# Is a random variable's support always a subset of its image?

Random variable X (as a measureable function from some probability space to the reals) has support S and image T. True or False:

i.) S is always a subset of T.

ii.) (S,E,dF) is the induced probability space with: sample space S, events E the intersections of S with the Borel sets, and dF the probability measure derived from F -the distribution of X.

(As S is the smallest closed subset of reals R with probability 1, it is a subset of the closure of T. The two issues come from having read somewhere that the support is, "loosely" speaking, the outcomes X can take on mapping the space to R. Started with case T finite.)

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What do you mean by induced probability space in ii)? –  Michael Greinecker Sep 14 '12 at 12:29
I asked this question as "ace091312". Thanks Rabee and Michael for your answers settling the issue. They actually led to another related question (key words: support, image and image closure) which I posted posted under "ace" yesterday. –  Ace Sep 17 '12 at 21:04

The answer is no to i). And the situation can be pathological.

Let the probability spaces $(\Omega,\Sigma,\mu)$ be $[0,1]$ with the Lebesgue measure. Let $V_\alpha \subseteq \Omega$ be a closed nowhere dense set with $\mu(V_\alpha)\ge\alpha$ that does not contain $0$ (such sets can be shown to exist).

Let $X: \Omega\to [0,1]$ be the random variable (measurable function) $X(\omega)=\omega$ if $\omega\notin V_\alpha$ and $X(\omega)=0$ otherwise.

Let $S$ be the support of $X$. It must contain the open set $[0,1]\setminus V_\alpha$, because $S$ is closed and each open subset of $[0,1]\setminus V_\alpha$ has positive probability of realizing under $X$. But any measurable set containing $[0,1]\setminus V_\alpha$ has probability one of realizing under $X$. Thus, the support $S$ of $X$ is the closure of $[0,1]\setminus V_\alpha$ which is $[0,1]$.

But $S$ contains $V_\alpha$ which is not in the range of $X$.

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For i), let $q_1,q_2,\ldots$ be an enumeration of the rationals. Let $\Omega=\mathbb{Q}$, $\Sigma=2^\mathbb{Q}$ and $\mu$ be given by $\mu\big(\{q_n\}\big)=1/2^n$. Let $X:\mathbb{Q}\to\mathbb{R}$ be the identity. Then $X(\Omega)=\mathbb{Q}$ which is smaller than the support of $X$, which is all of $\mathbb{R}$.

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