n(x)=\Phi\left(x-\frac{1}{n}\right), \qquad F(x)=\Phi(x). $$ The function $\Phi$ is continuous, so for every $x$, continuity of $\Phi$ gives $$ F_n(x)=\Phi\left(x-\frac{1}{n}\right)\longrightarrow\Phi(x)=F(x). $$ The distribution-function criterion therefore yields $\mu_n\Rightarrow\mu$. In this case every point is a permitted test point because the limiting distribution has no jumps.

Worked Example: Uniform Laws Shrinking to a Point Mass

Let $\mu_n$ be the uniform law on $[-1/n,1/n]$, and let $\mu=\delta_0$, the probability measure concentrated at $0$. The limiting distribution function is $$ F(x)= \begin{cases} 0,&x<0,\\ 1,&x\geq 0. \end{cases} $$ It is continuous at every $x\ne0$. Fix $x<0$. For all sufficiently large $n$, $x<-1/n$, and hence $F_n(x)=0=F(x)$. Fix $x>0$. For all sufficiently large $n$, $x>1/n$, and hence $F_n(x)=1=F(x)$. Thus $F_n(x)\to F(x)$ at every continuity point of $F$, so $\mu_n\Rightarrow\delta_0$. At the discontinuity $x=0$, however, $F_n(0)=1/2$ for every $n$, while $F(0)=1$. This does not contradict the criterion: convergence at $0$ is not required. The example shows how probability can concentrate near an atom while the distribution-function values at the atom itself fail to converge.

Worked Example: Bernoulli Laws with Varying Parameters

Let $\mu_n$ assign probability $p_n$ to $1$ and probability $1-p_n$ to $0$, where $p_n\to p\in[0,1]$. Let $\mu$ be the corresponding Bernoulli law with parameter $p$. Their distribution functions satisfy $$ F_n(x)= \begin{cases} 0,&x<0,\\ 1-p_n,&0\leq x<1,\\ 1,&x\geq1, \end{cases} \qquad F(x)= \begin{cases} 0,&x<0,\\ 1-p,&0\leq x<1,\\ 1,&x\geq1. \end{cases} $$ For $x<0$ and $x\geq1$, $F_n(x)=F(x)$. For $0\leq x<1$, $F_n(x)\to F(x)$ because $p_n\to p$. Thus convergence holds even at the possible jump points $0$ and $1$, and in particular at every continuity point. The criterion gives $\mu_n\Rightarrow\mu$. Directly, for any bounded continuous $f$, $$ \int f\,d\mu_n=(1-p_n)f(0)+p_nf(1) \longrightarrow (1-p)f(0)+pf(1)=\int f\,d\mu, $$ which verifies the conclusion from the definition as well.

How to Use the Criterion

When applying the theorem, first identify the proposed limiting distribution function and its continuity points. Then calculate or estimate $F_n(x)$ for a fixed point $x$ in each relevant region. At a jump of the limit, do not impose a convergence requirement unless a separate reason makes it useful.

Find the candidate limit.
Write its distribution function $F$, including the values at any jumps.

Locate continuity points.
Identify where $F$ is continuous; these, rather than all real numbers, are the required test points.

Check pointwise convergence there.
Show $F_n(x)\to F(x)$ for each continuity point $x$, then invoke the criterion to conclude weak convergence.

A common mistake is to demand convergence at every $x$. As the shrinking-uniform example shows, that stronger condition can fail even when weak convergence holds. The reverse mistake is to check only a few convenient points: the theorem requires convergence at every continuity point of the limit. In applications, formulas often divide the line into regions, as in the Bernoulli example, so one argument per region can establish the full condition.

This criterion is especially useful on the real line because it replaces a family of integral tests with a pointwise statement about distribution functions. Its continuity-point qualification records exactly where the limiting law has no atom at the endpoint being tested.

Check Your Understanding

Use the definitions and criterion to answer these questions.

What event does the distribution-function value $F(x)$ measure?
Why does weak convergence imply convergence of $F_n(x)$ at a continuity point of $F$?
In the shrinking-uniform example, why is convergence at $x=0$ not needed?
For a proposed limit with a jump at $c$, which points must be checked by the criterion?
Why does convergence of $F_n$ at all continuity points suffice to establish convergence of integrals against every bounded continuous function?