Why CDFs Are Right-Continuous

Why the Value at a Threshold Matches the Limit from Its Right

A cumulative distribution function records the probability that a random variable falls at or below a threshold: \(F(x)=P(X\leq x)\). If a threshold moves down toward \(x\) while staying above it, the corresponding events shrink toward \(\{X\leq x\}\). This simple relationship between thresholds and events explains an important property of every CDF: it is right-continuous.

The direction matters. As a threshold decreases toward \(x\), values just above \(x\) are still included in \(\{X\leq t\}\). The limiting event is therefore \(\{X\leq x\}\), not an event that excludes equality at \(x\). We will make that argument precise using continuity of probability for decreasing events, then check the result in discrete examples where the CDF has steps.

Definition: A function \(F\) is right-continuous at \(x\) if \(\lim_{t\downarrow x}F(t)=F(x)\), where \(t\downarrow x\) means that \(t\) approaches \(x\) from values greater than \(x\). A cumulative distribution function of a real-valued random variable \(X\) is \(F(x)=P(X\leq x)\).

Continuity of Probability for Decreasing Events

We need a probability result before applying the definition. A sequence of events \(A_1,A_2,\ldots\) is decreasing if \(A_{n+1}\subseteq A_n\) for every \(n\). Its intersection consists of outcomes that belong to every event in the sequence. For such a sequence, probability of the events converges to the probability of that intersection.

Theorem: If \(A_1\supseteq A_2\supseteq\cdots\) are events in a probability space, then

\lim_{n\to\infty}P(A_n)=P\left(\bigcap_{n=1}^{\infty}A_n\right).

Here is a proof from countable additivity. Let \(A=\bigcap_{n=1}^{\infty}A_n\) and define \(C_n=A_1\setminus A_n\). Because the \(A_n\) decrease, the \(C_n\) increase, and \(C_n\subseteq A_1\). Also, \(\bigcup_{n=1}^{\infty}C_n=A_1\setminus A\). To see the last equality, an outcome in \(A_1\) fails to belong to \(A\) exactly when it fails to belong to at least one \(A_n\).

Set \(D_1=C_1\), and for \(n\geq2\) set \(D_n=C_n\setminus C_{n-1}\). These events are pairwise disjoint, and \(\bigcup_{j=1}^{n}D_j=C_n\). By countable additivity,

P(C_n)=\sum_{j=1}^{n}P(D_j) \quad\text{and}\quad P\left(\bigcup_{j=1}^{\infty}C_j\right)=\sum_{j=1}^{\infty}P(D_j).

The finite partial sums on the left converge to the infinite sum, so \(P(C_n)\to P(A_1\setminus A)\). Since \(C_n=A_1\setminus A_n\) and all these events are contained in \(A_1\), subtraction from the finite quantity \(P(A_1)\) gives

P(A_n)=P(A_1)-P(C_n) \longrightarrow P(A_1)-P(A_1\setminus A)=P(A).

This proves the theorem. The finiteness needed for the subtraction holds because probabilities are at most 1. \(\square\)

Applying the Theorem to Threshold Events

For a random variable \(X\), each set \(\{X\leq t\}\) is an event. If thresholds \(t_n\) decrease to \(x\), then these events decrease as well. Their intersection is exactly \(\{X\leq x\}\): an outcome with \(X\leq x\) meets every threshold, while if \(X>x\), some threshold sufficiently close to \(x\) lies below \(X\), so that outcome is not in the intersection. Continuity of probability therefore gives the key sequence result.

Theorem: Let \(X\) be a real-valued random variable and \(F(x)=P(X\leq x)\). If \(t_n\downarrow x\), then \(F(t_n)\to F(x)\). Consequently, \(F\) is right-continuous.

For the sequence statement, the events \(\{X\leq t_n\}\) decrease, and their intersection is \(\{X\leq x\}\). Applying continuity of probability proves \(F(t_n)\to F(x)\). To verify right-continuity for an arbitrary sequence \(y_n>x\) with \(y_n\to x\), we need to handle the possibility that the \(y_n\) do not decrease. Define the tail supremum

u_n=\sup\{y_k:k\geq n\}.

Each \(u_n\) is finite because the convergent sequence \((y_n)\) is bounded. The sequence \((u_n)\) is nonincreasing, \(u_n\geq y_n>x\), and \(u_n\to x\): for any \(\varepsilon>0\), eventually every \(y_k\) in the tail lies between \(x\) and \(x+\varepsilon\), so the supremum of that tail does too. The defining threshold events are nested whenever their thresholds are ordered. Thus

\{X\leq x\}\subseteq\{X\leq y_n\}\subseteq\{X\leq u_n\}, \qquad F(x)\leq F(y_n)\leq F(u_n).

By the decreasing-threshold result, \(F(u_n)\to F(x)\). The lower and upper bounds therefore squeeze \(F(y_n)\) to \(F(x)\). The sequential criterion for limits of real-valued functions now gives \(\lim_{t\downarrow x}F(t)=F(x)\). In particular, the upper bound is essential: knowing only that \(F(y_n)\geq F(x)\) would not show that \(F(y_n)\) converges to \(F(x)\). \(\square\)

Checking the Property in Discrete Examples

For discrete random variables, CDFs are constant between possible values and change at those values. At a step location, right-continuity says that the value at the location agrees with the values approached from just above it. The following calculations verify that agreement directly.

