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  • Let $(\Omega,\mathcal A, P)$ be a probability space, and let $\Sigma$ be the set of sub $\sigma$-fields of $\mathcal A$, that is, $$ \Sigma = \{\mathcal F \subset \mathcal A:\; \mathcal F \; \text{is a $\sigma$-field} \}. $$ For $\mathcal F \in \Sigma$, let $L^2(\mathcal F)$ be the space of equivalence classes of random variables that are $\mathcal F$-measurable and have a finite second moment. We know that $L^2(\mathcal F)$ is closed linear subspace of $L^2(\mathcal A)$, the latter viewed as a Hilbert space with the usual $L^2$ norm.

  • Let $\text{Lin}(L^2(\mathcal A))$ be the set of all closed linear subspaces of $L^2(\mathcal A)$.

Now consider the map $\mathcal F \mapsto L^2(\mathcal F)$. Is this map a bijection from $\Sigma$ to $\text{Lin}(L^2(\mathcal A))$?

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For one thing, the map is not injective in general. For instance if $\mathcal{A}$ is the sigma algebra of Lebesgue measurable sets (and the probability measure e.g. the normal distribution), then both the Borel sigma algebra and the Lebesgue sigma algebra map to all of $L^2(\mathcal{A})$.

Furthermore, the map is also not surjective in general. To see this, let $$ V := \Big\{f \in L^2(\mathcal{A}) : \int f(x) dP(x) =0 \Big\}. $$ If we had $V = L^2(\mathcal{F})$, we would have $x \mapsto 1\in V$ (why?!), which is absurd.

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Assuming you can get access to it, look at theorem 1 in "On Relations Between Strict-Sense and Wide-Sense Conditional Expectations": https://epubs.siam.org/doi/10.1137/1102020

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