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add details concerning measurability
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coudy
coudy
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Let $(X, {\cal \tau}, \mu)$ be a probability space and $F$ some vector subspace of integrable functions $L^1(X)$(defined everywhere). I can look at the set $$\hat{F} = \{f \in L^1 \mid \forall \varepsilon>0, \ \exists \,g,h \in F \hbox{ such that } g \leq f \leq h \hbox{ and } \int h-g < \varepsilon\}$$$$\hat{F} = \{f \hbox{ integrable } \mid \forall \varepsilon>0, \ \exists \,g,h \in F \hbox{ such that } g \leq f \leq h \hbox{ and } \int h-g < \varepsilon\}$$

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.

Let $(X, {\cal \tau}, \mu)$ be a probability space and $F$ some vector subspace of $L^1(X)$. I can look at the set $$\hat{F} = \{f \in L^1 \mid \forall \varepsilon>0, \ \exists \,g,h \in F \hbox{ such that } g \leq f \leq h \hbox{ and } \int h-g < \varepsilon\}$$

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.

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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coudy
coudy
  • 18.7k
  • 5
  • 75
  • 135

Is the set of functions between two arbitrarily close L^1 functions closed?

Let $(X, {\cal \tau}, \mu)$ be a probability space and $F$ some vector subspace of $L^1(X)$. I can look at the set $$\hat{F} = \{f \in L^1 \mid \forall \varepsilon>0, \ \exists \,g,h \in F \hbox{ such that } g \leq f \leq h \hbox{ and } \int h-g < \varepsilon\}$$

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.