Proof of the law of total expectation

Proof

We apply the defining property of the conditional expectation to the whole space.

1. Let $Z = E[X \mid \mathcal{G}]$ be a conditional expectation of $X$ given $\mathcal{G}$. By definition, $Z$ is integrable and $\mathcal{G}$-measurable.
2. The definition also gives $E[Z \, 1_G] = E[X \, 1_G]$ for each $G \in \mathcal{G}$.
3. A σ-algebra on $\Omega$ contains $\Omega$. So $\Omega \in \mathcal{G}$, and step 2 applies with $G = \Omega$.
4. The indicator $1_\Omega$ is the constant $1$. So step 3 reads $E[Z] = E[X]$, where $E$ is the expected value.
5. Two versions of $E[X \mid \mathcal{G}]$ are equal almost surely, so they have the same expectation. Thus step 4 does not depend on the choice of $Z$, and $E[E[X \mid \mathcal{G}]] = E[X]$.
6. Let $Y$ be a random variable on the same space. By definition, $E[X \mid Y] = E[X \mid \sigma(Y)]$, and $\sigma(Y) \subseteq \mathcal{F}$ is a σ-algebra.
7. Step 5 with $\mathcal{G} = \sigma(Y)$ gives $E[E[X \mid Y]] = E[X]$.