Absorbing ball in the space of measurable functions

Question

We know that if $\alpha$ is a real number then $\exists n\in \mathbb{N}$ such that $|\alpha|\le n$. In other words, the unit ball is an aborbing set.

Let $(\Omega,\mathcal{F},P)$ be a probability space. Denote $L_0(\Omega)$ the space of random variables $X$ such that $P(\omega: |X(\omega)|<\infty)=1$. Is there a random variable $X_0\in L_0(\Omega)$ such that for all $X\in L_0(\Omega)$ there exists a natural number $n$ (depends on $X$) so that: $$|X(\omega)|\le n|X_0(\omega)|\; \text{almost surely?}$$

Answer 1

With sufficiently large probability spaces the claim is clearly false.

Suppose $\Omega $ is rich enough that one can split it into countable union of events with positive probability. Namely, assume there exists $E_1,E_2,... \in \mathcal{F}$ such that $\mathbb{P}(E_i) >0 $ for all $i=1,2,...$, $E_i\cap E_j = \emptyset$ if $i\neq j$ and $\Omega = \bigcup_{i=1}^\infty E_i$.

Now assume, that there is such $X_0$, and define the following random variable $$ X = (1+|X_0|) \sum_{n=1}^\infty n \chi_{E_n}. $$ In view of the definition of $E_n$ and $X_0$, we clearly have $X \in L_0(\Omega)$, and $X = (1+|X_0|)n$ on $E_n$. But since $\mathbb{P}(E_n)>0$ for all $n\in \mathbb{N}$ then $X$ cannot be bounded by $n|X|$ for any fixed $n$.

Answer 2

Suppose such a r.v. exists. For $Y=e^{|X_0|}$ there exists an integer $n$ such that $e^{|X_0|} \leq n|X_0|$ almost surely. Since $e^{a} \geq \frac {a^{2}} 2$ for all $a \geq 0$ this gives $|X_0| \leq 2n$. Thus $X_0$ is abounded random variable and this implies (by hypothesis) that all random variables are bounded. This is a contradiction except when the probability space is 'trivial'.