Atoms, Endpoints, and Support Closure

Why Support Is Not the Same as Positive Point Probability

A distribution’s support describes where its probability is concentrated in a neighborhood sense. That distinction matters because many familiar distributions assign probability zero to every individual point. For example, a continuous distribution can have support from $0$ to $1$ even though neither endpoint, nor any interior point, has positive probability on its own. By contrast, an atom—a point with positive probability—must belong to the support.

The useful question is therefore not simply “Does $X$ equal this point with positive probability?” It is “Does every neighborhood of this point have positive probability?” That question identifies support points, including endpoints, while distinguishing them from isolated points that happen to appear in a larger set used to describe possible values.

Definition: For a real-valued random variable $X$, its support is the set $$ \operatorname{supp}(X)=\{x\in\mathbb{R}:P(|X-x|<\varepsilon)>0\text{ for every }\varepsilon>0\}. $$ A point $x$ is an atom if $P(X=x)>0$. An atom is also called a point mass.

The interval $(x-\varepsilon,x+\varepsilon)$ in the definition is an open neighborhood of $x$. A point is in the support when no matter how tightly we look around it, that neighborhood retains positive probability. The probability need not be at the point itself; it can come from nearby values.

Atoms Belong to the Support, but Support Points Need Not Be Atoms

If $P(X=x)>0$, then every neighborhood of $x$ contains the event $\{X=x\}$. Thus every such neighborhood has probability at least $P(X=x)>0$, so $x$ is in the support. The converse does not hold: positive probability in every neighborhood does not require positive probability at the center.

Worked Example: Uniform Endpoints With Zero Point Probability

Let $X\sim \operatorname{Uniform}(0,1)$. Its density is $f(x)=1$ for $0<x<1$ and $f(x)=0$ otherwise. Determine its support and the probabilities at the endpoints.

For any $x\in(0,1)$ and any $\varepsilon>0$, the interval $(x-\varepsilon,x+\varepsilon)$ intersects $(0,1)$ in an interval of positive length. The uniform probability of that intersection is its length, so it is positive. Thus every interior point is in the support.

For $x=0$, the neighborhood $(-\varepsilon,\varepsilon)$ contains $(0,\min(\varepsilon,1))$, which has positive length for every $\varepsilon>0$. Therefore $0$ is in the support. Similarly, every neighborhood of $1$ contains an interval of positive length inside $(0,1)$, so $1$ is in the support. If $x<0$, choosing $\varepsilon<-x$ gives a neighborhood with no overlap with $(0,1)$; if $x>1$, choose $\varepsilon<x-1$. Those points are not in the support. Hence

\operatorname{supp}(X)=[0,1].

Yet $P(X=0)=P(X=1)=0$, because a continuous uniform distribution assigns probability equal to interval length, and a singleton has length zero. The endpoints are support points because every neighborhood reaches positive-probability values, not because either endpoint is an atom.

Support Is Closed and Has Probability One

The neighborhood definition has two important consequences: the support is a closed set, and $X$ lies in it with probability one. The second fact is not automatic merely because each point outside the support has some zero-probability neighborhood: there may be uncountably many such points. The countable-basis argument below handles that issue.

Theorem: For any real-valued random variable $X$, $\operatorname{supp}(X)$ is closed and $P(X\in\operatorname{supp}(X))=1$. Moreover, it is contained in every closed set $C\subseteq\mathbb{R}$ for which $P(X\in C)=1$.

Proof. Write $S=\operatorname{supp}(X)$. If $x\notin S$, some $\varepsilon>0$ satisfies $P(|X-x|<\varepsilon)=0$. For every $y$ sufficiently close to $x$, there is a neighborhood of $y$ contained in $(x-\varepsilon,x+\varepsilon)$; that neighborhood also has probability zero. Thus nearby points are outside $S$, so the complement of $S$ is open and $S$ is closed.

To establish that $X\in S$ with probability one, consider all intervals $(a,b)$ with rational endpoints for which $P(a<X<b)=0$. There are at most countably many such intervals, so their union $N$ has probability zero. If $x\notin S$, there is a zero-probability neighborhood $(x-\varepsilon,x+\varepsilon)$. Choose rational numbers $a,b$ such that

x-\varepsilon<a<x<b<x+\varepsilon.

Then $P(a<X<b)=0$, so $x\in N$. Consequently, $S^{c}\subseteq N$, and $P(X\notin S)=0$.

Finally, let $C$ be closed with $P(X\in C)=1$. For any $x\notin C$, closedness gives an open neighborhood of $x$ disjoint from $C$. Since $X\in C$ with probability one, that neighborhood has probability zero. Hence $x\notin S$, which proves $S\subseteq C$. $\square$

The last statement explains why support is not just any closed set that contains all the probability. A larger closed set can also have probability one, but it may include points that have a zero-probability neighborhood. The support excludes those unnecessary additions.

