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Let $(\Omega,\mathcal A)$, $(E,\mathcal E)$ be measurable spaces and $X:\Omega\to E$. How can we show that

  1. $X$ is constant;
  2. $\sigma(X)=\{\emptyset,\Omega\}$

are equivalent?

(1.) clearly implies (2.). Now, a proof of the other direction can be found in this answer. The argument is that if there are $\omega,\omega'\in\Omega$ with $x:=X(\omega)\ne X(\omega')$, then $\{X=x\}\not\in\sigma(X)$.

However, is this argument really correct? It seems like this is only the case if $\{x\}\in\mathcal E$. Can we fix this issue?

0xbadf00d
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  • @KaviRamaMurthy What do you mean? $\mathcal E$ is not a Borel $\sigma$-algebra. – 0xbadf00d May 27 '22 at 09:22
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    For general measurable spaces you will need some assumption on $(E, \mathcal{E})$. For example, if $\mathcal{E} = { \emptyset, E}$ then every function $X: \Omega \to E$ is measurable. – Rhys Steele May 27 '22 at 09:23
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    In your link they are considering real valued measurable functions. It is understood that $\mathcal E$ is the Borel $\sigma-$ algebra of $\mathbb R$. – Kavi Rama Murthy May 27 '22 at 09:28
  • @RhysSteele ${x}\in\mathcal E$ for all $x\in E$ should be a sufficient assumption, right? Or is there a weaker assumption (or a different argument then the one used in the proof mentioned in the question)? – 0xbadf00d May 27 '22 at 09:47
  • Yes, that assumption is sufficient but is stronger than is needed for the idea of this argument to work (and is satisfied as soon as $\mathcal{E}$ contains the Borel $\sigma$-algebra for a $T_1$-topology even). It is enough for $\mathcal{E}$ to have the property that for any pair $x,y \in E$ with $x \neq y$ there exists a set $E_{x,y} \in \mathcal{E}$ such that $x \in E_{x,y}, y \not \in E_{x,y}$. Then replacing ${X = x}$ with $E_{x, X(\omega')}$ would work in the proof; but I'm not sure of an example that naturally occurs that satisfies this weaker condition but not the first one. – Rhys Steele May 27 '22 at 12:38

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