I want to costruct a measure space $(X,\mathcal{F},\mu)$ and a $\mathcal{C}\subset\mathrm{m}\mathcal{F}$, where $\mathrm{m}\mathcal{F}$ be the set of extended real-valued measurable functions on $X$, with following properties:

  1. $\mathcal{C}\subset L^1 (X)$,
  2. for any $\epsilon>0$ there exists a $\delta>0$ such that for any $f\in\mathcal{C}$ and for any $A\in\mathcal{F}$ with $\mu (A)<\delta$, $\int_A |f|d\mu<\epsilon$ holds,
  3. there exists $\epsilon>0$ such that for any $K>0$, $\mu(|f|>K)\ge\epsilon$ holds for some $f\in\mathcal{C}$.

Any help?

  • $\begingroup$ What exactly do you want to ask? In the title you talk about non-tightness but 3) is tightness. $\endgroup$ – Kavi Rama Murthy Jun 12 at 5:46
  • $\begingroup$ @KaviRamaMurthy Exactly. I fixed it. Thank you. $\endgroup$ – Ichiko Jun 12 at 5:54
  • $\begingroup$ @saz Exactly. Some measure spaces have no such sets. So I want to costruct a suitable measure space $(X,\mathcal{F},\mu)$ and a $\mathcal{C}\subset\mathrm{m}\mathcal{F}$. $\endgroup$ – Ichiko Jun 12 at 6:49

The problems here can come from atoms. If $X$ has only one element, say $x_0$, say of measure one and $\mathcal C$ is the class of all the function from $X$ to $\mathbb R$ which map $x_0$ to some $c$, the first item is satisfied. The second as well, since $\delta=1/2$ is always a good choice. And also the third because we can choose $\varepsilon=1/2$.


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