Continuity Points and Zero Point Mass

Why Continuity Detects Point Probability

A cumulative distribution function records accumulated probability up to each threshold. Its value at $x$ is $F(x)=P(X\le x)$, so it includes any probability located exactly at $x$. The behavior immediately to the left of $x$ records the probability accumulated strictly below $x$. The difference between those two quantities is precisely the probability at the point.

This gives a useful way to distinguish two cases. At a threshold with no probability concentrated at that single value, the CDF has no jump and is continuous there. At a threshold with positive point probability, the CDF jumps by exactly that amount. A CDF is right-continuous, so the only possible obstruction to continuity at a specified point is a mismatch with its left-hand limit.

Definition: For a random variable $X$ with CDF $F(x)=P(X\le x)$, the point mass at $x$ is $P(X=x)$. The point $x$ is a continuity point of $F$ if $F$ is continuous at $x$; if $P(X=x)>0$, it is a jump point.

The Left-Hand Limit and the Point Mass

For a CDF, the left-hand limit at $x$ exists because $F$ is nondecreasing. Write it as $F(x-)=\lim_{t\uparrow x}F(t)$. To identify it probabilistically, take the increasing sequence $x-1/n$, which approaches $x$ from below. The events $\{X\le x-1/n\}$ increase to $\{X<x\}$: every value strictly below $x$ is eventually no greater than $x-1/n$, while $x$ itself is never included. Continuity of probability for increasing events therefore gives the left limit.

Theorem: For every real-valued random variable $X$ and every $x\in\mathbb{R}$, $$ F(x-)=P(X<x) \qquad\text{and}\qquad F(x)-F(x-)=P(X=x). $$ Thus $F$ has a jump at $x$ exactly when $P(X=x)>0$, and the jump size equals that point mass.

Proof. Let $A_n=\{X\le x-1/n\}$. The events $A_n$ increase with $n$, and their union is $\{X<x\}$. Indeed, if $X(\omega)<x$, then $x-X(\omega)>0$, so for all sufficiently large $n$, $1/n\le x-X(\omega)$ and $X(\omega)\le x-1/n$. Conversely, $X(\omega)\le x-1/n$ implies $X(\omega)<x$. By continuity of probability from below, $$ \lim_{n\to\infty}F(x-1/n) =\lim_{n\to\infty}P(A_n) =P(X<x). $$ Since $F$ is nondecreasing, this sequence gives its full left-hand limit: for any $t<x$, $x-1/n\ge t$ for all sufficiently large $n$, and hence $F(x-1/n)\ge F(t)$. Therefore $F(x-)=P(X<x)$. Now $\{X\le x\}$ is the disjoint union of $\{X<x\}$ and $\{X=x\}$, so $$ F(x)-F(x-)=P(X\le x)-P(X<x)=P(X=x). $$ $\square$

Continuity Exactly When the Point Mass Is Zero

The theorem identifies the possible jump from the left. On the right, every CDF is right-continuous: as thresholds decrease to $x$, the corresponding events decrease to $\{X\le x\}$, and continuity of probability from above gives the limit $F(x)$. This standard CDF property means that checking continuity at $x$ reduces to comparing $F(x)$ with $F(x-)$.

Theorem: A CDF $F$ is continuous at $x$ if and only if $P(X=x)=0$.

Proof. If $F$ is continuous at $x$, its left-hand limit equals its value, so $F(x)-F(x-)=0$. By the jump formula, $P(X=x)=0$. Conversely, suppose $P(X=x)=0$. The jump formula yields $F(x)=F(x-)$, so $F$ is left-continuous at $x$. It is also right-continuous there, as every CDF is. The left- and right-hand limits therefore both equal $F(x)$, which is continuity at $x$. $\square$

The statement concerns the probability of equality at one exact value, not the probability of being near that value. A random variable can place positive probability in every small interval around $x$ while still assigning probability zero to $x$ itself. Conversely, even a small positive point mass forces a discontinuity, however smooth the rest of the distribution may be.

Worked Classifications

Worked Example: A Finite Discrete Distribution

Let $X$ take the values $-1$, $0$, and $2$, with probabilities $0.2$, $0.5$, and $0.3$, respectively. Classify the thresholds $-1$, $-0.4$, $0$, $1$, and $2$.

The probabilities sum to $0.2+0.5+0.3=1$. At a possible value, the point mass is its listed probability: $P(X=-1)=0.2$, $P(X=0)=0.5$, and $P(X=2)=0.3$. Each is positive, so $-1$, $0$, and $2$ are jump points. Their jump sizes are $0.2$, $0.5$, and $0.3$, respectively.

