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In the literature the support, $S$, of a random variable $X$ is defined as the smallest closed subset of real line $\mathbb{R}$ with probability $1$. Looking to prove that $S$ is where the graph of $X$'s cdf, $F$, is not “flat”. More formally, with $O_x$ an open interval containing $x$ with probability $P(O_x)$, show that: $S=\{x: \forall O_x\quad P(O_x)\neq 0\}$.

To clarify, prove that $S=\{x: \forall O_x\quad P(O_x)\neq 0\}$ is closed with probability 1 and is the smallest such closed certain set.

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up vote 1 down vote accepted

If $x \notin S$, since $S$ is closed there is an open interval $O_x$ around $x$ that is disjoint from $S$, and since $P(S) = 1$ we must have $P(O_x) = 0$.

Conversely, if there is an open interval $O_x$ around $x$ such that $P(O_x) = 0$, then ${\mathbb R} \backslash O_x$ is a closed subset of $\mathbb R$ with $P({\mathbb R} \backslash O_x) = 1$, so $S \subseteq {\mathbb R} \backslash O_x$, and in particular $x \notin S$.

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I agree with the result you and others have posted. However, I was not clear on my intent, which was to show how my "set" description of support satisfied the definition. My "proof" made use of the second countability of the reals and closed sets containing their frontiers, etc. and seemed a bit long. – Ace Jun 12 '12 at 1:46
Call your set $R$. $R$ is a closed set (because its complement is the union of open sets $O_x$); if $C$ is any closed set with $P(C) = 1$, we must have $R \subseteq C$. The not-quite-obvious fact is that $P(R) = 1$. For this note that you can assume each $O_x$ has rational endpoints, and thus that $R$ is the intersection of countably many sets of probability $1$. – Robert Israel Jun 12 '12 at 18:06
Yes. I used second countability of the reals to show complement, R', covered by the Ox' has a countable sub-cover - implying P(R')=0, so that P(R)=1-P(R')=1. Also, if x is in R but not C, x is an exterior point of C. Thus, there's an Ox disjoint from C, implying P(Ox U C)>1. – Ace Jun 12 '12 at 19:53

Let $x\in S$ and suppose $P(O_x) = 0$ for some nontrivial open interval $O_x$ about $x.$ Then $S - O_x$ is a closed proper subset of $S,$ contradicting the minimality assumption. Conversely, if $x$ satisfies the aforementioned property for all such $O_x,$ then every neighborhood of $x$ intersects $S$ non-trivially - that is, $x\in S.$

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Oh, looks like Robert beat me... – user17794 Jun 11 '12 at 22:34

You can do this by contradiction.

If $S$ is a closed set with $\Pr(X \in S)=1$ and there is an $x \in S$ with an open set $O_x$ containing $x$ where $\Pr(X \in O_x)=0$ then $S\backslash O_x$ is a closed set smaller than $S$ [it does not contain $x$] and $\Pr(X \in S\backslash O_x) \ge \Pr(X \in S)-\Pr(X \in O_x) =1-0=1$.

So $S$ is not the smallest closed subset with probability $1$ and so $x$ cannot be in the support.

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