This is regarding the proof of a central theorem (Theorem 4.1) in Hassan Khalil's seminal Nonlinear systems book:
Theorem 4.1: Let $x = 0$ be an equilibrium point for $\dot{x} = f(x)$ where $f:D\rightarrow \mathbb{R}^n$ is a locally Lipschitz map from a domain $D \subset \mathbb{R}^n$ into $\mathbb{R}^n$. Let $V:D \rightarrow \mathbb{R}$ (the Lyapunov function) be a continuously differentiable function such that $$V(0) = 0 \text{ and } V(x) > 0 \text{ in } D - \{0\}$$ $$\dot{V}(x) \leq 0 \text{ in } D$$ Then, $x=0$ is stable.
My concern is that Khalil defines $\Omega_\beta = \{x \in B_r \mid V(x) \leq \beta\}$ and claims that a trajectory starting from $x(0) \in \Omega_\beta$ stays in $\Omega_\beta$ for all $t \geq 0$ just because $\dot{V}(x) \leq 0 \implies V(x(t)) \leq \beta,\ \forall t\geq 0$. But note that $\Omega_\beta$ is a restriction of the set $\{x \mid V(x) \leq \beta\}$ to $B_r$; $V(x(t)) \leq \beta$ does not rule out the possibility that the trajectory goes outside the ball $B_r$.
I feel like there's an argument about the trajectory not being able to "cross" the boundary $\|x\| = r$ that's required, but Khalil does not make? I can informally see why this must be true since the trajectory cannot lie on the boundary ($V(x) > \beta$ on the boundary), and it must "cross" the boundary if it needs to go from inside $B_r$ to outside.
I thought Khalil's book was pretty rigorous and complete, so I'm really doubting that I'm misunderstanding something here.