= Nested conditioning keeps the coarser information
{title2=$\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H]=\mathbb E[X\mid\mathcal H]\quad(\mathcal H\subseteq\mathcal G)$}
For <integrable random variable> $X$ and nested sub-<sigma-algebras> $\mathcal H\subseteq\mathcal G$, the <tower property> says
$$
\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H]
=\mathbb E[X\mid\mathcal H].
$$
The finer-information <conditional mean> can be averaged within each coarser group, recovering the coarser-information <conditional mean>. Conversely, conditioning an already $\mathcal H$-measurable <conditional mean> on $\mathcal G$ leaves it unchanged. Both orders therefore recover $\mathbb E[X\mid\mathcal H]$ when the information is nested.<Probability: Theory and Examples (fifth edition) by Rick Durrett/4.1 Conditional Expectation>[Theorem 4.1.13, p. 212]{ref}
For the finite example in <law of total expectation>, exact knowledge of $X$ gives the finer mean $X$ itself. Averaging within the groups $A,B$ recovers two and eight. Forgetting the group then gives $13/2$. The nesting condition is essential; <iterated conditional expectation over nonnested sigma-algebras> describes why arbitrary successive conditioning cannot simply be replaced by conditioning on an intersection.
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