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Let $(X, {\cal \tau}, \mu)$ be a probability space and $F$ some vector subspace of integrable functions (defined everywhere). I can look at the set $$\hat{F} = \{f \hbox{ integrable } \mid \forall \varepsilon>0, \ \exists \,g,h \in F \hbox{ such that } g \leq f \leq h \hbox{ and } \int h-g < \varepsilon\}$$

Is $\hat{F}$ the closure or the completion of $F$ for some topology? If so, which one? Does it come from a distance or a norm?

Note that this is not the completion with respect to the $L^1$ norm in general. Such extension occurs for example in Riemann integration, in the proof of the Weyl criterion for equidistribution or in the portmanteau theorem in probability.

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  • $\begingroup$ Are a.e. functions identified in $L^1(X)$? And if so, is the inequality $g \leq f \leq h$ understood a.e.? $\endgroup$
    – Willie Wong
    Commented Feb 18, 2022 at 16:23
  • $\begingroup$ Good point. The space $F$ should be a subspace of Borel integrable functions without identification, in order to define $\hat{F}$ with inequality holding everywhere. $\endgroup$
    – coudy
    Commented Feb 18, 2022 at 16:35
  • $\begingroup$ What if you define $d(f,f') = \inf \int h-g$ where the inf is taken over $\{(g,h)\in F^2: g \leq f-f' \leq h\}$? This probably defines a pseudometric. $\endgroup$
    – Willie Wong
    Commented Feb 18, 2022 at 16:42
  • $\begingroup$ This is a good idea. I will try to sort it out. Still I am surprised that it has not a well known answer. $\endgroup$
    – coudy
    Commented Feb 19, 2022 at 16:05
  • $\begingroup$ uh... my previous comment is a bit facetious. (It is not that useful since the pseudometric itself now depends on the set F, so is not external to the set F as what I am guessing you would prefer.) $\endgroup$
    – Willie Wong
    Commented Feb 20, 2022 at 20:27

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