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:
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Is capital $X$ supposed to be $A$?
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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 $P$?
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Is $C$ supposed to be a Borel set?
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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.
Best Answer
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 $\sigma(X)=\{\varnothing, \Omega\}$.