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We say that a sequence of random variables $X_n$ converges in $L^{\log}$ to $X$ if $$ \exp\left(\mathbb{E}\left[\log\left|X_n-X\right|\right]\right) \to 0\quad\text{as}\quad n\to\infty. $$ Is it true that, if $X_n$ converges in $L^{\log}$, then $X_n$ also converges in probability? I know how to show that this is true if $X_n$ converges in $L^p$ for $1<p<\infty$, but I couldn't figure out how to prove or disprove this when $p\searrow 0$ which corresponds to the notion of convergence in $L^{\log}$ as we can see in this proof. How should I approach this?

mathmd
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    $L_0$ has other metrics that coincide with convergence of probability. For example $|f|_0=\inf{\varepsilon>0: P[|f|>\varepsilon]\leq \varepsilon}$ or $g(X)=E[|X|\wedge1]$. The distances being $d(fog):=|f-g|_0$ and $\rho(f, g)=E[|f-g|\wedge1]$. – Mittens Oct 07 '22 at 00:20
  • Notice also that convergence in probability also follows from convergence in $L_p$ for $0<p<1$. However, $|f|p$ does not induce a metric in such $L_p$ spaces, $|f|^p_p$ does. Although $|f|_p\xrightarrow{p\rightarrow0+}\exp(E[\log(|f|)]$ for any $f\in \bigcap{r>0}L_r(\mathbb{P})$, the distance properties of $|f|^{\min(1,p)}_p$ are not preserved by the limit. – Mittens Oct 07 '22 at 00:36
  • I'm not familiar with the space $L^0$. Is it interpreted as the space of $f$ with $\exp(\mathbb{E}[\log |f|])<\infty$? Does it coincide with $\bigcap_{r>0} L^r(\mathbb{P})$? – mathmd Oct 07 '22 at 00:58
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    The space $L_0$ is the set of almost surely finite measurable functions (it contains alll $L_p$'s in particular. It just so happens that in this space, there is a distance that gives convergence in probability. Notice that at the other end of the spectrum one has $\bigcap_{r>0}L_p(\mathbb{P}$, where your limit operates. What I am saying is that the limit $\exp(E[\log(|f|)])$ is not closed to be a distance. – Mittens Oct 07 '22 at 01:03

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Counter-example: Let $X_n$ take the values $n$ and $\frac 1 {n^{2}}$ with probability $\frac 1 2$ each. Let $X=0$. Then the hypothesis holds but $P(|X_n-X| >\epsilon) \to \frac 1 2$.

geetha290krm
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