Given $ (X_n)_{n\geq0}$ is a non-negative supermartingale and $X_{\infty}$ is an almost sure limit, I want to show that $ \forall n\geq0$ $$\mathbb{E}(X_\infty \mid \mathcal{F}_n) \leq X_n$$
My approach:
First, for $(X_n)$ a supermartingale I would say we then have $$ \mathbb{E}(X_m\mid\mathcal{F}_n)\leq X_n \quad \forall m\geq n$$ Now, my assumption is that $X_m$ does not necessarily converge to $X_\infty$ in $L^1$, so $\mathbb{E}(X_m)$ does not necessarily converge to $\mathbb{E}(X_\infty)$. Hence my idea would be to show that $X_m$ is uniformly integrable, where then we could show, for $X_m \longrightarrow X_\infty$ a.s in $L^1$ then $\mathbb{E}(X_m\mid\mathcal{F}_n) = \mathbb{E}(X_\infty\mid\mathcal{F}_n)$.
However, I think my approach is way off. Could someone please help me out, it would be greatly appreciated.
By conditional Fatou, $E(\liminf_k X_{n+k}\mid\mathcal F_n)\leq \liminf_k E(X_{n+k}\mid\mathcal F_n)$, hence $$E(X_\infty\mid\mathcal F_n)\leq \liminf_k E(X_{n+k}\mid\mathcal F_n)$$
Since $\forall k, E(X_{n+k}\mid\mathcal F_n)\leq X_n$, we have $\liminf_k E(X_{n+k}\mid\mathcal F_n)\leq X_n$ and the result follows.