[proofplan] We verify the two defining conditions of conditional expectation for $\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \mid \mathcal{H}]$: that the candidate $\mathbb{E}[X \mid \mathcal{H}]$ is $\mathcal{H}$-measurable, and that it satisfies the [integral](/page/Integral)-matching condition on $\mathcal{H}$. The key insight is that every $A \in \mathcal{H}$ also belongs to $\mathcal{G}$ (since $\mathcal{H} \subset \mathcal{G}$), which allows chaining the integral-matching conditions for both conditional expectations. [/proofplan] [step:Verify $\mathcal{H}$-measurability of the candidate] The candidate is $\mathbb{E}[X \mid \mathcal{H}]$, which is $\mathcal{H}$-measurable by the definition of conditional expectation with respect to $\mathcal{H}$. [/step] [step:Chain the integral-matching conditions using the inclusion $\mathcal{H} \subset \mathcal{G}$] For every $A \in \mathcal{H}$, we need to show $\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A] = \mathbb{E}[\mathbb{E}[X \mid \mathcal{H}] \, \mathbb{1}_A]$. Since $\mathcal{H} \subset \mathcal{G}$, the [set](/page/Set) $A$ also belongs to $\mathcal{G}$. Applying the integral-matching condition for $\mathbb{E}[X \mid \mathcal{G}]$ (with test set $A \in \mathcal{G}$) and for $\mathbb{E}[X \mid \mathcal{H}]$ (with test set $A \in \mathcal{H}$): \begin{align*} \mathbb{E}\bigl[\mathbb{E}[X \mid \mathcal{H}] \, \mathbb{1}_A\bigr] = \mathbb{E}[X \mathbb{1}_A] = \mathbb{E}\bigl[\mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A\bigr]. \end{align*} The first equality uses the defining property of $\mathbb{E}[X \mid \mathcal{H}]$ (since $A \in \mathcal{H}$), and the second uses the defining property of $\mathbb{E}[X \mid \mathcal{G}]$ (since $A \in \mathcal{G}$). Since $\mathbb{E}[X \mid \mathcal{H}]$ is $\mathcal{H}$-measurable and satisfies $\mathbb{E}[\mathbb{E}[X \mid \mathcal{H}] \, \mathbb{1}_A] = \mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A]$ for all $A \in \mathcal{H}$, the uniqueness of conditional expectation ([Existence and Uniqueness of Conditional Expectation](/theorems/1147)) gives: \begin{align*} \mathbb{E}\bigl[\mathbb{E}[X \mid \mathcal{G}] \mid \mathcal{H}\bigr] = \mathbb{E}[X \mid \mathcal{H}] \quad \text{a.s.} \end{align*} [guided] The tower property states that "conditioning twice is the same as conditioning on the coarser information." The proof exploits the fact that the integral-matching condition for $\mathbb{E}[X \mid \mathcal{G}]$ is required to hold for *all* $A \in \mathcal{G}$, and since $\mathcal{H} \subset \mathcal{G}$, it holds in particular for all $A \in \mathcal{H}$. This means that $\mathbb{E}[X \mid \mathcal{G}]$ already "agrees with $X$" on the smaller $\sigma$-algebra $\mathcal{H}$ in the sense of integrals. The argument has a satisfying symmetry: $\mathbb{E}[X \mathbb{1}_A]$ serves as the common value linking the two conditional expectations. Reading left to right, $\mathbb{E}[X \mid \mathcal{H}]$ integrates to $\mathbb{E}[X \mathbb{1}_A]$ because $A \in \mathcal{H}$; reading right to left, $\mathbb{E}[X \mid \mathcal{G}]$ integrates to $\mathbb{E}[X \mathbb{1}_A]$ because $A \in \mathcal{G}$. No other [properties of conditional expectation](/theorems/1122) are needed. [/guided] [/step]