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I'm reading some lecture notes, and the author made the following assertions:

Convergence in probability is equivalent to convergence induced by the metric $d$ defined on $L^0$ such that $$ d(X,Y) := E\min\{|X-Y|,1\}. $$ I'm guessing there is a typo, because it's clear that this $d$ is not a metric, since $d(X,X) = 1$. My question is if anyone would know what is the actual metric to which he is referring to, and also, what does he mean by $L^0$. Would it be the space of measurable functions?

Mittens
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Davi Barreira
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  • In part, but not the part about which metric $d$ the author was intending. Or are you saying that this is the correct metric, but the author actually implied pseudo-metric? – Davi Barreira Sep 02 '21 at 16:03
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    It simply should be a $\min$, not a $\max$. See here. – nejimban Sep 02 '21 at 16:05
  • @DaviBarreira: as with $L_p$'s with $p>0$, one identifies functions that differ on sets of measure $0$ and then $d$ becomes a metric in $L_0\mod 0$. The gauge $|;|_0$ after identification also becomes a metric in $L_0\mod 0$. – Mittens Sep 02 '21 at 23:08

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