Let $\Omega=[0,1]$, $\mathcal{F}=\mathcal{B}([0,1])$, $P$ the lebesgue measure on $[0,1]$ and $X(\omega)=1_{[0,1/2]}(\omega)$ . I showed that $\mathcal{G}=\{A\subset \Omega : A\text{ countable} \, \vee \, A^C \text{ countable}\}$ is a Sub-$\sigma$-Algebra of $\mathcal{F}$ and that it is generated by $\{\{x\} : x\in \Omega\}$, i. e. $\sigma(\{\{x\} : x\in \Omega\})=\mathcal{G}$.
I can also compute $E(X)$:$$E(X)=\int X(\omega)\, \mathrm{d}P(\omega)=P([0,1/2])=1/2$$ But how do I compute $E(X|\mathcal{G})$?
First notice that if $A\in\mathcal{G}$ then $P(A)=0$ or $P(A)=1$ . Here $P=\lambda$ .
So , the co-countable sigma algebra is a trivial one with respect to $\lambda$.
Now show that $E(X|\mathcal{G})=E(X)$ for any random variable $X$ and a trivial sigma algebra $\mathcal{G}$ when we are working on any probability space $(\Omega,\mathcal{F},P)$ .
i.e. if $B\in \mathcal{G}$ and $P(B)=1$ , then $\int_{B}X\,dP=\int_{\Omega} X\, dP=E(X)=E(X)P(B)=\int_{B}E(X)\,dP$
and obviously if $P(B)=0$ , then again $\int_{B}X\,dP=0=E(X)P(B)=\int_{B}E(X)\,dP$ . This holds for all $B\in\mathcal{G}$. Thus $E(X|\mathcal{G})\stackrel{\text{a.s.}}{=}E(X)$