[proofplan] The conditional expectation $X_t=\mathbb E[Y\mid \mathcal F_t]$ is automatically $\mathcal F_t$-measurable and integrable, so the process is adapted and integrable. For the martingale identity, if $s\leq t$, then $\mathcal F_s\subseteq \mathcal F_t$, and the tower property collapses $\mathbb E[\mathbb E[Y\mid \mathcal F_t]\mid \mathcal F_s]$ to $\mathbb E[Y\mid \mathcal F_s]$. This is exactly $X_s$ by definition. [/proofplan] [step:Verify adaptedness and integrability] Fix $t \in T$. By the definition of conditional expectation, $X_t=\mathbb E[Y\mid\mathcal F_t]$ is $\mathcal F_t$-measurable. Hence $(X_t)_{t \in T}$ is adapted to $(\mathcal F_t)_{t \in T}$. Since $Y$ is integrable, the [Basic Properties of Conditional Expectation](/theorems/1148) give that $X_t$ is integrable and \begin{align*} \mathbb E[|X_t|] &\leq \mathbb E[|Y|] < \infty. \end{align*} Thus every $X_t$ is integrable. [/step] [step:Use the tower property to prove the martingale identity] Let $s,t \in T$ with $s \leq t$. Since $(\mathcal F_t)_{t\in T}$ is a filtration, $\mathcal F_s \subseteq \mathcal F_t$. Applying the [Tower Property of Conditional Expectation](/theorems/1150) to the integrable random variable $Y$ and the nested $\sigma$-algebras $\mathcal F_s \subseteq \mathcal F_t$ gives \begin{align*} \mathbb E[X_t \mid \mathcal F_s] &= \mathbb E[\mathbb E[Y\mid \mathcal F_t]\mid \mathcal F_s] \\ &= \mathbb E[Y\mid \mathcal F_s] \\ &= X_s \end{align*} almost surely. [/step] [step:Conclude that the conditional expectation process is a martingale] The process $(X_t)_{t \in T}$ is adapted and integrable at every time, and for every $s \leq t$ it satisfies \begin{align*} \mathbb E[X_t\mid \mathcal F_s] &= X_s \end{align*} almost surely. Therefore $(X_t)_{t \in T}$ is a martingale with respect to $(\mathcal F_t)_{t \in T}$. [/step]