Is it true that $L^2_0(\mathcal{G},\mathbb{P})\oplus L^2_0(\mathcal{H},\mathbb{P}) = L^2_0(\sigma(\mathcal{G},\mathcal{H}),\mathbb{P})$?

Question

If $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, define $L^2_0(\Omega,\mathcal{F},\mathbb{P}):=\{f\in L^2(\Omega,\mathcal{F},\mathbb{P})\ | \ \int_\Omega f\operatorname{d}\mathbb{P}=0\}$.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $\mathcal{G},\mathcal{H}$ sub-$\sigma$-algebras of $\mathcal{F}$ that are $\mathbb{P}$-independent. Then it is immediate that $$L^2_0(\Omega,\mathcal{G},\mathbb{P})\perp L^2_0(\Omega,\mathcal{H},\mathbb{P}),$$ so, denoting with $\sigma(\mathcal{G},\mathcal{H})$ the $\sigma$-algebra generated by $\mathcal{G}$ and $\mathcal{H}$, from $L^2_0(\Omega,\mathcal{G},\mathbb{P})\subset L^2_0(\Omega,\sigma(\mathcal{G},\mathcal{H}),\mathbb{P})$ and $L^2_0(\Omega,\mathcal{H},\mathbb{P})\subset L^2_0(\Omega,\sigma(\mathcal{G},\mathcal{H}),\mathbb{P})$, it follows that: $$L^2_0(\Omega,\mathcal{G},\mathbb{P})\oplus L^2_0(\Omega,\mathcal{H},\mathbb{P})$$ is a closed subspace of $$L^2_0(\Omega,\sigma(\mathcal{G},\mathcal{H}),\mathbb{P}).$$ So the question: is it also true that $$L^2_0(\Omega,\mathcal{G},\mathbb{P})\oplus L^2_0(\Omega,\mathcal{H},\mathbb{P}) = L^2_0(\Omega,\sigma(\mathcal{G},\mathcal{H}),\mathbb{P})?$$

Answer 1

In other words: is any $\sigma(\mathcal F, \mathcal G)$-measurable function the sum of a $\mathcal F$-measurable function and a $\mathcal G$-measuruable function. This is wildly false. Functions of two variables are much more complicated than that!

Even in the plane, where the two coordinates are independent for Lebesgue measure, it is not true that a (mean zero square-integrable) function $f(x,y)$ of two variables can be written as a sum $f_1(x)+f_2(y)$.

Explanation: Write $\mathcal B_1$ for the Borel sets in $\mathbb R$ and $\mathcal B_2$ for the Borel sets in $\mathbb R^2$. Let $$ \mathcal F = \{ A \times \mathbb R : A \in \mathcal B_1\} \\ \mathcal G = \{ \mathbb R \times A : A \in \mathcal B_1\} $$ These are independent with respect to $2$-dimensional Lebesgue measure. A function $f : \mathbb R^2 \to \mathbb R$ is $\mathcal F$-measurable if and only if there is a $\mathcal B_1$-measurable function $f_1 : \mathbb R \to \mathbb R$ with $f(x,y) = f_1(x)$ for all $x,y$. Similarly, $f : \mathbb R^2 \to \mathbb R$ is $\mathcal G$-measurable if and only if there is a $\mathcal B_1$-measurable function $f_2 : \mathbb R \to \mathbb R$ with $f(x,y) = f_2(y)$ for all $x,y$.

Answer 2

What follows is a counterexample.

• $\Omega:=\{a,b,c,d\}$;
• $\mathcal{G}:=\{\emptyset , \{a,d\}, \{b,c\}, \Omega\}$;
• $\mathcal{H}:=\{\emptyset , \{b,d\}, \{a,c\}, \Omega\}$;
• $\mathbb{P} :2^\Omega\rightarrow[0,1]$ the uniform probability on $\Omega$.

Then from $$\mathbb{P}(\{a,d\}\cap\{b,d\})=\mathbb{P}(\{d\})=\frac{1}{4}=\frac{1}{2}\frac{1}{2}=\mathbb{P}(\{a,d\})\mathbb{P}(\{b,d\}),$$ also the other relations to check the independence follows, and then $\mathcal{G}$ and $\mathcal{H}$ are $\mathbb{P}$-independent. Also, notice that $\sigma(\mathcal{G},\mathcal{H})=2^\Omega$.

Now:

• $L^2_0(\Omega,\mathcal {G},\mathbb{P})=\{\lambda 1_{\{a,d\}}-\lambda 1_{\{b,c\}}\ |\ \lambda\in\mathbb{R}\}$;
• $L^2_0(\Omega,\mathcal {H},\mathbb{P})=\{\mu 1_{\{b,d\}}-\mu 1_{\{a,c\}}\ |\ \mu\in\mathbb{R}\}$;
• $L^2_0(\Omega,\mathcal {G},\mathbb{P})\oplus L^2_0(\Omega,\mathcal {H},\mathbb{P})=\{\lambda 1_{\{a,d\}}-\lambda 1_{\{b,c\}} + \mu 1_{\{b,d\}}-\mu 1_{\{a,c\}}\ |\ \lambda,\mu\in\mathbb{R}\}$;
• $1_{\{a\}}-1_{\{c\}}\in L^2_0(\Omega,\sigma(\mathcal {G},\mathcal {H}),\mathbb{P})$,

but: $$1_{\{a\}}-1_{\{c\}}\notin L^2_0(\Omega,\mathcal {G},\mathbb{P})\oplus L^2_0(\Omega,\mathcal {H},\mathbb{P}).$$ In fact, if this is not the case, than the following system has a solution: \begin{cases} \lambda - \mu =1\\ -\lambda + \mu = 0 \\ -\lambda-\mu = -1 \\ \lambda + \mu = 0, \\ \end{cases} however, this system has not a solution.