# Mapping $\sigma$-fields to $L^2$

• 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 $$\text{Lin}(L^2(\mathcal A))$$?

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.