When Pointwise Limits of CDFs Define a CDF

What Can Go Wrong With a Pointwise Limit?

A sequence of distribution functions may converge at every real number without its limit being a distribution function. The issue is that the CDF conditions include behavior at the two tails and right-continuity, and pointwise convergence alone does not preserve all of these. Monotonicity, in contrast, is preserved. A careful check separates the properties that follow automatically from those that must be verified.

Recall that the CDF of a real-valued random variable $X$ is $F(x)=P(X\le x)$. It is nondecreasing and right-continuous, with limits $0$ as $x\to-\infty$ and $1$ as $x\to+\infty$. These conditions are not merely convenient: together they characterize CDFs.

Definition: A function $F:\mathbb{R}\to[0,1]$ is a CDF if and only if it is nondecreasing, right-continuous, and satisfies $$ \lim_{x\to-\infty}F(x)=0 \qquad\text{and}\qquad \lim_{x\to+\infty}F(x)=1. $$

Properties Retained by Pointwise Limits

Suppose $F_n$ is a CDF for each positive integer $n$, and $F_n(x)\to F(x)$ for every real $x$. Since each $F_n(x)$ lies in $[0,1]$, so does $F(x)$. If $x\le y$, then $F_n(x)\le F_n(y)$ for every $n$; taking limits gives $F(x)\le F(y)$. Thus the pointwise limit is nondecreasing.

Result: A pointwise limit of CDFs takes values in $[0,1]$ and is nondecreasing. Since it is bounded and nondecreasing, it has limits at both tails. However, those limits need not be $0$ and $1$, and the limit need not be right-continuous.

The existence of tail limits follows from bounded monotonicity, but their required values do not. Also, right-continuity concerns the values just to the right of each point, whereas pointwise convergence checks each fixed point separately. These distinctions account for the two main ways the pointwise limit can fail to be a CDF.

Worked Example: A Pointwise Limit That Is a CDF

Let $X_n$ take the value $0$ with probability $p_n$ and the value $1$ with probability $1-p_n$, where $p_n\to p$ and $0\le p\le1$. Find the pointwise limit of the CDFs and check its properties.

For each $n$, the CDF is $$ F_n(x)= \begin{cases} 0, & x<0,\\ p_n, & 0\le x<1,\\ 1, & x\ge 1. \end{cases} $$

Taking the limit at each fixed $x$ gives $$ F(x)= \begin{cases} 0, & x<0,\\ p, & 0\le x<1,\\ 1, & x\ge 1. \end{cases} $$ This function is nondecreasing because $0\le p\le1$. It is right-continuous at $0$, where both $F(0)$ and the limit from the right equal $p$, and at $1$, where both $F(1)$ and the limit from the right equal $1$. It is constant, hence right-continuous, elsewhere. Its left and right tail limits are $0$ and $1$. Therefore $F$ is a CDF, including when $p=0$ or $p=1$.

The CDF Characterization and Its Justification

The criterion above includes a sufficiency claim: any nondecreasing, right-continuous function with the correct tail limits really is the CDF of some random variable. The following construction verifies that claim rather than assuming a distribution exists.

Theorem: Let $F:\mathbb{R}\to[0,1]$ be nondecreasing and right-continuous, with limits $0$ and $1$ at $-\infty$ and $+\infty$, respectively. Then $F$ is the CDF of a real-valued random variable. Consequently, a pointwise limit of CDFs is a CDF exactly when it is right-continuous and has these two tail limits.

Proof. Work on the probability space $(0,1)$ with the uniform distribution, and let $U$ be its coordinate random variable. For $0<u<1$, define $$ Q(u)=\inf\{x\in\mathbb{R}:F(x)\ge u\}. $$ The set in this definition is nonempty because $F(x)\to1$ as $x\to+\infty$. It is bounded below because $F(x)\to0$ as $x\to-\infty$. Thus $Q(u)$ is finite.

For each real $x$, we claim that $Q(u)\le x$ if and only if $u\le F(x)$. If $u\le F(x)$, then $x$ belongs to the set defining $Q(u)$, so $Q(u)\le x$. Conversely, suppose $Q(u)\le x$. By the definition of an infimum, for each positive integer $k$ there is a point $y_k$ in that set with $y_k<Q(u)+1/k\le x+1/k$. Monotonicity gives $F(x+1/k)\ge F(y_k)\ge u$. Right-continuity at $x$ implies $F(x+1/k)\to F(x)$, so $F(x)\ge u$. This proves the claim.

