An open set $\Omega \subset \mathbb{R}^n$ ($n>1$). Let $x' = (x_1,\dots, x_{j-1}, x_{j+1}, \dots, x_n)$ and $x={x',x_j}$. Assume functions $a, b $ maps $\mathbb{R}^{n-1} \rightarrow \mathbb{R}$ s.t. $\{ x' \} \times[a(x'), b(x')] \subset \Omega$.
Let $u \in L^1_{loc}(\Omega)$ and $u$ has weak derivative $\frac{\partial u}{\partial x_j} \in L^1_{loc}(\Omega)$. Prove that function $\tilde{u}(x_j) = u(x', x_j)$ is absolutely continuous on $[a(x'), b(x')]$ for a.e. $x'$.
I already know that function $f:\mathbb{R} \rightarrow \mathbb{R}$ is absolutely continuous if and only if $f$ has an integrable weak derivative. So I tried to find the weak derivative to $\tilde{u}(x_j)$.
Because $\frac{\partial u}{\partial x_j}$ is the weak derivative of $u(x)$, we have
$$
\int_{\Omega'} \int_{a(x')}^{b(x')} \frac{\partial u}{\partial x_j} \eta dx_j dx' = - \int_{\Omega'}\int_{a(x')}^{b(x')} u \frac{\partial \eta}{\partial x_j} dx_j dx'
$$
for all $\eta \in C_c^\infty(\Omega)$.
But how can I use the above equality to arrive at the following equality?
$$ \int_{a(x')}^{b(x')} \frac{d \tilde{u}}{d x_j} \eta dx_j = -\int_{a(x')}^{b(x')} \tilde{u} \frac{d \eta}{d x_j} dx_j $$
for all $\eta \in C_c^\infty([a(x'),b(x')])$.