Let $I := (0,1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.3 Assume that $(u_n)$ is a bounded sequence in $W^{1, 1}(I)$. Show that there is a subsequence $(n_k)$ such that $(u_{n_k} (x))$ converges to a limit for every $x \in [0, 1]$.
My approach is different from the author's guideline. There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it?
By Theorem 8.8 (in the same book), the injection $W^{1, 1}(I) \subset L^1 (I)$ is compact because $I$ is bounded. Then there is a subsequence $(n_k)$ and $v \in L^1 (I)$ such that $\|u_{n_k}-v\|_{L^1} \xrightarrow{k \to \infty} 0$. Up to a further subsequence, we can assume $u_{n_k} \xrightarrow{k \to \infty} v$ a.e. on $I$.
Let $E$ be a null subset of $I$ such that $u_{n_k} (x) \xrightarrow{k \to \infty} v(x)$ for all $x \in E^c := I \setminus E$. Clearly, $E^c$ is dense in $I$.
Fix $x \in E$. It suffices to prove that $(u_{n_k} (x))_k$ is a Cauchy sequence. Fix $\varepsilon >0$. There is $y \in E^c$ such that $|x-y| < \varepsilon$. We have $(u_{n_k} (y))_k$ is a Cauchy sequence. Let $C := \sup_n \|u_n\|_{W^{1, 1}} < \infty$ $$ \begin{align*} | u_{n_i} (x) - u_{n_j} (x) | &= | (u_{n_i} (x) - u_{n_i} (y)) - (u_{n_j} (x) - u_{n_j} (y)) + (u_{n_i} (y) - u_{n_j} (y))| \\ &\le | u_{n_i} (x) - u_{n_i} (y) | + | u_{n_j} (x) - u_{n_j} (y) |+ |u_{n_i} (y) - u_{n_j} (y)| \\ &\le (\| u_{n_i}' \|_{L^1} + \| u_{n_j}' \|_{L^1} ) |x-y|+ |u_{n_i} (y) - u_{n_j} (y)| \\ &\le 2\varepsilon C+ |u_{n_i} (y) - u_{n_j} (y)|. \end{align*} $$
The claim then follows.
