3

Consider $\mathcal{F}\subset L^2[0,1]$ such that every element in $F$ can be represented by a weakly increasing function (i.e., equals to a weakly increasing function a.e.), then is $\mathcal{F}$ closed? I'm very tempted to say yes but haven't got a clue for a proof.

Ecthelion
  • 135

1 Answers1

3

Fix representatives $f_n$ and $f$. If $f_n\to f$ in $L^2[0,1]$, there is a subsequence $f_{n_k}$ such that $f_{n_k}\to f$ a.e.

Let $E$ be a negligible set such that, on $[0,1]\setminus E$, all the $f_n$-s are defined and weakly increasing, $f$ is defined and $f_n\to f$ pointwise.

Let $x,y\in [0,1]\setminus E$ and $x<y$. The inequalities $\forall n,\ f_n(x)\le f_n(y)$ pass to the limit, and thus $f(x)\le f(y)$.

Now, $g(x)=\begin{cases} f(x)&\text{if }x\notin E\\ \inf\{f(z)\,:\, z\notin E\wedge z>x\}&\text{if }x\in E\end{cases}\quad$ is a weakly increasing function such that $\left.g\right\rvert_{[0,1]\setminus E}=\left.f\right\rvert_{[0,1]\setminus E}$.

  • Thanks a lot! Yeah the subsequence argument definitely does the trick! – Ecthelion Nov 05 '17 at 17:08
  • It does not actually matter that $[0,1]$ is a finite measure space to get a subsequence converging almost everywhere. $L^p$ convergence implies convergence in measure implies some subsequence converges almost everywhere. In a finite measure space, convergence almost everywhere also implies convergence in measure. – nullUser Nov 05 '17 at 17:10
  • @nullUser Ah, ok. I recalled there were subtleties, but I did not remember which one. Thank you. –  Nov 05 '17 at 17:15