Is a random variable constant iff it is trivial sigma-algebra-measurable?

I found a proof here for a measurable function (instead of probability theory's random variable) being constant if and only if the sigma-algebra generated by it is the trivia sigma-algebra, I think (If so, I believe it is the same in the probabilistic version since the poster actually says "probability space"). I copied the proof below.

Here are my questions:

1. Is capital $X$ supposed to be $A$?

2. Is $A$ supposed to be a sample space rather than a probability space? So we say $f$ is a random variable/measurable function on probability space/measure space ($X, F, P$) for some probability measure $P$?

3. Is $C$ supposed to be a Borel set?

4. What is the relevance of $c_1$ being a closed set?

Here is the proof:

In reply to "probability", posted by alex on May 10, 2004:

Suppose that $A$ is probability space and $f$ is a any real-valued function on $A$. Prove that If $F=\{\emptyset, A\}$, then $f$ is $F$-measurable $\iff f$ is a constant

If $f==c$ is constant it is ALWAYS measurable (for any sigma-algebra). This holds as $f^{-1}[C]$ is $X$ if $c \in C$ and empty if $c \notin C$. And both sets are in any sigma-algebra.

On the other hand, if $f$ is $F$-measurable and non-constant, then it assumes at least two values $c_1$ and $c_2$. The set $f^{-1}[{c_1}]$ must be in $F$ (by being $F$-measurable, as ${c_1}$ is a closed set) but this set is non-empty (as $c_1$ IS a value of $f$) and not $X$ (as the points $x$ where $f$ assumes the value $c_2$ are not in it). So this set cannot be in $F$, and so $f$ must be constant.

• As an interesting side, this fact also parallels how a function is constant if and only if it is continuous with respect to the trivial topology, with the proof in both cases essentially identical. – Chill2Macht Feb 6 '18 at 13:46
• @Chill2Macht anyone who knows that would I guess know the answer to my question so I guess I wouldn't know any of those terms you just say that? – BCLC Feb 6 '18 at 14:02
• Literally yesterday I knew that fact about the trivial topology, but didn't realize/understand that it also applied to the trivial $\sigma$-algebra, so I thought it was something interesting to mention. You are right that it isn't a particularly useful observation, that's why I thought it made sense only as a comment. – Chill2Macht Feb 6 '18 at 14:26

Suppose $\sigma(X)=\{\varnothing,\Omega\}$, and there exists $a,b\in\Omega$ such that $X(a)\ne X(b)$. Then $X^{-1}(\{X(a)\})\notin\sigma(X)$, a contradiction. It follows that $X$ is constant.
For the converse, suppose $X(\omega)=c$ for all $\omega\in\Omega$. Then $X^{-1}(B)=\varnothing$ if $c\notin B$ and $X^{-1}(B)=\Omega$ if $c\in B$. Hence $\sigma(X)=\{\varnothing, \Omega\}$.
• Technically, $P(X=c)=1$ doesn't mean that $X$ is constant. It depends on the probability measure. For example, let $\Omega=\{a,b\}$ and $\mathbb P(\{a\})=1$. Then you could have a random variable $X(a)=0, X(b)=1$. In this case, $\mathbb P(X=0)=1$, but it isn't true that $X(\omega)=0$ for all $\omega\in\Omega$. – Math1000 May 14 '15 at 19:53
• Right so, 1 you're not really answering the title question and 2 you didn't answer the questions in the OP intersection $title^c$? – BCLC May 14 '15 at 19:55
• Is the claim, and your proof, assuming that the sigma algebra on $\mathbb{R}$ is the Borel sigma algebra? Because if we take the trivial sigma algebra on $\mathbb{R}$, then any function from a measurable space is measurable, right? – Joe Mar 19 at 13:21