Helley's selection theorem

Question

I was doing Brezis functional analysis Sobolev space PDE textbook,in exercise 8.2 needs to prove the Helly's selection theorem:As shown below:

Let $\left(u_{n}\right)$ be a bounded sequence in $W^{1,1}(0,1) .$ The goal is to prove that there exists a subsequence $\left(u_{n_{k}}\right)$ such that $u_{n_{k}}(x)$ converges to a limit for every $x \in[0,1]$.

1. Show that we may always assume in addition that (1) $\forall n, u_{n}$ is nondecreasing on $[0,1]$.

[Hint: Consider the sequences $v_{n}(x)=\int_{0}^{x}\left|u_{n}^{\prime}(t)\right| d t$ and $\left.w_{n}=v_{n}-u_{n} .\right]$ In what follows we assume that (1) holds.

2. Prove that there exist a subsequence $\left(u_{n_{k}}\right)$ and a measurable set $E \subset[0,1]$ with $|E|=0$ such that $u_{n_{k}}(x)$ converges to a limit, denoted by $u(x)$, for every $x \in[0,1] \backslash E$

[Hint: Use the fact that $W^{1,1} \subset L^{1}$ with compact injection.]

3. Show that $u$ is nondecreasing on $[0,1] \backslash E$ and deduce that there are a countable set $D \subset(0,1)$ and a nondecreasing function $\bar{u}:(0,1) \backslash D \rightarrow \mathbb{R}$ such that $\bar{u}(x+0)=\bar{u}(x-0) \forall x \in(0,1) \backslash D\ \ $ and $\bar{u}(x)=u(x) \forall x \in(0,1) \backslash(D \cup E)$

4. Prove that $u_{n_{k}}(x) \rightarrow \bar{u}(x) \forall x \in(0,1) \backslash D$.

I was stuck in step 4.What confuse me is that :

If convergence in $L^1$ ,implies almost everywhere convergence subsequence,i.e. convergence everwhere on $(0,1)\setminus E$ ,which limit of the subsequence is not defined for $x\in E$

in step 4 it's possible for $x\in E$

I know the rough idea is that $g = f$ a.e for two continuous function then $f = g$ everywhere,the point is that we need to prove that $u_{n_k}(x) \to u(x)$ for $x \in (0,1)\setminus D$, such that $u(x)$ is continuous.then we are done

Answer 1

You can use the continuity of $\bar{u}$ on $(0,1) \setminus D$ and the non-decreasing property.

Take $x \in E$ and $y,z$ such that $u_n(z) \to \bar{u}(z)$ and $u_n(y) \to \bar{u}(y)$ (up to subsequence).

You want to prove that $u_n(x) \to \bar{u}(x)$.

Express continuity: for $\varepsilon >0$ there is $\delta >0$ such that if $|x-y|<\delta$ then $|\bar{u}(x)-\bar{u}(y)|<\varepsilon$. Say $z$ is such that $x-\delta<y<x<z<x+\delta$ Thus, for $n$ large enough $$ \bar{u}(y)-\varepsilon<u_n(y)<u_n(x)<u_n(z)<\bar{u}(z)+\varepsilon $$ And by continuity: $\bar{u}(x)-2\varepsilon < \bar{u}(y)-\varepsilon$ and: $\bar{u}(z)+\varepsilon < \bar{u}(x)+2\varepsilon$.

Finally $$ |u_n(x)-\bar{u}(x)|<2\varepsilon. $$

So the result follows by the non-decreasing properties that allows you to "squeeze" $u_n(x)$.