Any measurable function $f:X \to \mathbb R, \Sigma=\{X,\emptyset\}$(where $X$ is a non empty set, $\Sigma$ is a sigma algebra on $X$) must be constant.
My attempt:-
We know that $f:X \to \mathbb R$ is measurable. Assume on contrary that $f$ is not constant. I could write the proof till here. I don't know how to proceed? Please help me.
Best Answer
Assume that $f$ takes at least two different values $a<b$, $f(x)=a,f(y)=b$. Then $A=f^{-1}((a,+\infty))$ must be in the $\sigma$-algebra $\Sigma$ because $f$ is measurable. But this means that $A=∅$ or $A=X$. We definitely cannot have $A=∅$ because $y \in A$. Thus, $A= X$. But this would imply $f(x) = a \in (a,+\infty)$, which is a contradiction.