[proofplan]
We exploit the relationship between local martingales and Fatou's lemma for conditional expectations. Let $(T_n)$ be a reducing sequence for $X$. Since $X$ is non-negative, $X_t = \liminf_{n \to \infty} X_{t \wedge T_n}$ almost surely (by right-continuity of paths and $T_n \to \infty$). Each stopped process $X^{T_n}$ is a genuine martingale, so conditional Fatou's lemma — applied to the non-negative sequence $(X_{t \wedge T_n})_{n \geq 1}$ — yields the supermartingale inequality $\mathbb{E}[X_t \mid \mathcal{F}_s] \leq X_s$. A separate integrability check via Fatou confirms $\mathbb{E}[X_t] < \infty$ for every $t$.
[/proofplan]
[step:Verify integrability of $X_t$ for each $t$ via Fatou's lemma]
Let $(T_n)_{n \geq 1}$ be a reducing sequence for $X$: each $T_n$ is a stopping time, $T_n \leq T_{n+1}$, $T_n \to \infty$ a.s., and $X^{T_n}$ is a uniformly integrable martingale for each $n$.
Since $X^{T_n}$ is a martingale, the optional sampling theorem gives $\mathbb{E}[X_{t \wedge T_n}] = \mathbb{E}[X_0]$ for every $t \geq 0$ and $n \geq 1$. As $T_n \to \infty$ a.s. and $X$ is càdlàg, we have $t \wedge T_n \to t$ and therefore $X_{t \wedge T_n} \to X_t$ a.s. Since $X_{t \wedge T_n} \geq 0$ for all $n$, Fatou's lemma gives
\begin{align*}
\mathbb{E}[X_t] = \mathbb{E}\!\left[\liminf_{n \to \infty} X_{t \wedge T_n}\right] \leq \liminf_{n \to \infty} \mathbb{E}[X_{t \wedge T_n}] = \mathbb{E}[X_0] < \infty.
\end{align*}
Hence $X_t \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ for every $t \geq 0$.
[guided]
Before we can talk about $\mathbb{E}[X_t \mid \mathcal{F}_s]$, we need $X_t$ to be integrable. A local martingale is only required to have $X_{t \wedge T_n} \in L^1$ for each $n$ and $t$, not $X_t \in L^1$ itself. So integrability is not automatic and must be established.
Let $(T_n)_{n \geq 1}$ be a reducing sequence for $X$: each $T_n$ is a stopping time, $T_n \leq T_{n+1}$, $T_n \to \infty$ a.s., and the stopped process $X^{T_n}$ is a uniformly integrable martingale for each $n$.
Since $X^{T_n}$ is a martingale, the optional sampling theorem applied at the deterministic time $t$ gives
\begin{align*}
\mathbb{E}[X_{t \wedge T_n}] = \mathbb{E}[X_{0 \wedge T_n}] = \mathbb{E}[X_0],
\end{align*}
where the second equality holds because $T_n > 0$ a.s. (we may assume this WLOG by replacing $T_n$ with $T_n \vee \varepsilon$ and sending $\varepsilon \to 0$, or by noting that $X_{0 \wedge T_n} = X_0$ since $X$ is càdlàg and $T_n$ is a.s. positive for large $n$).
As $n \to \infty$, $T_n \to \infty$ a.s., so $t \wedge T_n \to t$ and the càdlàg property of $X$ gives $X_{t \wedge T_n} \to X_t$ a.s. The non-negativity $X_{t \wedge T_n} \geq 0$ for all $n$ is what allows us to invoke Fatou's lemma (which requires a.e. non-negativity of the sequence):
\begin{align*}
\mathbb{E}[X_t] = \mathbb{E}\!\left[\liminf_{n \to \infty} X_{t \wedge T_n}\right] \leq \liminf_{n \to \infty} \mathbb{E}[X_{t \wedge T_n}] = \mathbb{E}[X_0] < \infty.
\end{align*}
Therefore $X_t \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ for every $t \geq 0$, and in fact $\mathbb{E}[X_t] \leq \mathbb{E}[X_0]$, which already hints at the supermartingale inequality.
