I am studying this book where Theorem 4.2.12. on page 222 is the following:
If $X_n \geq 0$ is a supermartingale then as $n \to \infty$, $X_n \to X$ almost surely and $\mathbb{E}[X] \leq \mathbb{E}[X_0]$. This theorem needs the sequence of random variable be non-negative.
Question:
Is it possible to replace the nonnegativity of $X_n \geq 0$ to $X_n \geq \alpha$ for some $\alpha \in \mathbb{R}$. If so, how the proof goes?
I'd generalize the theorem like this:
Corollary
If $X_n \ge \alpha \in \mathbb{R}$ is a supermartingale, then as $n \to + \infty, \; X_n \to X$ a.s. and $\mathbb{E}[X] \le \mathbb{E}[X_0]$.
Proof:
Let $Y_n := X_n - \alpha$. This process is a non-negative supermartingale. Indeed:
$$ \bullet \quad Y_n \ge 0 \\ \bullet \quad \mathbb{E}[Y_{n+1}|\mathcal{F}_n] = \mathbb{E}[X_{n+1}|\mathcal{F}_n] - \alpha \overset{\mathrm{X_n supermart}}{\ge} X_n - \alpha = Y_n $$ Hence we can apply the theorem you provided and conclude that $Y_n$ converges to some $Y$ a.s. So, $X_n \to \alpha + Y =: X$.
For the inequality you have the same steps that in the old theorem