What is the difference between $\mathbb{E}^{\mathscr{F}}[x]<\infty$ and $\mathbb{E}^{\mathscr{F}}[\,|x|^2]<\infty$?

Question

Let $$\mathcal{M}=\{x|x\text{ is a }\mathscr{F}\text{-measurable variable which satisfies }\mathbb{E}^{\mathscr{F}}[x]<\infty\}$$ $$\mathcal{N}=\{x|\;x\text{ is a }\mathscr{F}\text{-measurable variable which satisfies }\mathbb{E}^{\mathscr{F}}[\;|x|^2]<\infty\}$$ I want to know what is the relationship between $\mathcal{M}$ and $\mathcal{N}$. If possible, please give some heuristic examples as well. Thanks.

Answer 1

If you are integrating with respect to a probability, then $\mathcal N\subseteq \mathcal M$. In fact, let $x\in\mathcal N$ and $A=\{\omega \in\Omega\,\mid\,x(\omega)\ge 1\}$. Then $$\lvert x\rvert\cdot 1_A\le \lvert x\rvert^2\cdot 1_A\le\lvert x\rvert^2\\ \lvert x\rvert\cdot 1_{\Omega\setminus A}\le 1_{\Omega\setminus A}\le 1$$

and $$\mathbb E^{\mathscr F}\left[\lvert x\rvert\right]=\mathbb E^{\mathscr F}\left[\lvert x\rvert\cdot 1_A\right]+\mathbb E^{\mathscr F}\left[\lvert x\rvert\cdot 1_{\Omega\setminus A}\right]\le \mathbb E^{\mathscr F}\left[\lvert x\rvert^2\right]+\mathbb E^{\mathscr F}\left[ 1\right]=\\=1+\mathbb E^{\mathscr F}\left[\lvert x\rvert^2\right]<\infty$$

In general, the inclusion is strict. For instance, $\Omega=(0,1)$, Lebesgue measure and $x(\omega)=\omega^{-1/2}\in \mathcal M\setminus \mathcal N$.

For infinite measures, both $L^1(\Omega,\mathscr F,\mu)\setminus L^2(\Omega,\mathscr F,\mu)$ and $L^2(\Omega,\mathscr F,\mu)\setminus L^1(\Omega,\mathscr F,\mu)$ might be non-empty simultaneously: for instance, this happens when $\Omega=\Bbb R$ with Lebesgue measure.