Use $C^\infty$ function to approximate $W^{1,\infty}$ function in finite domain

This is exercise 10.21 from Leoni's book.

The exercise asks me to prove that for any $u\in W^{1,\infty}(\Omega)$ where $\Omega$ is open FINITE, there exists a sequence $(u_n)\subset C^\infty(\Omega)$ such that $$\|u_n-u\|_{L^\infty(\Omega)}\to 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ and $$\|\nabla u_n\|_{L^\infty(\Omega)}\to \|\nabla u\|_{L^\infty(\Omega)}\,\,\text{with }\,\,\nabla u_n\to\nabla u \,\,\text{a.e.} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$ The exercise gives a hint that I should try to modify the prove of Meyers-Serrin theorem that shows every $W^{1,p}$ function can be approximated by $C^\infty$ function in $W^{1,p}$ norm.

This is what I did so far:

I first make a claim. Assume $U\subset R^n$ is open and $U'\subset\subset U$. Hence, for $\epsilon>0$ small enough we can define $u_\epsilon:=\eta_\epsilon \ast u$ inside $U'$ where $\eta_\epsilon$ is the standard mollifier. Then I claim that there is a subsequence of $u_\epsilon$, still denote as $u_\epsilon$, such that $u_\epsilon$ satisfies $(1)$ and $(2)$ in $U'$ (with $\Omega$ replaced by $U'$).

To show $u_\epsilon\to u$ uniformly in $U'$, I use ascoli-arzela theorem. Clearly, from the definition of mollification, we have $$\sup_{\epsilon>0}\|u_\epsilon\|_{L^{\infty}(U')}\leq \|u\|_{L^{\infty}(U)},\,\sup_{\epsilon>0}\|\nabla u_\epsilon\|_{L^{\infty}(U')}\leq \|\nabla u\|_{L^{\infty}(U)} \,\,\,\,\,\,\,\,\,\,(3)$$ Hence, ascoli-arzela theorem states that, up to a subsequence, that $u_\epsilon\to u$ uniformly, i.e., in $L^\infty(U')$ because $U'$ is compact. This gives $(1)$ in $U'$.

To show $(2)$, we deduce from $(3)$ that, up to a subsequence,$\nabla u_\epsilon\to \nabla u$ in weak star sense, and hence we have $\liminf \|\nabla u_\epsilon\|\geq \|\nabla u\|$, together with $(3)$ again, we have $\lim \|\nabla u_\epsilon\|=\|\nabla u\|$. In the end, we also have $\nabla u_\epsilon\to \nabla u$ a.e., because $u_\epsilon\to u$ in $W^{1,p}$ on any compact domain.

Now go back to our original question. We will use partition of unity as it be used in the prove of Meyers-Serrin theorem. The prove is on page 285. It is too long I can not type everything here. We use the same partition of unity $\phi_i$ as it used in the book. Briefly, spt$\phi_i\subset\subset \Omega_{i+1}\setminus\Omega_{i-1}$ and $\Omega=\cup \Omega_i$ for $\Omega_i\subset\subset \Omega_{i+1}$. Then by my claim I could choose $\epsilon_i$ for each $i$ such that, for arbitrary $\eta>0$, $$\|(\phi_iu)_{\epsilon_i}-\phi_iu\|_{L^\infty(\Omega)}< \frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$ and $$\left|\|\nabla (\phi_iu)_{\epsilon_i}\|_{L^\infty(\Omega)}- \|\nabla (\phi_iu)\|_{L^\infty(\Omega)}\right|<\frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$

So far I am comfortable and confident with my proof. But the rest I am NOT sure at all.

Now we define $v:=\sum_{i=1}^\infty (\phi_iu)_{\epsilon_i}$ and by $u=\sum_{i=1}^\infty (\phi_iu)$ I want to conclude $$\|v-u\|_{L^\infty(\Omega)}\leq \eta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)$$ and $$\left|\|\nabla v\|_{L^\infty(\Omega)}- \|\nabla u\|_{L^\infty(\Omega)}\right|<\eta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7)$$

And here is my CONCERN.

1: I am NOT sure at all about my conclusion of $(6)$ and $(7)$. It should go like this way but I can not prove it formally. After all, $L^\infty$ norm is so different then $L^p$ norm. I never have monotone convergence thm in $L^\infty$ nor LDCT.

Thx for @sanjab's answer, but my concern is still remain.

More specific, we never have $$\lim_{n\to \infty}\|\chi_{\Omega_n}u\|_{L^\infty(\Omega)} = \|u\|_{L^\infty(\Omega)}$$ That is, I accept @sanjab's answer that

$$\|v_l-u_l\|_{L^\infty(\Omega_l)} \leq \sum_{i=1}^l \|(\phi_iu)_{\epsilon_i}-\phi_iu\|_{L^\infty(\Omega)}< \sum_{i=1}^l \frac{\eta}{2^i} \leq \eta$$

and hence $$\limsup_{l\to\infty}\|v_l-u_l\|_{L^\infty(\Omega_l)} \leq \eta$$

However, it may happen that $$\limsup_{l\to\infty}\|v_l-u_l\|_{L^\infty(\Omega_l)}<\|v-u\|_{L^\infty(\Omega)}$$ but not $$\limsup_{l\to\infty}\|v_l-u_l\|_{L^\infty(\Omega_l)}=\|v-u\|_{L^\infty(\Omega)}$$ so we lose the upper bound.

I understand this case won't happen if $u$ is continuous. But as we have no control of domain, we can not use embedding theorem to conclude that $u$ is actually Lipschitz.

2: So far I NEVER use the fact the $\Omega$ is finite. So either I prove a stronger version then this exercise, or I make some mistake in my prove. (I think the second assumption will hold :))

Please help! Thx!

$$\|(\phi_iu)_{\epsilon_i}-\phi_iu\|_{L^\infty(\Omega)}< \frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$ and $$\left|\|\nabla (\phi_iu)_{\epsilon_i}\|_{L^\infty(\Omega)}- \|\nabla (\phi_iu)\|_{L^\infty(\Omega)}\right|<\frac{\eta}{2^i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$

So far I am comfortable and confident with my proof. But the rest I am NOT sure at all.

Now we define $v:=\sum_{i=1}^\infty (\phi_iu)_{\epsilon_i}$ and by $u=\sum_{i=1}^\infty (\phi_iu)$ I want to conclude $$\|v-u\|_{L^\infty(\Omega)}\leq \eta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)$$ and $$\left|\|\nabla v\|_{L^\infty(\Omega)}- \|\nabla u\|_{L^\infty(\Omega)}\right|<\eta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7)$$

The proof from here should work the same way as in the book. $$v_l:=\sum_{i=1}^l (\phi_iu)_{\epsilon_i}$$ $$u_l=\sum_{i=1}^l (\phi_iu)$$

$$\|v_l-u_l\|_{L^\infty(\Omega_l)} \leq \sum_{i=1}^l \|(\phi_iu)_{\epsilon_i}-\phi_iu\|_{L^\infty(\Omega)}< \sum_{i=1}^l \frac{\eta}{2^i} \leq \eta$$

Those are all real numbers. The sequence is bounded by $\eta$ and monotone, therefore converges by Lebesgue monotone convergence (Convergence of a monotone sequence of real numbers).