Worked Example: A Closed Set With an Unneeded Isolated Point

Let $X\sim\operatorname{Uniform}(1,2)$, and consider the closed set $C=\{0\}\cup[1,2]$. The set $C$ is closed and $P(X\in C)=1$, but the point $0$ is isolated in $C$. Is $0$ in the support of $X$?

No. Take $\varepsilon=\tfrac12$. Then $(-\tfrac12,\tfrac12)$ does not intersect $[1,2]$, so $P(|X-0|<\tfrac12)=0$. Thus $0\notin\operatorname{supp}(X)$. In fact, the same neighborhood argument as for the uniform endpoints gives $\operatorname{supp}(X)=[1,2]$.

This example separates two claims: an isolated point may occur in a closed set that carries all the probability, but that does not make it a support point. Support membership is decided by neighborhood probabilities.

What Can Happen at an Endpoint or an Isolated Support Point?

A finite endpoint of the support belongs to the support. For example, if a nonempty support is bounded below, its infimum belongs to the support: choose support points arbitrarily close to that infimum and use closedness. The same reasoning applies to a finite supremum. But endpoint membership does not determine the probability at the endpoint. That probability can be zero or positive, depending on the law.

For an isolated point, it is important to say what set the point is isolated in. An isolated point in a larger candidate set need not be in the support, as the previous example showed. But if a point is isolated in the support itself, it must be an atom.

Theorem: Every isolated point of $\operatorname{supp}(X)$ is an atom. An isolated point of some other set is not necessarily in the support.

Proof. Suppose $x\in S=\operatorname{supp}(X)$ is isolated in $S$. Then some $\varepsilon>0$ satisfies $S\cap(x-\varepsilon,x+\varepsilon)=\{x\}$. Since $P(X\in S)=1$, whenever $X$ falls in $(x-\varepsilon,x+\varepsilon)$, it equals $x$, except on the probability-zero event that $X\notin S$. Therefore

P(|X-x|<\varepsilon)=P(X=x).

Because $x\in S$, the left-hand side is positive. Thus $P(X=x)>0$, and $x$ is an atom. $\square$

Worked Example: Two Isolated Support Points

Suppose $P(X=0)=0.3$ and $P(X=2)=0.7$. Determine the support and check whether either point is an atom.

Every neighborhood of $0$ contains $0$, so its probability is at least $0.3$. Every neighborhood of $2$ contains $2$, so its probability is at least $0.7$. Hence $0$ and $2$ are both in the support.

If $x$ is different from both $0$ and $2$, choose a neighborhood around $x$ small enough to exclude both values. That neighborhood has probability zero. Therefore

\operatorname{supp}(X)=\{0,2\}.

Both support points are isolated, and both are atoms: $P(X=0)=0.3$ and $P(X=2)=0.7$. This agrees with the theorem. It also illustrates why support is not restricted to intervals: it can contain separated points.

A support endpoint may likewise carry an atom or no atom. For instance, a distribution concentrated at $0$ has support $\{0\}$ and $P(X=0)=1$, so its endpoint is an atom. The uniform example has the same support endpoints as its range limits, but $P(X=0)=P(X=1)=0$. These are different probability laws, even though both examples illustrate endpoint membership.

Common Mistake: Treating Support as the Set of Positive-Mass Points

A frequent error is to define support as the set of points $x$ for which $P(X=x)>0$. That definition works for identifying atoms, but it fails for continuous distributions: for $X\sim\operatorname{Uniform}(0,1)$, it would produce the empty set even though $X$ lies in $[0,1]$ with probability one.

The correction is to check all neighborhoods. A point is in the support if every open interval around it has positive probability. This includes atoms, continuous regions, and boundary points whose own point probabilities are zero. An isolated point in the support is a special case: because a sufficiently small neighborhood contains no other support points, it must carry positive probability itself.

Key takeaway: Support is the closed set of points whose every neighborhood has positive probability. Every atom belongs to it, but a support point need not be an atom. Finite endpoints belong to a nonempty closed support when finite, while their point probabilities depend on the distribution.

Check Your Understanding

Use neighborhood probabilities to answer each question.

For a uniform distribution on $(2,5)$, what is the support, and what are the probabilities at $2$ and $5$?
Why must every atom belong to the support?
A closed set $C$ has probability one but includes an isolated point $z$. What additional check determines whether $z$ is in the support?
Why must an isolated point of the support have positive point probability?
Can a finite endpoint of a support have probability zero? Give a distribution that demonstrates your answer.