Space of $L^0$ finite random variables

Question

I have a quite clear idea of the definition of $L^p(\Omega, \mathcal{F}, P)$ spaces, for $0 < p \leq \infty$. But, I don't understand the definition (from the appendix of this book) of the space $L^0(\Omega, \mathcal{F}, P)$ as the set of all $P$-almost surely finite random variables.

My understanding is that the difference between $L^1(\Omega, \mathcal{F}, P)$ is that for some $\omega \in \Omega$, a function $Z \in L^1$ CAN be $+\infty$, as long as $\|Z\|_1 = \mathbb{E}[|Z|^1|]^{1/1} \leq \infty$ (I am being just pedantic in the notation to highlight that $p=1$). While if a function is in $L^0$, we have that $Z(\omega)\leq \infty \forall \omega \in \Omega$ where $P(\Omega)>0$. (We can have $Z(\omega)=\infty$ if $P(\omega)=0$)

Am I right?

Answer 1

Presumably $P$ is a probability measure on $(\Omega,\mathscr{F})$. In any case, in what follows, it suffices to assume that $P$ is a finite measure.

• $\mathcal{L}_0(\Omega,\mathscr{F},P)$, as a set, is the space of all $\mathscr{F}$-measurable functions $f:(\Omega,\mathscr{F})\rightarrow(\overline{R},\mathscr{B}(\overline{R}))$ such that $P(|f|=\infty)=0$ (here $\overline{R}=\mathbb{R}\cup\{\pm\infty\}$). The space $\mathcal{L}_0(\Omega,\mathscr{F},P)$ admits a pseudo-metric $d$ under which convergence in $d$ is the same as convergence in probability.

• Once $P$-almost surely equal functions are identified ($f\sim g$ iff $P(f=g)=1$) the resulting space $(L_0,d)$ becomes a (metrizable) topological vector space, that is, sum $(f,g)\mapsto f+g$ and scalar product $(r, f)\mapsto rf$ are continuous as maps from $(L_0,d)\times(L_0,d)$ into $(L_0,d)$ and from $(\mathbb{R},|\,|)\times(L_0,d)$ into $(L_0,d)$ respectively. With some additional effort one can show that $(L_0,d)$ is a complete metric linear space.

Here is one way to metrize $L_0$ to obtain the so called convergence in measure topology. Define $$\|f\|_0:=\inf\{\varepsilon>0: \mu(|f|>\varepsilon)\leq\varepsilon\}$$

It is easy to check that

1. $P(|f|>\|f\|_0)\leq \|f\|_0\leq P(\Omega)$
2. If $P(f=\pm\infty,g=\mp\infty)=0$, then $\|f+g\|_0\leq\|f\|_0+\|g\|_0$
3. If $|f|\leq |g|$, then $\|f\|_0\leq \|g\|_0$
4. $\|r f\|_0\leq (r\vee1)\|f\|_0$ for all $r\in[0,\infty)$
5. If $P(|f|=\infty)=0$, then $\lim_{r\rightarrow0}\|rf\|_0=0$

• All these relations show that $d(f,g)=\|f-g\|_0$ is a (translation invariant) pseudo-metric, that $\|f\|_0=0$ iff $f=0$ $P$-almost surely. Furthermore, after identification of $P$-almost surely functions, the sum and the scalar multiplication operations are continuous.

• The most important part of this discussion is that of convergence in measure.

$f_n\rightarrow f$ in measure iff $\|f_n-f\|_0\xrightarrow{n\rightarrow\infty}0$.

Proof: Suppose $\|f_n-f\|_0\rightarrow0$ as $n\rightarrow\infty$. Let $\delta>0$ be fixed. For any $0<\varepsilon\leq\delta $, there is an integer $N$ such that $\|f_n-f\|_0<\varepsilon$ whenever $n\geq N$, and so \begin{align} P\big(|f_n-f|>\delta\big)\leq P(|f_n-f|>\varepsilon)\leq P\big(|f_n-f|>\|f_n-f\|_0\big)\leq\|f_n-f\|_0<\varepsilon \end{align} for all $n\geq N$. Hence $f_n$ converges to $f$ in $P$-measure.

Conversely, suppose $f_n$ converges to $f$ in $P$-measure. Then, for any $\varepsilon>0$ there exists and integer $N$ such that $P(|f_n-f|>\varepsilon)<\varepsilon$ for all $n\geq N$. Hence, $\|f_n-f\|_0\leq\varepsilon$ whenever $n\geq N$.

• To show that $d$ is complete, suppose $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence. There is an increasing sequence of integers $n_\ell$ such that \begin{align} \sup_{m,k\geq n_\ell}\mu(|f_k-f_m|>2^{-\ell})\leq \sup_{m,k\geq n_\ell}\|f_m-f_k\|_0< 2^{-\ell}. \end{align} Hence, $\sum_l\mu\big(|f_{n_{\ell+1}}-f_{n_\ell}|>2^{-\ell}\big)<\infty$, and by the Borel--Cantelli theorem, for $\mu$-almost all $\omega\in\Omega$, there is an integer $N$ (depending on $\omega$) such that $|f_{n_{\ell+1}}(\omega)-f_{n_\ell}(\omega)|\leq 2^{-\ell}$ for all $\ell\geq N$. Consequently, $f:=\lim_{\ell\rightarrow\infty}f_{n_\ell}=\sum_k(f_{n_k}-f_{n_{k-1}}) +f_{n_1}$ exists $\mu$--a.s. This means that $\{f_n\}$ has subsequence $\{f_{n_\ell}\}$ that converges $\mu$--a.s, and so in $L_0$.

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• There are other equivalent metrics on $L_0$. For any bounded metric $\rho$ on $\mathbb{R}$, $d_\rho(f,g)=E[\rho(f,g)]$ also defines a metric on $L_0$ that coincides with the convergence in measure topology. One of the most common metrics used is $d'(f,g):=E[|f-g|\wedge1]$.

• One big difference between $L_0$ and $L_p$ for $p\geq1$ every such $L_p$, being a Banach space, is locally convex. $L_0$ is not in general locally convex. In this respect, $L_0$ has more similarities with the $L_p$ spaces with $0<p<1$.

• If $P$ has no atoms, then the dual of $L_0$ is $\{0\}$; the dual space of $L_p$, for $p\geq1$ is $L_q$ with $\frac1p+\frac1q=1$.

Answer 2

No, that's not correct in several ways. First, $P(\Omega)$ is never zero, it is $1$ by definition of a probability measure. Second, $Z$ being finite almost everywhere means that $P(\{\omega\in \Omega:|Z(\omega)|=\infty\})=0$. This is really a condition about the collection of all elementary events for which $Z$ takes an infinite value, not about the individual elementary events. Third you seem to think that being in $L^0$ is a stronger conditition than being in $L^1$, but it is the other way round: If $Z\in L^1(P)$, then $\infty>\mathbb E[|Z|]\geq \infty\cdot P(\{\omega\in\Omega: |Z(\omega)|=\infty\}$, hence $P(\{\omega\in \Omega: |Z(\omega)|=\infty\})=0$. Finally, neither $L^1$ nor $L^0$ consist of random variables, but of equivalence classes of random variables (with random variables being called equivalent if they coincide almost everywhere).