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Consider the trivial $\sigma$-algebra on $\Omega$ given by $$ \mathcal{A} = \{C \subseteq \Omega: \mathbb{P}(C) = 0 \text{ or } 1\} $$ I want to show that if $X$ is measurable with respect to $\mathcal{A}$, then $X$ is almost surely constant, i.e. for some $c\in\mathbb{R}$ $$\mathbb{P}(X = c) = 1$$ Note that $\mathcal{A}$ is not "trivial" in the sense that it only contains $\emptyset$ and $\Omega$. It contains all subsets of $\Omega$ with probability $0$ or $1$.

I've seen a proof that uses the distribution function as a "stepping stone", but I would like to see if it's possible to complete the proof without the use of distribution functions. That is, a proof using "vanilla" set reasoning.

Similar questions:

  1. Case where $\mathcal{A} = \{\emptyset, \Omega\}$: Is a random variable constant a.e. iff it is trivial sigma-algebra-measurable?
  2. Proof using distribution function: If $P(A)=0$ or $1$ for all $A\in\mathcal{G}$, then any $\mathcal{G}$-measurable random variable is a.s. constant
Neon9357
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  • I think the statement as you've written it is false. If $X$ is measurable with respect to $A$, I don't think that means $X$ is almost surely constant. I think you mean that if this is the largest $\sigma$-algebra with respect to which $X$ is measurable, then it is almost surely constant? For example if $X \sim N(0,1)$, then $X$ should be measurable with respect to $A$ – Alan Chung Oct 14 '24 at 19:17

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