[proofplan] We approximate the unbounded stopping time $T$ by the bounded stopping times $\min\{T, t\}$, use the supermartingale property to bound $\mathbb{E}[X_{\min\{T,t\}}] \leq \mathbb{E}[X_0]$, and pass to the [limit](/page/Limit) $t \to \infty$ using [Fatou's Lemma](/theorems/510). The non-negativity of $X$ is essential: it ensures that Fatou's lemma applies (providing the inequality in the correct direction) and that the stopped process remains a supermartingale. [/proofplan] [step:Verify that the stopped process $X_{\min\{T,t\}}$ satisfies $\mathbb{E}[X_{\min\{T,t\}}] \leq \mathbb{E}[X_0]$] The stopped process $(X_{\min\{T,n\}})_{n \geq 0}$ is a supermartingale: the argument from the [Optional Stopping Theorem](/theorems/1153), part (i), adapts to supermartingales by replacing the equality $\mathbb{E}[X_t \mid \mathcal{F}_{t-1}] = X_{t-1}$ with the inequality $\mathbb{E}[X_t \mid \mathcal{F}_{t-1}] \leq X_{t-1}$, yielding $\mathbb{E}[X_{\min\{T,t\}} \mid \mathcal{F}_{t-1}] \leq X_{\min\{T,t-1\}}$. In particular, taking expectations iteratively: \begin{align*} \mathbb{E}[X_{\min\{T,t\}}] \leq \mathbb{E}[X_{\min\{T,t-1\}}] \leq \cdots \leq \mathbb{E}[X_0]. \end{align*} [/step] [step:Pass to the limit using Fatou's lemma] Since $T < \infty$ a.s., we have $\min\{T, t\} \to T$ a.s. as $t \to \infty$, hence $X_{\min\{T,t\}} \to X_T$ a.s. The random variables $X_{\min\{T,t\}}$ are non-negative (since $X$ is a non-negative supermartingale), so [Fatou's Lemma](/theorems/510) applies: \begin{align*} \mathbb{E}[X_T] = \mathbb{E}\bigl[\liminf_{t \to \infty} X_{\min\{T,t\}}\bigr] \leq \liminf_{t \to \infty} \mathbb{E}[X_{\min\{T,t\}}] \leq \mathbb{E}[X_0]. \end{align*} [guided] Why is non-negativity needed? Fatou's lemma requires $X_{\min\{T,t\}} \geq 0$ a.s. Without this, the $\liminf$ inequality can fail. For example, if $(X_n)$ is a simple symmetric random walk (a martingale taking negative values) and $T = \inf\{n : X_n = -1\}$, then $\mathbb{E}[X_T] = -1 \neq 0 = \mathbb{E}[X_0]$, and neither Fatou nor DCT can be applied without further integrability conditions. The argument also fails without $T < \infty$ a.s.: if $T = \infty$ with positive probability, then $X_T$ is not well-defined on the event $\{T = \infty\}$ (or requires a convention about $X_\infty$, which introduces complications about the existence of the a.s. limit). [/guided] [/step]