Support as the Region Where a Law Lives

Introduction

A probability law describes where a random variable can place probability, but the phrase “where it lives” needs a precise meaning. A law may assign probability zero to a particular value while still assigning positive probability arbitrarily close to it. Such a value can be an essential boundary or accumulation point of the distribution, even though it is never observed exactly.

The support captures this local behavior. It is the smallest closed set that contains the random variable with probability one. Equivalently, a point belongs to the support exactly when every open neighborhood of that point has positive probability. The neighborhood formulation works uniformly for discrete, continuous, and other probability laws; it does not require a probability mass function or density.

Definition of Support

Let $P$ be a probability law on the Borel subsets of $\mathbb{R}$. For $x\in\mathbb{R}$ and $\varepsilon>0$, write $B(x,\varepsilon)=(x-\varepsilon,x+\varepsilon)$ for the open neighborhood of radius $\varepsilon$ around $x$.

Definition: The support of $P$, written $\operatorname{supp}(P)$, is the set of all $x\in\mathbb{R}$ such that every open neighborhood of $x$ has positive probability: $$ \operatorname{supp}(P)=\{x\in\mathbb{R}: P(B(x,\varepsilon))>0\text{ for every }\varepsilon>0\}. $$ If $X$ has law $P$, this is also called the support of the distribution of $X$.

The word “every” matters. A support point cannot be separated from the probability law by even a very small interval of probability zero. A point outside the support has at least one open neighborhood whose probability is zero.

Why This Is the Smallest Closed Set of Probability One

The neighborhood definition is local, while “smallest closed set carrying probability one” is a global description. For laws on the real line, these descriptions agree. The result below establishes both the agreement and the basic properties of support.

Theorem: For any probability law $P$ on the Borel subsets of $\mathbb{R}$, its support $S=\operatorname{supp}(P)$ is closed and $P(S)=1$. Moreover, if $C$ is any closed set with $P(C)=1$, then $S\subseteq C$. Thus $S$ is the smallest closed set, under set inclusion, that has probability one.

Proof. First, the complement of $S$ is open. If $x\notin S$, the definition gives some $\varepsilon>0$ for which $P(B(x,\varepsilon))=0$. For any $y\in B(x,\varepsilon/2)$, choose $\delta=\varepsilon/2-|y-x|>0$. Then $B(y,\delta)\subseteq B(x,\varepsilon)$, so $P(B(y,\delta))=0$, and hence $y\notin S$. Therefore $B(x,\varepsilon/2)$ lies in the complement of $S$, proving that complement is open and $S$ is closed.

To show $P(S)=1$, use intervals with rational endpoints. For each $x\notin S$, there is an open interval around $x$ of probability zero. By density of the rationals, there are rational numbers $a<b$ such that $x\in(a,b)$ and $(a,b)$ lies inside that zero-probability interval. Thus every point outside $S$ belongs to a rational-endpoint interval of probability zero. There are only countably many such intervals, so $\mathbb{R}\setminus S$ is contained in a countable union of probability-zero intervals. Countable subadditivity gives $P(\mathbb{R}\setminus S)=0$, and therefore $P(S)=1$.

Finally, suppose $C$ is closed and $P(C)=1$. If $x\notin C$, the complement of $C$ is open, so some open interval $B(x,\varepsilon)$ is disjoint from $C$. Consequently, $P(B(x,\varepsilon))\leq P(\mathbb{R}\setminus C)=0$, which means $x\notin S$. Thus every point outside $C$ is outside $S$, or equivalently $S\subseteq C$. Since $S$ itself is closed and has probability one, it is the smallest such closed set. $\square$

Worked Examples

Worked Example: A Uniform Law on a Closed Interval

Let $X$ be uniform on $[0,1]$. Determine its support directly from neighborhood probabilities.

For any $x\in[0,1]$ and any $\varepsilon>0$, the intersection $B(x,\varepsilon)\cap[0,1]$ contains an interval of positive length. For example, if $x=0$, it contains $[0,\min(\varepsilon,1))$; if $x=1$, it contains $(\max(0,1-\varepsilon),1]$. At an interior point, a sufficiently small interval around $x$ lies inside $[0,1]$. Since the uniform probability of an interval within $[0,1]$ equals its length, in every case $P(B(x,\varepsilon))>0$.

If $x<0$, choose $\varepsilon=-x/2$. Then $x+\varepsilon=x/2<0$, so $B(x,\varepsilon)$ does not meet $[0,1]$ and has probability zero. If $x>1$, choose $\varepsilon=(x-1)/2$; then $x-\varepsilon=(x+1)/2>1$, so this neighborhood also has probability zero. Hence $\operatorname{supp}(P)=[0,1]$.

