I want to solve the following problem:

Get $\omega \in \Omega \subset \mathbb{R}$, $x \in D \subset \mathbb{R}^2$ and $0<a_i\leq a(.,.)\leq a_x<\infty$.

Let $a( x;. )$ and $f(x;.)$ be $\mathcal{F}$-measurable and consider the variational problem: find $u \in H:= L^2(\Omega;\mathcal{F}; H^1_0(D))$ satisfying $$ \mathbb{E}[\int_{D} \alpha (x,\omega) \nabla u \nabla v dx] =\mathbb{E}[\int_{D}f(x,\omega) v dx],~~~ \forall v \in H $$ and let $\mathcal{G} \subset \mathcal{F}$ be a sub $\sigma$-algebra and $a(x,.)$ and $f(x,.)$ are also $\mathcal{G}$-measurable. Now consider the alternative variational problem: find $w \in W:= L^2(\Omega;\mathcal{G}; H^1_0(D))$ satisfying $$ \mathbb{E}[\int_{D} \alpha (x,\omega) \nabla w \nabla v' dx] =\mathbb{E}[\int_{D}f(x,\omega) v' dx],~~~ \forall v' \in W $$ show that $u=w$ and hence $u(x,.)$ is $\mathcal{G}$-measurable.

I know that for given sample $\omega \in \Omega$ (noting that $(\Omega, \mathcal{F},\rho)$ and $(\Omega, \mathcal{G},\rho')$ are complete measure spaces and $\Omega$ is sample space) both variational problems has solution and $u(.,\omega) = w(.,\omega),~~ \forall \omega \in \Omega$. (from uniqueness of the solution of variational problem comes from Lax-Milgiram Lemma), but I can't figure out the rest of the problem.

thanks a lot for your attention



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