is a point $z_1 \in |\gamma_j| \cap \zeta_1$. In this case it follows that $0 = \inf_{z \in \partial_* U_j} u(z) = u(z_1) = \min_{z \in |\gamma_j|} u(z)$. On the other hand, if $\zeta_1 \cap \partial_* U_j = \emptyset$, then by Lemma \[lemm:maximum\_principle\] we again have $\inf_{z \in \partial_* U_j} u(z) = \min_{z \in |\gamma_j|} u(z)$.

The same argument shows that if $\zeta_3 \cap \partial_* U_j \neq \emptyset$, then there exists $y_1 \in |\gamma_j| \cap \zeta_3$ such that $1 = \sup_{y \in \partial_* U_j} u(y) = u(y_1) = \max_{y \in \gamma_j} u(y)$. In general, we likewise have $\sup_{y \in \partial_* U_j} u(y) = \max_{y \in |\gamma_j|} u(y)$. This establishes the equality $$\sup_{y,z \in \partial_* U_j} |u(y) - u(z)| = \max_{y,z \in |\gamma_j|} |u(y) - u(z)| .$$ By the upper gradient inequality , $$\max_{y,z \in |\gamma_j|} |u(y) - u(z)| \leq \int_{\gamma_j} \rho\,d\mathcal{H}^1.$$ Finally, combining the estimates gives $$\mathcal{H}^1(u(B(x,r) \cap Q)) \leq \sum_j \operatorname{diam}u(U_j) \leq \sum_j \int_{\gamma_j} \rho\, d\mathcal{H}^1 \leq \int_{S(x,s)} \rho\, d\mathcal{H}^1.$$ Observe that this estimate is the same independent of our choice of $s$. Inequality then follows from integrating over $s$ from $r$ to $2r$ and applying Proposition \[prop:coarea\].

The same argument also verifies inequality , since for each choice 