Worked Example: A Bernoulli Random Variable

Let \(X\sim\operatorname{Bernoulli}(p)\), with \(0<p<1\), so \(P(X=0)=1-p\) and \(P(X=1)=p\). From the definition \(F(t)=P(X\leq t)\),

F(t)= \begin{cases} 0, & t<0,\\ 1-p, & 0\leq t<1,\\ 1, & t\geq1. \end{cases}

At \(x=0\), every \(t\) with \(0<t<1\) has \(F(t)=1-p\). Therefore \(\lim_{t\downarrow0}F(t)=1-p=F(0)\). At \(x=1\), every \(t>1\) has \(F(t)=1\), so \(\lim_{t\downarrow1}F(t)=1=F(1)\). At a point strictly between 0 and 1, a sufficiently small move to the right stays in that interval, where \(F\) is constant. These checks agree with right-continuity at every real \(x\).

Worked Example: The Outcome of a Fair Die

Let \(X\) be the number shown by a fair six-sided die. Each of \(1,2,\ldots,6\) has probability \(1/6\). For \(k=1,\ldots,5\), if \(k\leq t<k+1\), the outcomes not exceeding \(t\) are \(1,\ldots,k\), so \(F(t)=k/6\). Also, \(F(t)=0\) for \(t<1\) and \(F(t)=1\) for \(t\geq6\).

Consider a step location \(k\in\{1,2,3,4,5\}\). Whenever \(k<t<k+1\), the CDF is \(F(t)=k/6\). Hence

\lim_{t\downarrow k}F(t)=\frac{k}{6}=F(k).

For instance, immediately to the right of \(k=3\), the CDF is \(3/6=1/2\), and \(F(3)=P(X\leq3)=3/6=1/2\). At the largest outcome, \(F(t)=1\) for every \(t\geq6\), so \(\lim_{t\downarrow6}F(t)=1=F(6)\). The calculation uses the values on the right of each step, not the values just below it.

Worked Example: A Three-Point Distribution

Suppose \(X\) takes the values \(-1\), \(2\), and \(4\), with probabilities \(1/3\), \(1/2\), and \(1/6\), respectively. These probabilities are valid because

\frac{1}{3}+\frac{1}{2}+\frac{1}{6} =\frac{2}{6}+\frac{3}{6}+\frac{1}{6}=1.

The CDF is \(0\) for \(t<-1\), \(1/3\) for \(-1\leq t<2\), \(5/6\) for \(2\leq t<4\), and \(1\) for \(t\geq4\). At \(x=2\), every threshold \(t\) just to the right of 2 but below 4 has \(F(t)=5/6\). At the threshold itself, \(F(2)=P(X\leq2)=1/3+1/2=5/6\). Thus \(\lim_{t\downarrow2}F(t)=5/6=F(2)\). At \(x=4\), the CDF is 1 both at 4 and for every threshold above 4, so \(\lim_{t\downarrow4}F(t)=1=F(4)\). The probability values on each interval and at each endpoint follow directly from the specified three possible outcomes.

Why Right-Continuity Matters

Right-continuity is not an additional assumption on a distribution. It follows from the definition of a CDF and the continuity property of probability. Its source is the way threshold events fit together: as a threshold decreases to \(x\) from above, the decreasing events \(\{X\leq t\}\) have intersection \(\{X\leq x\}\). The event at the limiting threshold still includes equality.

This property is useful when a CDF is described by a formula with different pieces on neighboring intervals. The value assigned at a boundary must agree with the limiting values from the right. The examples show this directly: at each step location, the CDF’s value is the value on the interval immediately to its right.

Common Mistake

A frequent proof error is to obtain only a lower bound and treat it as if it established convergence. For example, from \(x\leq y_n\leq u_n\), monotonicity gives \(F(x)\leq F(y_n)\leq F(u_n)\). The first inequality alone gives no upper control: a sequence can stay above \(F(x)\) and still fail to approach it. The second inequality matters because \(F(u_n)\to F(x)\). Together, the two bounds force \(F(y_n)\to F(x)\).

Another way to avoid losing that upper bound is to keep the event inclusions visible. A threshold \(y_n\) lies below the supremum \(u_n\) of its tail, so \(\{X\leq y_n\}\subseteq\{X\leq u_n\}\). Reversing this inclusion would reverse the probability bound and would not establish the squeeze used in the proof.

Key takeaway: If thresholds decrease to \(x\), their CDF events decrease to \(\{X\leq x\}\). Continuity of probability then gives the CDF value at \(x\) as the limit from the right.

Check Your Understanding

Use the event definition of a CDF and the decreasing-event argument to answer these questions.

For thresholds \(t_n\downarrow x\), identify the intersection of the events \(\{X\leq t_n\}\) and explain why equality at \(x\) is included.
Why is countable additivity relevant to proving continuity of probability for a decreasing sequence of events?
In the fair-die example, what is \(F(4)\), and what value does \(F(t)\) have when \(t\) is just greater than 4 but less than 5?
Why does \(F(x)\leq F(y_n)\) by itself not prove that \(F(y_n)\to F(x)\)?
In the tail-supremum argument, why does \(F(y_n)\leq F(u_n)\) hold, and what limiting fact about \(F(u_n)\) completes the proof?

What You'll Learn