Conditional expectation I'm still stuck on conditional probabilities/expectations given a sigma field. Everyone here has been extremely helpful, and I'm hoping someone can do this simple proof, using the definition of conditional expectation. 
$E(E(X|\mathcal{G}))=E(X)$.
 A: By definition, $\mathbb E[X\mid \mathcal G]$ satisfies $$\mathbb E[\mathsf 1_A\mathbb E[X\mid\mathcal G]]=\mathbb E[\mathsf 1_AX] $$ for all $A\in\mathcal G$. Since $\Omega\in\mathcal G$ and $\mathsf 1_\Omega\equiv 1$, we have
$$\mathbb E[\mathbb E[X\mid \mathcal G]] = \mathbb E[\mathsf 1_\Omega\mathbb E[X\mid\mathcal G]] = \mathbb E[\mathsf 1_\Omega X] = \mathbb E[X].$$
A: By definition, $E(X\mid\mathcal{G})$ satisfies $E(XZ)=E\left(E(X\mid\mathcal{G})Z\right)$ whenever $Z$ is $\mathcal{G}$-measurable.
Now use this with $Z=1$, which is $\mathcal{G}$-measurable, as $Z^{-1}(A)\in\{\emptyset,\Omega\}\subset\mathcal{G}$, for any borel set $A$.
A: Of course, as noted in the other answers, $\Omega\in\mathcal{F}$ for any $\sigma$-algebra. However, I will try to spell out the calculation with more detail. 
If $X$ is $\mathcal{G}$-measurable, and $\mathcal{F}\subset\mathcal{G}$, then for any $A\in\mathcal{F}$, $E(\mathbf{1}_AE(X\mid \mathcal{F})=E(\mathbf{1}_AX)$. That's the definition of conditional expectation. Now we'll perform the calculation in the question.
$$
\begin{aligned}
E(E(X|\mathcal{F}))&=\int_\Omega E(X\mid\mathcal{F}) \ P(d\omega)\quad \text{ (by definition of expectation)}\\
&=E(\mathbf{1}_\Omega E(X\mid\mathcal{F})) \\
&=E(\mathbf{1}_\Omega X) \quad \text{ (by definition of conditional expectation since $\Omega\in\mathcal{F}$)}\\
&=\int_\Omega X \ P(d\omega)\\
&=E(X).
\end{aligned}
$$
However, a simple example is always helpful. 
Example:
Let $X$ be uniform on $\{-1,0,1\}$, and $\mathcal{F}=\big\{\emptyset,\{-1\},\{0,1\},\Omega\big\}.$ Let $\Omega=\Omega_{-1}\cup\Omega_0\cup\Omega_1$ such that $X(\omega_i)=i$ for $\omega_i\in\Omega_i$. Let $Y=E(X\mid\mathcal{F})$. Because $E(\mathbf{1}_AY)=E(\mathbf{1}_AX)$ for any $A\in\mathcal{F}$, we deduce that $Y(\omega)=-1$ for any $\omega\in\Omega_{-1}$ and $Y(\omega)=1/2$ for any $\omega\in\Omega_{0}\cup\Omega_{1}$ (a simple but instructive exercise).
Thus, somewhat trivially, $E(Y)=0=E(X)$. However, we'll just go ahead and compute $E(Y)$. 
$$
\begin{aligned}
E(Y)&=\frac{1}{3}\left(Y(\omega_{-1})+Y(\omega_0)+Y(\omega_1)\right) \\
&=\frac{1}{3}\left(-1+\frac{1}{2}+\frac{1}{2}\right)=0
\end{aligned}
$$