The endpoints are included even though $P(X=0)=P(X=1)=0$: every neighborhood of either endpoint reaches into the interval and has positive probability.

Worked Example: A Law Concentrated at Two Values

Let $X$ equal $-2$ with probability $0.3$ and $3$ with probability $0.7$. Determine its support.

Every neighborhood of $-2$ contains the outcome $-2$, so its probability is at least $0.3$. Every neighborhood of $3$ contains the outcome $3$, so its probability is at least $0.7$. Both points therefore belong to the support.

Now take any $x\notin\{-2,3\}$. Its distances to the two possible values are both positive. Choose $\varepsilon=\tfrac12\min\{|x+2|,|x-3|\}>0$. Then $B(x,\varepsilon)$ contains neither $-2$ nor $3$, and so $P(B(x,\varepsilon))=0$. No other point is in the support, giving $\operatorname{supp}(P)=\{-2,3\}$.

Worked Example: An Accumulation Point with Zero Point Probability

Let $X$ take the value $1/n$ with probability $2^{-n}$, for each integer $n\geq1$. These probabilities sum to one because $\sum_{n=1}^{\infty}2^{-n}=1$. Determine the support.

For each $k\geq1$, every neighborhood of $1/k$ contains that value, which has probability $2^{-k}>0$. Thus every $1/k$ belongs to the support.

Every neighborhood of $0$ also has positive probability, despite $P(X=0)=0$. Given $\varepsilon>0$, choose an integer $N\geq0$ large enough that $1/(N+1)<\varepsilon$. For all $n\geq N+1$, $1/n<\varepsilon$, so these outcomes lie in $B(0,\varepsilon)$. Their total probability is $$ \sum_{n=N+1}^{\infty}2^{-n}=2^{-N}>0. $$ Therefore $0$ belongs to the support.

It remains to exclude any $x\notin A$, where $A=\{0\}\cup\{1/n:n\geq1\}$. This set is closed: the only limit of distinct terms $1/n$ is $0$, which is already included. Since $x\notin A$ and $A$ is closed, there is an open neighborhood of $x$ disjoint from $A$. The random variable takes values only in $A$, so that neighborhood has probability zero. Consequently, $\operatorname{supp}(P)=\{0\}\cup\{1/n:n\geq1\}$.

Interpreting Support Correctly

Support is a property of a law, not a list of outcomes that each have positive point probability. In the third example, $0$ has probability zero as an exact outcome, yet every interval around it has positive probability. In the uniform example, the same is true at both endpoints. Neighborhoods, rather than the probabilities of single points, determine membership.

The definition also explains why the support is closed. If points of the support approach a limit, no neighborhood around that limit can have probability zero: such a zero-probability neighborhood would also exclude support points close enough to the limit. The theorem proves this more generally by showing that the support is closed and carries the entire law.

A useful consequence is that a random variable lies in its support with probability one. This does not mean every support point must occur with positive probability, nor does it mean every point in a probability-one set must be in the support. The support is specifically the smallest closed probability-one set, which makes its boundary and accumulation behavior part of the description.

Common Mistake

Wrong reasoning: “A point has probability zero, so it cannot be in the support.” This confuses the probability of a single value with the probability of neighborhoods around that value.
Correction: Check whether every positive-radius neighborhood has positive probability. In the sequence example, $P(X=0)=0$, but every interval around $0$ includes a tail of the sequence with total probability $2^{-N}>0$. Thus $0$ is in the support.
Another pitfall: Naming a set that has probability one is not enough to identify the support. The definition requires the smallest closed probability-one set; the support includes limit points needed to make that set closed.

Key takeaway: A real number is in the support of a probability law exactly when every open interval around it has positive probability. The resulting set is closed, has probability one, and is contained in every other closed set of probability one.

Check Your Understanding

Use the neighborhood definition, not just point probabilities, to answer these questions.

For a uniform law on $[2,5]$, why does every neighborhood of $2$ have positive probability even though $P(X=2)=0$?
A law assigns probability $0.4$ to $-1$ and $0.6$ to $4$. Why can no point strictly between $-1$ and $4$ belong to its support?
In the sequence example, calculate the probability contributed by outcomes $1/n$ with $n\geq 6$. What does this show about neighborhoods of zero?
State the smallest-closed-set characterization of support precisely.
Can a point have probability zero and still be in the support? Give an example and explain it using neighborhoods.