pointwise convergence of mollification combined with a cut-off function

Question

i want to show that for $x^m>2\varepsilon$ $$u^{\varepsilon}(x)=[\zeta(x^m)\cdot u]\ast\eta_{\varepsilon}(x), \qquad x=(x^1,x^2,\ldots,x^m)\in \mathbb{R}^m_+:=\lbrace x\in \mathbb{R}|x^m\geq0\rbrace $$ converges pointwise a.e. to $u$. Here, $\zeta(x^m)\in C^{\infty}([0,\infty),[0,1])$ is a cut-off function with $\zeta=0$ on $[0,2\varepsilon]$ and $\zeta=1$ on $[3\varepsilon,\infty)$. My attempt: \begin{align*} \vert u^{\varepsilon}(x)-u(x)\vert&=\bigg\vert \int_{B_{\varepsilon}(x)}\zeta(z^m)\cdot u(z)\cdot \eta_{\varepsilon}(z-x)d z-u(x)\bigg\vert\\ &= \bigg\vert \int_{B_{\varepsilon}(x)}[\zeta(z^m)\cdot u(z)-u(x)]\cdot \eta_{\varepsilon}(z-x)d z\bigg\vert\\ &\leq\bigg\vert \int_{B_{\varepsilon}(x)}(\zeta(z^m)-1)\cdot u(z)\cdot \eta_{\varepsilon}(z-x)d z\bigg\vert\\ &\quad+\bigg\vert \int_{B_{\varepsilon}(x)}[u(z)-u(x)]\cdot \eta_{\varepsilon}(z-x)d z\bigg\vert\\ &\leq \Vert \eta\Vert_{\infty}\varepsilon^{-m}\int_{B_{\varepsilon}(x)}\vert \zeta(z^m)-1\vert \cdot \vert u(z)\vert d z\\ &\quad+\Vert \eta\Vert_{\infty}\varepsilon^{-m} \int_{B_{\varepsilon}(x)} \vert u(z)-u(x)\vert d z\\ &=\Vert \eta\Vert_{\infty}\varepsilon^{-m}\int_{B_{\varepsilon}(x)}\vert \zeta(z^m)-1\vert \cdot \vert u(z)\vert d z\\ &\quad+\mathcal{L}^m(B_1(x))\Vert \eta\Vert_{\infty}\dfrac{1}{\mathcal{L}^m(B_{\varepsilon}(x))}\int_{B_{\varepsilon}(x)} \vert u(z)-u(x)\vert d z \end{align*}
Now, when $\varepsilon \searrow0$ then the second integral vanishes by Lebesgue's differentiation theorem. How can i argue that the first integral vanishes?