[proofplan] We verify the two defining conditions for $\mathbb{E}[XY \mid \mathcal{G}]$: that the candidate $Y \, \mathbb{E}[X \mid \mathcal{G}]$ is $\mathcal{G}$-measurable (as a product of $\mathcal{G}$-[measurable functions](/page/Measurable%20Functions)), and that the [integral](/page/Integral)-matching condition holds. The integral-matching step uses the characterisation of conditional expectation via bounded $\mathcal{G}$-measurable [test functions](/page/Test%20Function): since $Y \mathbb{1}_A$ is bounded and $\mathcal{G}$-measurable for $A \in \mathcal{G}$, it may replace $\mathbb{1}_A$ in the defining identity. [/proofplan] [step:Verify $\mathcal{G}$-measurability of the candidate $Y \, \mathbb{E}[X \mid \mathcal{G}]$] Both $Y$ and $\mathbb{E}[X \mid \mathcal{G}]$ are $\mathcal{G}$-measurable by hypothesis. Their product is therefore $\mathcal{G}$-measurable. [/step] [step:Verify the integral-matching condition using bounded $\mathcal{G}$-measurable test [functions](/page/Function)] For $A \in \mathcal{G}$, the product $Y \mathbb{1}_A$ is bounded (since $Y$ is bounded) and $\mathcal{G}$-measurable. The integral-matching condition for $\mathbb{E}[X \mid \mathcal{G}]$ extends from indicators $\mathbb{1}_A$ to all bounded $\mathcal{G}$-measurable functions $Z$ (by linearity for [simple functions](/page/Simple%20Function), then approximation by bounded convergence): $\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \cdot Z] = \mathbb{E}[X \cdot Z]$ for all bounded $\mathcal{G}$-measurable $Z$. Applying this extended characterisation with $Z = Y \mathbb{1}_A$: \begin{align*} \mathbb{E}\bigl[Y \, \mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A\bigr] = \mathbb{E}\bigl[\mathbb{E}[X \mid \mathcal{G}] \cdot (Y \mathbb{1}_A)\bigr] = \mathbb{E}[X \cdot Y \mathbb{1}_A] = \mathbb{E}[XY \, \mathbb{1}_A]. \end{align*} By the uniqueness of conditional expectation ([Existence and Uniqueness of Conditional Expectation](/theorems/1147)), $\mathbb{E}[XY \mid \mathcal{G}] = Y \, \mathbb{E}[X \mid \mathcal{G}]$ a.s. [guided] The key technical point is the extension of the integral-matching condition from indicators to bounded $\mathcal{G}$-measurable functions. The defining property states $\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \, \mathbb{1}_A] = \mathbb{E}[X \mathbb{1}_A]$ for all $A \in \mathcal{G}$. By linearity, this extends to all $\mathcal{G}$-measurable simple functions $Z = \sum_{i=1}^n c_i \mathbb{1}_{A_i}$. For a general bounded $\mathcal{G}$-measurable function $Z$, approximate by simple functions $Z_n \to Z$ pointwise with $|Z_n| \leq \|Z\|_\infty$. Since $|\mathbb{E}[X \mid \mathcal{G}] \cdot Z_n| \leq \|Z\|_\infty \cdot |\mathbb{E}[X \mid \mathcal{G}]|$ and $|X \cdot Z_n| \leq \|Z\|_\infty \cdot |X|$, both dominated by integrable functions, the [Dominated Convergence Theorem](/theorems/4) gives: \begin{align*} \mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \cdot Z] = \lim_{n \to \infty} \mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \cdot Z_n] = \lim_{n \to \infty} \mathbb{E}[X \cdot Z_n] = \mathbb{E}[X \cdot Z]. \end{align*} With this extension established, the proof reduces to observing that $Y \mathbb{1}_A$ is a bounded $\mathcal{G}$-measurable function (it is bounded because $Y$ is bounded by hypothesis, and $\mathcal{G}$-measurable because both $Y$ and $\mathbb{1}_A$ are $\mathcal{G}$-measurable). Why is the boundedness of $Y$ needed? Without it, the product $XY$ might fail to be integrable, and the integral $\mathbb{E}[\mathbb{E}[X \mid \mathcal{G}] \cdot Y \mathbb{1}_A]$ might not converge. The result extends to $Y \in L^\infty(\mathcal{G})$ or more generally to $Y$ that is $\mathcal{G}$-measurable with $XY \in L^1$, but the bounded case is the cleanest and most frequently applied. [/guided] [/step]