[proofplan] Each property is established by exhibiting a candidate for $\mathbb{E}[\cdot \mid \mathcal{G}]$ and verifying the two defining conditions: $\mathcal{G}$-measurability and the [integral](/page/Integral)-matching identity. The arguments are short and self-contained, relying only on the uniqueness established in the [Existence and Uniqueness of Conditional Expectation](/theorems/1147). [/proofplan] [step:Prove the averaging property by testing on $A = \Omega$] **(i)** Taking $A = \Omega \in \mathcal{G}$ in the integral-matching condition $\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A] = \mathbb{E}[X \mathbb{1}_A]$ gives: \begin{align*} \mathbb{E}[\mathbb{E}[X \mid \mathcal{G}]] = \mathbb{E}[X]. \end{align*} [/step] [step:Verify that $\mathcal{G}$-[measurable functions](/page/Measurable%20Functions) condition onto themselves] **(ii)** If $X$ is $\mathcal{G}$-measurable, we claim $\mathbb{E}[X \mid \mathcal{G}] = X$ a.s. The candidate $Y = X$ is $\mathcal{G}$-measurable by hypothesis, and the integral-matching condition is the tautology $\mathbb{E}[X \mathbb{1}_A] = \mathbb{E}[X \mathbb{1}_A]$ for all $A \in \mathcal{G}$. By the uniqueness of conditional expectation ([Existence and Uniqueness of Conditional Expectation](/theorems/1147)), $\mathbb{E}[X \mid \mathcal{G}] = X$ a.s. [/step] [step:Show that conditioning on an independent $\sigma$-algebra yields the expectation] **(iii)** If $X$ is independent of $\mathcal{G}$, the candidate is the constant $Y = \mathbb{E}[X]$. This is $\mathcal{G}$-measurable (every constant is measurable with respect to any $\sigma$-algebra). For the integral-matching condition, independence gives $\mathbb{E}[X \mathbb{1}_A] = \mathbb{E}[X] \, \mathbb{P}(A)$ for all $A \in \mathcal{G}$, and: \begin{align*} \mathbb{E}[\mathbb{E}[X] \cdot \mathbb{1}_A] = \mathbb{E}[X] \, \mathbb{P}(A) = \mathbb{E}[X \mathbb{1}_A]. \end{align*} By uniqueness, $\mathbb{E}[X \mid \mathcal{G}] = \mathbb{E}[X]$ a.s. [/step] [step:Deduce positivity from the existence proof] **(iv)** This was established in the [Existence and Uniqueness of Conditional Expectation](/theorems/1147): if $X \geq 0$, the $L^2$ construction yields $Y \geq 0$ a.s. (by testing on $\{Y < 0\} \in \mathcal{G}$), and the monotone extension preserves this property. [/step] [step:Verify linearity by checking the defining conditions for $\alpha \mathbb{E}[X \mid \mathcal{G}] + \beta \mathbb{E}[Y \mid \mathcal{G}]$] **(v)** The candidate $Z = \alpha \, \mathbb{E}[X \mid \mathcal{G}] + \beta \, \mathbb{E}[Y \mid \mathcal{G}]$ is $\mathcal{G}$-measurable as a linear combination of $\mathcal{G}$-measurable [functions](/page/Function). For any $A \in \mathcal{G}$: \begin{align*} \mathbb{E}[Z \mathbb{1}_A] &= \alpha \, \mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A] + \beta \, \mathbb{E}[\mathbb{E}[Y \mid \mathcal{G}] \, \mathbb{1}_A] \\ &= \alpha \, \mathbb{E}[X \mathbb{1}_A] + \beta \, \mathbb{E}[Y \mathbb{1}_A] \\ &= \mathbb{E}[(\alpha X + \beta Y) \mathbb{1}_A]. \end{align*} By uniqueness, $\mathbb{E}[\alpha X + \beta Y \mid \mathcal{G}] = \alpha \, \mathbb{E}[X \mid \mathcal{G}] + \beta \, \mathbb{E}[Y \mid \mathcal{G}]$ a.s. [/step] [step:Derive the conditional triangle inequality from positivity and linearity] **(vi)** Since $X \leq |X|$, the positivity property (iv) applied to $|X| - X \geq 0$ gives $\mathbb{E}[|X| - X \mid \mathcal{G}] \geq 0$ a.s. By linearity (v): \begin{align*} \mathbb{E}[X \mid \mathcal{G}] \leq \mathbb{E}[|X| \mid \mathcal{G}] \quad \text{a.s.} \end{align*} Applying the same argument to $-X \leq |X|$ yields $-\mathbb{E}[X \mid \mathcal{G}] \leq \mathbb{E}[|X| \mid \mathcal{G}]$ a.s. Combining the two inequalities: \begin{align*} |\mathbb{E}[X \mid \mathcal{G}]| \leq \mathbb{E}[|X| \mid \mathcal{G}] \quad \text{a.s.} \end{align*} [/step]