Completeness of the space of random variables with bounded conditional first moment with respect t0 $\left\Vert \cdot\right\Vert _{2} $

Question

Consider a probability space $\left(\Omega,\mathcal{F},P\right) $, and a sub-sigma-algebra $\mathcal{G}\subseteq\mathcal{F} $. As usual, let $L^{2}\left(\Omega,\mathcal{F},P\right) $ be the space of $\mathcal{F} $-measurable, square integrable random variable and $L^{\infty}\left(\Omega,\mathcal{G},P\right) $ be the space of $\mathcal{G} $-measurable, essentally bounded random variables. Define:$$L^{2,\infty}=\left\{ f\in L^{2}\left(\Omega,\mathcal{F},P\right):\, E\left(f^{2}\mid\mathcal{G}\right)\in L^{\infty}\left(\Omega,\mathcal{G},P\right)\right\} . $$Define also the norm $\left\Vert \cdot\right\Vert _{2} $ on $L^{2}\left(\Omega,\mathcal{F},P\right) $ by$$\left\Vert f\right\Vert _{2}=\sqrt{E\left(f^{2}\right)}\qquad\forall f\in L^{2}\left(\Omega,\mathcal{F},P\right). $$My question is the following: is $L^{2,\infty} $ $\left\Vert \cdot\right\Vert _{2} $-complete?

Observe that, being $L^{2}\left(\Omega,\mathcal{F},P\right) $ complete, it would be sufficient to prove that, for every sequence $\left(f_{n}\right)_{n\in\mathbb{N}}\subseteq L^{2,\infty} $ such that $f_{n}\stackrel{\left\Vert \cdot\right\Vert _{2}}{\longrightarrow}f\in L^{2}\left(\Omega,\mathcal{F},P\right) $, we have $E\left(f\mid\mathcal{G}\right)\in L^{\infty}\left(\Omega,\mathcal{G},P\right) $. Also notice that, if $f_{n}\stackrel{\left\Vert \cdot\right\Vert _{2}}{\longrightarrow}f\in L^{2}\left(\Omega,\mathcal{F},P\right) $, by a standard probability theoretic argument we have that there exists a subsequence $\left(f_{n_{k}}\right)_{k\in\mathbb{N}} $ such that $f_{n_{k}}\stackrel{\left\Vert \cdot\right\Vert _{2}}{\longrightarrow}f $ almost surely; thus, it would be sufficient to prove that there extists an $M>0 $ such that $f_{n_{k}}<M $ almost surely for every $k $.

Any suggestion? Thank you!

Answer 1

Actually, taking $\mathcal G:=\mathcal F$, the question reduces to ask whether $L^\infty$ is complete when it is endowed with the $L^2$ norm. Take $X\in L^2\setminus L^\infty$ and define $X_n:=X\chi\{|X|\lt n\}$, which belongs to $L^\infty$. Then the sequence $(X_n)$ is Cauchy for the $L^2$-norm and it converges to $X$ (for this norm) which is not in $L^\infty$.

Answer 2

I think I worked out a simple counterexample. Actually I have to say that its simplicity shows that my question was not so interesting. I apologize for that. The example is as follows: consider the probability space $\left(\left[0,1\right],\mathcal{B}\left(\left[0,1\right]\right),\lambda\right) $, where $\lambda $ is the lebesgue measure on $\mathcal{B}\left(\left[0,1\right]\right) $. Let also $\mathcal{G}=\mathcal{B}\left(\left[0,1\right]\right) $ Consider the sequence of sets $\left(E_{k}\right)\subseteq\mathcal{B}\left(\left[0,1\right]\right) $ defined by$$E_{k}=\left[0,\frac{1}{8^{k}}\right]\qquad n\in\mathbb{N}. $$For every $n\in\mathbb{N} $ define:$$f_{n}\left(\omega\right)=\begin{cases} 2^{k} & \qquad\omega\in E_{k},\, k\leq n\\ 0 & \qquad\mbox{otherwise}. \end{cases} $$Define also$$f\left(\omega\right)=\begin{cases} 2^{k} & \qquad\omega\in E_{k},\, k\geq1\\ 0 & \qquad\omega\in\left(\cup_{k=1}^{\infty}E_{k}\right)^{c}. \end{cases} $$Clearly, for every $n $, $f_{n}\in L^{2,\infty} $, since $P\left(\left|E\left(f_{n}\mid\mathcal{G}\right)\right|>2^{n}\right)=0 $ for every $n $. Also, $f_{n}\stackrel{\left\Vert \cdot\right\Vert _{2}}{\longrightarrow}f $, since$$E\left(\left(f_{n}-f\right)^{2}\right)=\sum_{k=n+1}^{\infty}\left(2^{k}\right)^{2}\lambda\left(E_{k}\right)=\sum_{k=n+1}^{\infty}\frac{1}{2^{k}}\longrightarrow0. $$Take $M>0 $. There exists a $N $ such that $2^{2N}>M $. Then$$\lambda\left[\left|f^{2}\right|>M\right]=\lambda\left[\cup_{k=N}^{\infty}E_{k}\right]=\lambda\left(\left[0,\frac{1}{2^{N}}\right]\right)=\frac{1}{2^{N}}>0. $$Then, $f^{2}\notin L^{\infty}\left(\left(\left[0,1\right],\mathcal{B}\left(\left[0,1\right]\right),\lambda\right)\right) $. Also, $E\left(f^{2}\mid\mathcal{G}\right)=f^{2} $, and so $f\notin L^{2,\infty} $. We conclude that $L^{2,\infty} $ is not complete with repect to $\left\Vert \cdot\right\Vert _{2} $.