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I'm trying to make sense of conditional expectation of a $\sigma-$algebra, beginning with the finite case.

Let $X$ be a random variable on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

The conditional expectation of $X$ for some event $A \in \mathcal{F}$ is $E(X|A) = \sum_{\omega \in A} X(\omega) \frac{\mathbb{P}(w)}{\mathbb{P}(A)}$ and $X$ is $\mathcal{F}$-measurable if $X^{-1}= \{ \omega \in \Omega | X(\omega) = x_i \} \in \mathcal{F}, \forall x_i$. The conditional expectation with respect to a $\sigma-$algebra $\mathcal{G}$ is $E[X|\mathcal{G}](\omega) = \sum_{A \in \mathcal{G}} E(X|A)1_{A}(\omega)$.

Claim 1: If $X$ is $\mathcal{G}-$measurable, then $E[X|\mathcal{G}](\omega) = X$.

Proof. $X$ is $\mathcal{G}-$measurable iff $X$ is constant on each set in $\mathcal{G}$. Hence

$$E[X|\mathcal{G}](\omega) = \sum_{A \in \mathcal{G}} E(X|A)1_{A}(\omega) = \sum_{A \in \mathcal{G}}\sum_{\omega \in A} \alpha_i \frac{\mathbb{P}(\omega)}{\mathbb{P}(A)}1_{A}(\omega)=\sum_{A \in \mathcal{G}} \alpha_i 1_{A}(\omega) = X(\omega)$$

Claim 2. Let $\mathcal{F}_0 = \{ \emptyset, \Omega \}$ be the smallest sigma-algebra of $\Omega$. Then $E[X|\mathcal{F}_0](\omega) = E(X)$. $E(X|\emptyset)= 0$ since we're summing over zero elements. Thus

$$E[X|\mathcal{G}](\omega) = E(X|\Omega)1_{\Omega}(\omega)=\sum_{\omega \in \Omega} X(\omega) \frac{\mathbb{P}(\omega)}{\mathbb{P}(\Omega)}1_{\Omega}(\omega) = \sum_{\omega \in \Omega} X(\omega) \mathbb{P}(\omega) = E(X)$$

Claim 3. Let $\mathcal{F} = 2^{\Omega}$ be the powerset of $\Omega$. Then $X$ is $\mathcal{F}-$measurable iff $X$ is constant.

Proof. Since $\Omega \in \mathcal{F}$,and $X$ is constant on each set in $\mathcal{F}$ if $X$ is $\mathcal{F}$-measurable, $X$ is constant on $\Omega$.

Are my claims and proofs correct? And what is the interpretation of $E(X|2^{\Omega})(\omega)$?

Edit:

About the claim that $X$ is $\mathcal{G}$-measurable is wrong. What I confused it with was that on a finite sample space, every $\sigma-$algebra is generated by some partition. If $X$ is $\mathcal{G}$-measurable, then it's constant on each set generating the $\sigma-$algebra. Thanks for the comments!

Oskar
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    Claim 3 is wrong. Every map $X:\Omega\to\mathbb R$ is measurable w.r.t. the power set. Proof: every preimage $X^{-1}(B)$ of a Borel set is a subset of $\Omega,.$ Hence, it is in the power set. – Kurt G. May 05 '23 at 09:57
  • @KurtG. What is the argument that it's invalid in the finite case? – Oskar May 05 '23 at 09:59
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    The proof of claim 1 starts with a wrong statement. If $X$ is $\mathcal G$ measurable $X$ is not necessarily constant on each set on $\mathcal G$. However, $X$ is constant on some set in $\mathcal G$, mainly sets of the form $X^{-1}({x})$ for some $x$ in the state space. – Small Deviation May 05 '23 at 10:10
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    You make the same false statement in the proof of claim 3, which explains why the proof of claim 3 is wrong, as @KurtG. pointed out. – Small Deviation May 05 '23 at 10:11
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    Same argument in finite case. In fact, the power set is the $\sigma$-algebra that makes every r.v. measurable. If you want to enforce $X$ being constant you should take the trivial $\sigma$-algebra ${\emptyset,\Omega},.$ View it this way: the coarser the $\sigma$-algebra (i.e. smaller) the closer $X$ is to a constant r.v. The idea is quite similar to continuity of a function w.r.t. various topologies. – Kurt G. May 05 '23 at 10:27

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