Random variable constant on a sigma-algebra. Let $\Sigma$ be a sigma-algebra of $\Omega$ and define a relation $\sim$ on $\Omega$ by $$\omega \sim \omega' \Leftrightarrow \mathbb{1}_E(\omega) = \mathbb{1}_E(w') \text{ for all } E \in \Sigma.$$ Let $\mathcal{E}$ to be the set of containing the equivalence classes determined by $\sim$.

*

*If $\Sigma$ is finite then $\Sigma = \sigma(\mathcal{E})$.


*If $\Sigma$ is finite then $X:\Omega: \rightarrow \mathbb{R}$ is $\Sigma$-measurable if and only if $X$ is constant on elements of $\mathcal{E}$.
Me work:

*

*Showing that the set $\mathcal{E}$ is contained is $\Sigma$ is not difficult. This is becase if $[w] \in \mathcal{E}$, then write $$[w] =[ \bigcap_k E_k] \cap [\bigcap_j E_j^c],$$ with $k$ going through the sets such that $\omega \in E_k$ and $j$ going through the sets such that $\omega \notin E_j$. But how do I prove the other implication?


*For this I know we are supposed to use (1) somehow but I am not sure how to even start. One idea for the front implication, is that going by contradiction somehow. I know that $X$ is constant if it is measurable with respect to $\{\emptyset,\Omega\}$ and I was attempting to use the same technique .. For the back implication I am not sure ...
 A: In (1), as you say, to show that $\sigma(\mathcal{E}) \subseteq \Sigma$, it is enough to see that for any $\omega \in \Omega$,
$$
  [\omega] = \bigcap_{\substack{E \ni \omega} \\ E \in \Sigma} E,
$$
(the part with complements are unnecessary) which is a finite intersection and well-defined by the definition of $\sim$.
To see the reverse inclusion, we have that, for any $E \in \Sigma$,
$$
  E = \bigcup_{\omega \in E} [\omega].
$$
The $\subseteq$ direction is clear, and to show the reverse inclusion, if $\omega'$ is in the right-hand union, then $\omega' \in [\omega]$ for some $\omega \in E$ so that $\omega' \sim \omega$ and thus $\omega' \in E$.
Moreover, we just saw that $\mathcal{E} \subseteq \Sigma$, and in particuar is finite.
Thus, the union, which doesn't necessarily look countable at first, is actually just a union of finitely many elements of $\mathcal{E}$, and in particular is in $\sigma(\mathcal{E})$.
Hence, $E \in \sigma(\mathcal{E})$, and we are done.

For (2), if $X$ is constant on the elements of $\mathcal{E}$, then (since $\mathcal{E}$ is finite) we can write $X = \sum_{i=1}^n \alpha_i \mathbf{1}_{E_i}$ for some $\alpha_i \in \mathbf{R}$ and elements $E_i \in \mathcal{E} \subseteq \Sigma$.
This is a simple function, and so measurable wrt $\Sigma$.
On the other hand, let $X$ be measurable, and chose $\omega_1, \omega_2$ such that $X(\omega_1) \neq X(\omega_2)$.
But then the set $X^{-1}(\{X(\omega_1)\}) \in \Sigma$ contains $\omega_1$ but not $\omega_2$, and so $\omega_1 \nsim \omega_2$.
The contrapositive of what we just showed is that if $\omega_1 \sim \omega_2$, then necessarily $X(\omega_1) = X(\omega_2)$, ie that $X$ is constant on the elements of $\mathcal{E}$.