The sublevel sets of $Q$ satisfy $$ \{u\in(0,1):Q(u)\le x\}=\{u\in(0,1):u\le F(x)\}. $$ They are measurable intervals, so $Q$ is measurable. The claim also gives, for every real $x$, $$ P(Q(U)\le x)=P(U\le F(x))=F(x), $$ because $U$ is uniform on $(0,1)$ and $F(x)\in[0,1]$. Hence $F$ is the CDF of $Q(U)$. Necessity of the stated properties is part of the defining properties of a CDF. For a pointwise limit of CDFs, nondecreasingness and the range $[0,1]$ already follow from the preceding argument, so right-continuity and the two tail limits complete the test. $\square$

Examples of Failed Conditions

Each of the following sequences consists of genuine CDFs and converges at every real $x$. In each case, the pointwise limit retains monotonicity but fails one of the remaining CDF requirements.

Worked Example: Mass Escaping to Positive Infinity

Let $X_n=n$ with probability one. Its CDF is $F_n(x)=0$ for $x<n$ and $F_n(x)=1$ for $x\ge n$. Fix any real $x$. For all integers $n>x$, $F_n(x)=0$; consequently the pointwise limit is $F(x)=0$ for every finite $x$.

The limit is nondecreasing and right-continuous, and its left tail limit is $0$. But its right tail limit is also $0$, not $1$: $$ \lim_{x\to+\infty}F(x)=0. $$ Thus it is not a CDF. At each finite $n$, the entire probability mass is located at $n$; as $n$ increases, that mass leaves every fixed bounded region. Pointwise evaluation at finite $x$ no longer detects it.

Worked Example: Mass Escaping to Negative Infinity

Let $Y_n=-n$ with probability one. Then $G_n(x)=0$ for $x<-n$ and $G_n(x)=1$ for $x\ge -n$. For any fixed real $x$, eventually $-n\le x$, so $G_n(x)\to1$. The pointwise limit is $G(x)=1$ for all $x$.

This limit is nondecreasing and right-continuous, but $$ \lim_{x\to-\infty}G(x)=1, $$ rather than $0$. It is not a CDF. Here the unit mass moves left beyond every fixed finite threshold, so the limiting function incorrectly assigns probability one below every such threshold.

Worked Example: Failure of Right-Continuity

Let $Z_n=1/n$ with probability one. Its CDF is $H_n(x)=0$ for $x<1/n$ and $H_n(x)=1$ for $x\ge1/n$. If $x<0$, then $H_n(x)=0$ for every $n$. At $x=0$, $H_n(0)=0$ for every $n$. If $x>0$, then $1/n\le x$ for all sufficiently large $n$, so $H_n(x)\to1$. Therefore $$ H(x)= \begin{cases} 0, & x\le0,\\ 1, & x>0. \end{cases} $$

The tail limits of $H$ are correct, and $H$ is nondecreasing. But $H(0)=0$, while $\lim_{x\downarrow0,\ x>0}H(x)=1$. Thus $H$ is not right-continuous at $0$, and is not a CDF. The discontinuity has the wrong value at the boundary: a CDF with a jump at $0$ must take its value after the jump at $0$, not before it.

A Practical Validity Check

When a pointwise limit $F$ is proposed, check its properties in a useful order. First, monotonicity and the range $[0,1]$ follow if it is truly the pointwise limit of CDFs; still, a calculation error can make them worth checking explicitly. Then verify right-continuity at every point and compute both tail limits. Because $F$ is nondecreasing and bounded, the tail limits exist, but the values must be checked.

Confirm pointwise convergence.
For each fixed real $x$, find the limit of $F_n(x)$; do not let $x$ vary with $n$.

Check order and range.
A genuine pointwise limit of CDFs must be nondecreasing and take values in $[0,1]$.

Check right-continuity and tails.
Verify $\lim_{y\downarrow x}F(y)=F(x)$ at each $x$, and verify the limits at $-\infty$ and $+\infty$ are $0$ and $1$.

A common mistake is to argue, “Every $F_n$ is a CDF, so its pointwise limit must be a CDF.” The examples show why this is false: pointwise convergence at each finite argument can miss probability mass moving toward either infinity, and can also produce a value at a jump that violates right-continuity. The correction is to invoke the CDF characterization and check all of its conditions. In particular, do not confuse convergence at every real point with convergence only at the continuity points of a candidate limit, as used in the definition of convergence in distribution.

Key takeaway: Pointwise limits preserve the range and monotonicity of CDFs, but not necessarily right-continuity or the required tail values. A pointwise limit is a CDF precisely when the right-continuity and tail conditions also hold.

Check Your Understanding

Use the CDF characterization to answer each question.

Why must a pointwise limit of nondecreasing functions remain nondecreasing?
For $X_n=n$ with probability one, what are the two tail limits of the pointwise limiting function, and which one fails the CDF condition?
For $Z_n=1/n$ with probability one, what is the value of the limiting function at $0$, and what is its right-hand limit there?
Can a pointwise limit of CDFs fail to have a limit as $x\to+\infty$? Explain using boundedness and monotonicity.
State the conditions that remain to be checked after establishing that a pointwise limit of CDFs is nondecreasing and takes values in $[0,1]$.