u)  \circ \phi_{\alpha}^{-1}\|_{M^p_1(\mathbb{R}^n, \mathcal{L}^n)}.$$ Gather inequalities (\[ru\]), (\[niermodularfunkc\]), (\[eq:1\]) and (\[eq:2\]) to get $$\begin{aligned}
 \| (\eta_{\alpha} u) \circ \phi_{\alpha}^{-1} \|_{L^p(\mathbb{R}^n,\mathcal{L}^n)} + \| h_{\alpha} \|_{L^p(\mathbb{R}^n,\mathcal{L}^n)} - \epsilon &\leq&  C \| (\eta_{\alpha} u) \circ \phi_{\alpha}^{-1} \|_{L^p_1(\mathbb{R}^n,\mathcal{L}^n)} \\
& \leq& C \left( 2^{\frac{n+p}{2p}}C_0 + 2^{\frac{n}{2p}} \right) \| u \|_{L^p_1(X,V_g)}.\end{aligned}$$

Observe that for each $\alpha$ the function $\sqrt{2} h_{\alpha} \circ \phi_{\alpha} =: \tilde{h}_{\alpha}: U_{\alpha} \rightarrow \mathbb{R}$ is a Hajłasz gradient for $ (\eta_{\alpha} u) |_{U_{\alpha}}$. Indeed, since $h_{\alpha}$ is a Hajłasz gradient for $(\eta_{\alpha} u) \circ \phi_{\alpha} ^{-1}\  \mathbf{1}_{\phi_{\alpha} (U_{\alpha})}$, there exists a subset $E_{\alpha} \subset \mathbb{R}^n$ such that $\mathcal{L}^n(E_{\alpha})=0$, and such that for every pair $x,y \in \phi_{\alpha} (U_{\alpha}) \thicksim E_{\alpha}$ $$|\eta_{\alpha} u(\phi_{\alpha}^{-1}(x)) - \eta_{\alpha} u(\phi_{\alpha}^{-1}(y))| \leq \left( h_{\alpha}( x ) + h_{\alpha}(y) \right) |x - y|.$$ Therefore, for each pair $x,y$ in $U_{\alpha} \thicksim \phi_{\alpha} ^{-1}(E_{\alpha})$ $$\begin{aligned}
   \label{gradient}
|\eta_{\alpha}(x) u(x) - \eta_{\alpha}(y) u(y)| &=& |\eta_{\alpha} u(\phi_{\alpha}^{-1}( \phi_{\alpha}(x) )) - \eta_{\alpha} u(\phi_{\alpha}^{-1}( \phi_{\alpha}(y))|  \nonumber \\
&\leq &\left( h_{\alpha}( \phi_{\alpha}(x) ) + h_{\alpha}(\phi_{\alpha}(y)) \right) |\phi_{\alpha}(x) - \phi_{\alpha} (y)|  \nonumber \\
& \leq &\left( \tilde{h}_{\alpha}(x) + \tilde{h}_{\alpha}(y) \right) d_g(x,y).\end{aligned}$$

Our next goal is to prove that $$h:= \sum_{\alpha \in A} \tilde{h}_{\alpha} \mathbf{1}_{U_{\alpha}}$$ is a local Hajłasz gradient for $u$.\
Fix $z\in X$ and define

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