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From Jacod / Protter: "Probability Essentials", Springer:

Note that even if the state space (or range space) $T$ is not countable, the image $T'$ of $\Omega$ under $X$ (that is, all points $\{i\}$ in $T$ for which there exists an $\omega\in\Omega$ such that $X(\omega) = i$ ) is either finite or countably infinite.

(where $X$ is a function (random variable) from $\Omega$ into a set $T$)

I do not understand this. If $T$ is the uncountable set $\bf R$ (the real numbers), could the image also be uncountably infinite?

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Is $\Omega$ countable? Then the image $T' = T(\Omega)$ is also. If $\Omega = \{\omega_i \mid i \in \mathbb N\}$, then $T' = \{T(\omega_i)\mid i\in \mathbb N\}$. – martini Mar 25 '12 at 8:09
Doh. I mixed up $\Omega$ and $T$, I meant: $T' = X(\Omega) = \{ X(\omega_i) \mid i \in \mathbb N\}$. – martini Mar 25 '12 at 8:20
Yes, $\Omega$ is assumed countable. – lodhb Mar 25 '12 at 8:32
@lodhb: Then $X(\Omega)$ is also countable. If $\omega_1, \omega_2, \ldots$ is an enumeration for $\Omega$ (not nessarily injective), $X(\omega_1)$, $X(\omega_2)$, $\ldots$ enumerates $T'$. – martini Mar 25 '12 at 8:35
@martini: makes perfect sense, thank you! I think I got confused, thanks for clearing everything up. – lodhb Mar 25 '12 at 8:50
up vote 2 down vote accepted

Generally speaking, if $f$ is a function then $f$ is always onto its range. If the domain of $f$ is countable (or generally can be well ordered) then $f$ has a right inverse, and therefore $|\operatorname{Rng}(f)|\le|\operatorname{Dom}(f)|$.

By this property we have that if the domain of $f$ is countable then its range is at most countable.

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Range is a very ambiguous term.You may want to say image instead so that it is clear that you do not mean that the range is the codomain. – Nate Iverson Jun 28 '12 at 13:05
@Nate: I suppose it depends on your mathematical education and background. In my part of the woods there is no ambiguity when range is used. – Asaf Karagila Jun 28 '12 at 14:30

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