Iterated conditioning I understand how this holds:
$\mathbb{E}(\mathbb{E}(X|\mathit{G})|\mathit{H}) = \mathbb{E}(X|\mathit{H})$
where $\mathit{H} \subset \mathit{G} $. 
Then, how does this hold:
$\mathbb{E}(\mathbb{E}(X|\mathit{G})) = \mathbb{E}(X)$
as here, $\mathit{H} = \Omega$ and now $\mathit{G} \subset \Omega$ not $\Omega \subset \mathit{G}$.
 A: Let $(\Omega,\mathcal{F},P)$ be a probability space.
Taking ordinary expectation does not correspond to taking the conditional expectation conditioned on $\mathcal{F}$ (which is probably what you intended to write, since conditioning on $\Omega$ does not make sense). Instead, it corresponds to conditioning on the trivial sigma-field $\{\varnothing,\Omega\}$, i.e.
$$
{\rm E}[X]={\rm E}[X\mid\{\varnothing,\Omega\}]
$$
for any integrable variable $X$. 
In particular ${\rm E}[{\rm E}[X\mid\mathcal{G}]]={\rm E}[X]$ does not contradict the statement that
$$
{\rm E}[{\rm E}[X\mid\mathcal{G}]\mid\mathcal{H}]={\rm E}[X\mid\mathcal{H}]
$$
whenever $\mathcal{H}\subseteq\mathcal{G}$ since the trivial sigma-field $\{\varnothing,\Omega\}$ is a subset of every sigma-field.
A: 
Here $H$ and $G$ are sigma-algebras, not events. 

For example $E(E(X\mid G))=E(X)$ for every sigma-algebra $G$ is the case $H=\{\varnothing,\Omega\}$ of the preceding identity [$E(E(X\mid G)\mid H)=E(X\mid H)$ if $H\subseteq G$], since for this sigma-algebra $H$, $H\subseteq G$ is always true and $E(Y\mid H)=E(Y)$ for every (integrable) random variable.