[/guided]
[/step]
[step:Apply conditional Fatou's lemma to establish the supermartingale inequality]
Fix $0 \leq s \leq t$. Since $X^{T_n}$ is a martingale and $s \wedge T_n \leq t \wedge T_n$ are bounded stopping times,
\begin{align*}
\mathbb{E}[X_{t \wedge T_n} \mid \mathcal{F}_s] = X_{s \wedge T_n} \quad \text{a.s.}
\end{align*}
for each $n \geq 1$. As $n \to \infty$, $X_{t \wedge T_n} \to X_t$ a.s. (by $T_n \to \infty$ and càdlàg paths). Since $X_{t \wedge T_n} \geq 0$ for all $n$, the conditional version of Fatou's Lemma gives
\begin{align*}
\mathbb{E}[X_t \mid \mathcal{F}_s] = \mathbb{E}\!\left[\liminf_{n \to \infty} X_{t \wedge T_n} \;\Big|\; \mathcal{F}_s\right] \leq \liminf_{n \to \infty} \mathbb{E}[X_{t \wedge T_n} \mid \mathcal{F}_s] = \liminf_{n \to \infty} X_{s \wedge T_n} = X_s,
\end{align*}
where the final equality holds because $s \wedge T_n \to s$ as $T_n \to \infty$, and $X_{s \wedge T_n} \to X_s$ a.s. by right-continuity of paths. Combined with the integrability established in the previous step, this shows $X$ is a supermartingale.
[guided]
Fix $0 \leq s \leq t$. We want to show $\mathbb{E}[X_t \mid \mathcal{F}_s] \leq X_s$ a.s. The strategy is to use the martingale property of $X^{T_n}$ at the finite level and then pass to the limit using Fatou.
Since $X^{T_n}$ is a martingale with respect to the filtration $(\mathcal{F}_t)_{t \geq 0}$, the tower property of conditional expectation gives
\begin{align*}
\mathbb{E}[X_{t \wedge T_n} \mid \mathcal{F}_s] = X_{s \wedge T_n} \quad \text{a.s.}
\end{align*}
Why does this hold? The stopped process $X^{T_n}$ is a martingale, and $s \leq t$ are deterministic times, so the martingale property directly gives $\mathbb{E}[X^{T_n}_t \mid \mathcal{F}_s] = X^{T_n}_s$, i.e., $\mathbb{E}[X_{t \wedge T_n} \mid \mathcal{F}_s] = X_{s \wedge T_n}$.
Now we want to pass $n \to \infty$. As $T_n \to \infty$ a.s., we have $X_{t \wedge T_n} \to X_t$ a.s. by the càdlàg property. The question is: can we push the limit inside the conditional expectation? This is where the non-negativity hypothesis $X_t \geq 0$ is crucial. The conditional version of Fatou's lemma states: if $(Y_n)_{n \geq 1}$ are non-negative random variables and $\mathcal{G}$ is a sub-$\sigma$-algebra, then
\begin{align*}
\mathbb{E}\!\left[\liminf_{n \to \infty} Y_n \;\Big|\; \mathcal{G}\right] \leq \liminf_{n \to \infty} \mathbb{E}[Y_n \mid \mathcal{G}] \quad \text{a.s.}
\end{align*}
The hypothesis for conditional Fatou is $Y_n \geq 0$ a.s. for all $n$. Since $X_{t \wedge T_n} \geq 0$ by assumption, we may apply conditional Fatou with $Y_n = X_{t \wedge T_n}$ and $\mathcal{G} = \mathcal{F}_s$:
\begin{align*}
\mathbb{E}[X_t \mid \mathcal{F}_s] &= \mathbb{E}\!\left[\liminf_{n \to \infty} X_{t \wedge T_n} \;\Big|\; \mathcal{F}_s\right] \\
&\leq \liminf_{n \to \infty} \mathbb{E}[X_{t \wedge T_n} \mid \mathcal{F}_s] \\
&= \liminf_{n \to \infty} X_{s \wedge T_n} \\
&= X_s,
\end{align*}
where the last equality uses $s \wedge T_n \to s$ and $X_{s \wedge T_n} \to X_s$ a.s. (right-continuity). Note that the $\liminf$ on the right-hand side is actually a limit, since $X_{s \wedge T_n} \to X_s$ pointwise.
This proves the supermartingale inequality $\mathbb{E}[X_t \mid \mathcal{F}_s] \leq X_s$ a.s. for all $0 \leq s \leq t$. Together with integrability ($X_t \in L^1$ for every $t$, established in the previous step) and adaptedness (inherited from the local martingale), this shows $X$ is a supermartingale.
What would fail without non-negativity? Without the assumption $X_t \geq 0$, we could not apply Fatou's lemma, because Fatou requires non-negative (or at least uniformly bounded below) random variables. A general local martingale need not be a supermartingale — the inequality can fail when $X$ takes large negative values and the sequence $X_{t \wedge T_n}$ oscillates.
[/guided]
[/step]
