Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space and $\mathcal{F}$ a sub-$\sigma$-algebra. Let $\mathcal{F}$ be trivial, i.e. $\forall A\in\mathcal{F}: \mathbb{P}(A)\in\left\{0,1\right\}$. Show that $\mathbb{E}(f|\mathcal{F})=\mathbb{E}(f)$.
One criterion to prove that is to show that $$ \forall A\in\mathcal{F}: \int_A\mathbb{E}(f)\, d\mathbb{P}=\int_Af\, d\mathbb{P}. $$
Do not know exactly how to show that in common.
For the special case, that $\mathcal{F}=\left\{\Omega,\emptyset\right\}$ it is $$ \int_{\emptyset}\mathbb{E}(f)\, d\mathbb{P}=0=\int_{\emptyset}f\, d\mathbb{P},~~~~~\int_{\Omega}\mathbb{E}(f)\, d\mathbb{P}=\mathbb{E}(f)\mathbb{P}(\Omega)=\mathbb{E}(f)=\int_{\Omega}f\, d\mathbb{P}. $$
But how can I prove that if $\mathcal{F}$ is trivial, but not $\mathcal{F}=\left\{\Omega,\emptyset\right\}$?
Hint: For every $A$ in $\mathcal A$ such that $\mathbb P(A)=1$, one has $\displaystyle\int_Af\,\mathrm d\mathbb P=\mathbb E(f)$. For every $A$ in $\mathcal A$ such that $\mathbb P(A)=0$, one has $\displaystyle\int_Af\,\mathrm d\mathbb P=0$.