+ {\mathtt{ub}}')/2
  \\
  &> ({\mathtt{ub}}' + {\mathtt{ub}}')/2 
  \\
  &= {\mathtt{ub}}'.\end{aligned}$$ If $v = z$, then ${\mathtt{w}}'(z) = {\mathtt{w}}'(v) > {\mathtt{ub}}'/2$, and thus $$\begin{aligned}
  {\mathtt{w}}''(z) + {\mathtt{w}}''(v)
  &= 2{\mathtt{w}}'(z) - \varepsilon
  \\
  &\ge 2{\mathtt{w}}'(z) - |{\mathtt{w}}'(z) - {\mathtt{ub}}'/2|
  = 2{\mathtt{w}}'(z) - ({\mathtt{w}}'(z) - {\mathtt{ub}}'/2)
  \\
  &= {\mathtt{w}}'(z) + {\mathtt{ub}}'/2
  \\
  &> {\mathtt{ub}}'.\end{aligned}$$

Finally, we show that $\{v \mid {\mathtt{lb}}/2 \le {\mathtt{w}}(v) \le {\mathtt{ub}}/2\} = \{v \mid {\mathtt{lb}}'/2 \le {\mathtt{w}}''(v) \le {\mathtt{ub}}'/2\}$. To this end, it suffices to show that ${\mathtt{lb}}'/2 < {\mathtt{w}}'(z) < {\mathtt{ub}}'/2$ if and only if ${\mathtt{lb}}'/2 < {\mathtt{w}}''(z) < {\mathtt{ub}}'/2$ as $2{\mathtt{w}}'(z), 2{\mathtt{w}}''(z) \notin \{{\mathtt{lb}}', {\mathtt{ub}}'\}$. If ${\mathtt{lb}}'/2 < {\mathtt{w}}'(z) < {\mathtt{ub}}'/2$, then ${\mathtt{w}}''(z) < {\mathtt{w}}'(z) < {\mathtt{ub}}'/2$ and $${\mathtt{w}}''(z)
  = {\mathtt{w}}'(z) - \varepsilon/2 
  > {\mathtt{w}}'(z) - \varepsilon
  \ge  {\mathtt{w}}'(z) - |{\mathtt{w}}'(z) - {\mathtt{lb}}'/2| 
  = {\mathtt{w}}'(z) - ({\mathtt{w}}'(z) - {\mathtt{lb}}'/2) 
  = {\mathtt{lb}}'/2.$$ If ${\mathtt{w}}'(z) < {\mathtt{lb}}'/2$, then ${\mathtt{w}}''(z) < {\mathtt{w}}'(z) < {\mathtt{lb}}'/2$. If ${\mathtt{w}}'(z) > {\mathtt{ub}}'/2$, then we have $${\mathtt{w}}''(z)
  = {\mathtt{w}}'(z) - \varepsilon/2 
  > {\mathtt{w}}'(z) - \varepsilon
  \ge {\mathtt{w}}'(z) - |{\mathtt{w}}'(z) - {\mathtt{ub}}'/2|
  = {\mathtt{w}}'(z) - ({\mathtt{w}}'(z) - {\mathtt{ub}}'/2) 
  = {\mathtt{ub}}'/2.$$

We have shown that ${\mathtt{w}}''$, ${\mathtt{lb}}'$, and ${\mathtt{ub}}'$ define $G$, $\{v \mid {\mathtt{lb}}/2 \le {\mathtt{w}}(v) \le {\mathtt{ub}}/2\} = \{v \mid {\mathtt{lb}}'/2 \le {\mathtt{w}}''(v) \le {\mathtt{ub}}'/2\}$, and 