The values $-0.4$ and $1$ are not among the possible outcomes. Thus $P(X=-0.4)=P(X=1)=0$, and both are continuity points. For instance, $F(-0.4)=P(X\le -0.4)=0.2$, but the CDF does not jump there: the positive value of $F(-0.4)$ is accumulated at $-1$, not at $-0.4$.

Worked Example: A Continuous Uniform Distribution

Suppose $X\sim \operatorname{Uniform}(0,1)$, with density $f(t)=1$ for $0\le t\le1$ and zero otherwise. Decide whether $F$ is continuous at $0$, $0.4$, and $1$.

For any single point $x$, its probability is the integral of the density over the singleton $\{x\}$. A singleton has length zero, so $$ P(X=x)=\int_{\{x\}}f(t)\,dt=0. $$ This holds at all three thresholds, including the endpoints $0$ and $1$. The theorem therefore shows that $F$ is continuous at each.

The CDF values at the endpoints do not change this conclusion. In particular, $F(0)=0$ and $F(1)=1$, while both point probabilities are zero. At $x=0.4$, $F(0.4)=0.4$, which is also not the point probability: $P(X=0.4)=0$. In each case the point probability is determined by the jump, not by the height of the CDF.

Worked Example: A Mixed Distribution

Let $X$ equal $1$ with probability $0.4$. With the remaining probability $0.6$, let it have the uniform distribution on $[0,2]$. Classify $0$, $1$, $1.5$, and $2$.

The uniform component contributes $0.6(x/2)$ to $F(x)$ for $0\le x\le2$, and the point mass contributes $0.4$ when $x\ge1$. Thus, for thresholds inside the support, $$ F(x)= \begin{cases} 0.6x/2, & 0\le x<1,\\ 0.6x/2+0.4, & 1\le x\le2. \end{cases} $$ Also $F(x)=0$ for $x<0$ and $F(x)=1$ for $x>2$.

At $x=1$, the uniform component gives zero probability to the singleton, while the point component gives probability $0.4$. Hence $P(X=1)=0.4$: $1$ is a jump point of size $0.4$. At $0$, $1.5$, and $2$, neither the uniform component nor the point component places mass at that exact threshold. Each has point probability zero, so all three are continuity points. In particular, the support endpoint $2$ is not a jump point merely because the uniform component stops there.

A Reliable Classification Method

When asked to classify a threshold, look for probability concentrated exactly at that value. For a discrete variable, this means checking whether the value is one of its possible outcomes with positive probability. For a continuous density, a singleton has probability zero. For a mixed distribution, check both the discrete and continuous parts: the continuous part contributes no point mass, but a separate atom may still make the threshold a jump point.

Identify the threshold.
Keep the question focused on the exact value $x$, not an interval around it.

Find its point probability.
Determine $P(X=x)$ from the model, accounting for all components if the distribution is mixed.

Classify the CDF.
If $P(X=x)=0$, $F$ is continuous at $x$. If $P(X=x)>0$, $F$ has a jump of that size.

Common Mistake: Reading the CDF Height as a Point Mass

A frequent error is to treat $F(x)$ itself as $P(X=x)$. But $F(x)=P(X\le x)$ includes probability at every value at or below $x$, whereas $P(X=x)$ concerns only the exact value $x$. The correct comparison is between the CDF value and its left-hand limit: $$ P(X=x)=F(x)-F(x-). $$ For example, in the uniform model, $F(0.4)=0.4$, but $P(X=0.4)=0$; there is no jump at $0.4$. The positive CDF value represents accumulated probability below the threshold, not probability concentrated at it.

A related mistake is to assume that any endpoint of a distribution’s support must be a jump point. An endpoint can be a continuity point when there is no atom there, as the uniform and mixed examples show at their upper endpoints. The decisive question is always whether the point probability is positive.

Key takeaway: For every real threshold $x$, $F(x)-F(x-)=P(X=x)$. Since a CDF is right-continuous, it is continuous at $x$ exactly when this point mass is zero.

Check Your Understanding

Use the relationship between the left-hand limit and point probability to answer each question.

A random variable has probabilities $0.3$ at $-2$, $0.4$ at $1$, and $0.3$ at $4$. Which of $-2$, $0$, $1$, and $3$ are continuity points?
If $F(5)=0.82$ and $F(5-)=0.77$, what is $P(X=5)$, and is $F$ continuous at $5$?
For a variable with a density and no additional discrete component, why is every individual threshold a continuity point?
A mixed distribution has an atom of probability $0.15$ at $x=2$, as well as a continuous component. What is the jump size at $2$?
Explain why a threshold can have $F(x)>0$ while still satisfying $P(X=x)=0$.