{\text{vol}}_G(S_i).$$ Recall also that $U_t = U_0 \cup (\bigcup_{i=0}^{t-1}S_i)$. Using that volume is additive, i.e. ${\text{vol}}_{G}(U_t) = {\text{vol}}_G(U_0) + \sum_{i=0}^{t-1}{\text{vol}}_{G}(S_i)$, we have $$\begin{aligned}
            e_G(U_t,\overline{U_t}) 
= e_G(U_0, \overline{U_t}) + \sum_{i=0}^{t-1}e_G&(S_i,\overline{U_t})
\leq {\text{vol}}_G(U_0) + \sum_{i=0}^{t-1}e_G(S_i, U_i \setminus S_i) \\
            &\overset{\eqref{eq:volU0}, \eqref{eq:vol2}}{\leq} \nu n'^2 + \sum_{i=0}^{t-1}\phi \; {\text{vol}}_{G}(S_i) \\
            &\leq \nu n'^2 + \phi \; {\text{vol}}_G(U_t) \leq  \nu n'^2 + \phi |U_t|  n.    
            \end{aligned}$$ Therefore $$\begin{aligned}
  e_G(U_t,\overline{U_t}) \leq \nu n'^2 + \phi |U_t|  n   \overset{\eqref{eq:Ut}}{\leq} \nu n'^2 +  \frac{2}{3}\phi  |U|  n 
&\overset{{\rm Prop}\, \ref{Claim1}}{\leq} \nu n'^2 + \frac{\phi }{\alpha-\sqrt{\rho}}n'^2 \\
&\overset{\nu, \phi \ll \rho'}{\leq} \frac{1}{2}\rho'n'^2.\end{aligned}$$ Hence $e_G(U_t,V\setminus U_t)\leq \frac{1}{2} \rho' n'^2+\rho n^2\leq \rho' n^2$ since $U_t \subseteq U$ and $U$ is a $\rho$-component. Similarly $e_G(\overline{U_t},V\setminus\overline{U_t})\leq \rho' n^2$. Also, $|U_t|$, $|\overline{U_t}| \geq \frac{1}{4}\rho'n$ by (a) and . However, by Proposition \[Claim1\], we in fact have $|U_t|$, $|\overline{U_t}| \geq (\alpha-\rho'^2)n \geq \sqrt{\rho'}n$, so $U_t$ and $\overline{U_t}$ are $\rho'$-components.

First some observations. Since case (b) holds, $|U_t|\leq \frac{1}{4}\rho'n'\leq \frac{1}{2}\tau n'\leq \frac{1}{3}|U|$ and $\phi_t = \Phi_{G_t}(S_t)>\phi$.

Also, $\delta(G_t) = \delta(G[\overline{U_t}])
        \geq \min_{x \in \overline{U_t}} d_U(x) - |U_t| \geq  \frac{1}{2}\alpha n - \frac{1}{2}\tau n' \geq \frac{1}{3} \alpha n$, where the penultimate inequality follows from our choice of $U_0$. By Theorem \[th:cheeger\], for all $R \subseteq V(G_t)=U\setminus U_t$ we 