In the book *Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I*, page 186 ([MR4436025](https://mathscinet.ams.org/mathscinet-getitem?mr=4436025), [Zbl 1497.35005](https://zbmath.org/1497.35005)),  the authors proves a unique continuation theorem which is the argument of this question.<br>
In the following $P=\sum_{i,j=1}^{n}\partial_{i}(p_{ij}\partial_{j}\cdot)$ is a second order partial differential operator and the coefficient $p_{ij}$ is smooth.

**Theorem 5.2.** Let $\Omega$ be a connected open subset in $\mathbb{R}^{d}$ and let $\omega\subset\Omega$, with $\omega\neq\emptyset$. If $u\in H^{2}(\Omega)$ satisfies$$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$
for some $C>0$ and $u(x)=0$ in $\omega$, then $u$ vanishes in $\Omega$.

**Sketch of Proof**: the authors set $F=\operatorname{supp}(u)$ (so $F$ is a closed set according the definition of support set of function), and then they prove that $F$ is also open to get $F=\emptyset$ and thus $u=0$.<br> 
They take a point $x^{(1)}$ in $F\setminus \operatorname{int}(F)$ and work in order to get a contradiction by the following proposition (Proposition 5.1 in the book). Precisely, their proof of by contradiction is worked out by constructing a family of balls
$$
\mathscr{B}_{t}=B\big(x^{(0)},(1-t)r_{1}+tr_{2}\big)
$$ where $x^{(0)}\in\Omega\setminus F$, and then proving (according to the said proposition) that **if u vanishes in $\mathscr{B}_{t}$ with $0\leq t\leq1$, then there exists $\epsilon>0$ such that $u$ vanishes in $\mathscr{B}_{t+\epsilon}$**.<br> 
Here, we just know $B(x^{(0)};r_{1})\subset\Omega\setminus F$ in which $u=0$: now the question is **how can we push the radius from $r_{1}$ to $r_{2}$**?<br> 
The $\epsilon$ in each step is different, so how does there exist such a situation, that in each step, the increase of radius is$\frac{1}{10},\frac{1}{10^{2}},\frac{1}{10^{3}},\dots$ so that we can not intersect $B(x^{(0)};r_{2})$?<br>
Is it just as in ordinary differential equation, when we study the extension of classical solution of $\dot{x}(t)=f(x(t))$?<br> Everytime we extend a little, but how the total step is finite?

**Proposition 5.1.** Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\text{ a.e.  in }\Omega$$ for some $C>0$ and $u(x)=0$ in $\{x\in V;\phi(x)\geq\phi(x^{(0)})\}$. Then $u$ vanishes in a neighborhood of $x^{(0)}$.<br>
**Remark, the author didn't tell us how large of the neighborhood is.**