times the usual Hausdorff 2-measure. For each $m$, let $j_m$ be the largest integer such that $2r_m \leq 10^{-j_m}$. Observe that $10^{-j_m} < 20r_m \leq 20 \cdot 10^{-j}$ and hence that $j_m \geq j-1$.

From Lemma \[lemm:oscillation\] and we deduce $$\begin{aligned}
            r_m \mathcal{H}^1(u(B_m))& \leq \frac{4}{\pi} \int_{2B_m} \rho\,d\mathcal{H}^2 \leq \frac{4}{\pi} \int_{B(x,10^{-j_m})}\rho\,d\mathcal{H}^2  \\ 
            & \leq \frac{4}{\pi} \cdot \frac{1}{200} \int_{B(x,10^{-j_m+1})}\rho\,d\mathcal{H}^2 \\ 
            & \leq \cdots \leq \frac{4}{\pi} \cdot \frac{1}{200^{j_m-k}} \int_{B(x,10^{-k})} \rho\, d\mathcal{H}^2.
            \end{aligned}$$ In particular, $$\label{eq:oscillation}
            r_m \mathcal{H}^1(u(B_m)) \leq \frac{4}{\pi} \cdot \frac{200^k}{200^{j_m}} \int_Q \rho\,d\mathcal{H}^2.$$

Similar to the first step of the proof, for each $m$ fix a measurable $A_m \supset u(B_m)$ such that $\mathcal{H}^1(A_m)=\mathcal{H}^1(u(B_m))$ and define $g_j(t) =  \sum_m r_m\chi_{A_m}(t)$. Then, as before, the definition of $\mathcal{H}_{1/j}^1$ gives $$\label{nakki} 
            \mathcal{H}_{1/j}^1(\overline{u^{-1}(t)} \cap F_k) \leq 2g_j(t)$$ for all $t \in [0,1]$. Integrating gives $$\int_0^1 g_j(t)\,dt \leq \sum_m r_m\mathcal{H}^1(u(B_m)).$$ Applying and using the relationships $1 < 20\cdot 10^{j_m} r_m$ and $j_m \geq j-1$ gives $$\begin{aligned}
            \sum_m r_m\mathcal{H}^1(u(B_m)) & \leq \sum_m \frac{3200}{\pi}\cdot 200^k r_m^2 \left( \frac{100}{200} \right)^{j_m} \int_Q \rho\,d\mathcal{H}^2 \\
            & \leq \frac{3200}{\pi}\cdot 200^k \left( \frac{100}{200} \right)^{j} \left(\int_Q \rho\,d\mathcal{H}^2 \right) \sum_m r_m^2 \\
            & \leq \frac{3200}{\pi}\cdot 200^k \left( \frac{100}{200} \right)^{j} \left(\int_Q \rho\,d\mathcal{H}^2 \right)\left(\mathcal{H}^2(F_k)+1/j \right) .
            \end{aligned}$$ From this we obtain $$\lim_{j \to \infty} \int_0^1 g_j(t) \, dt \leq 