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Let $(\Omega, \Sigma, \mu)$ be a probability space.

Let $L_p (\mu)$ with fixed $p \geq 1$ is defined to be the collection of all ($\mu$- equivalence classes of) Borel measurable functions $f \colon \Omega \to \mathbb{R}$ for which $\int_{\Omega} |f|^{p} \mathrm{d} \mu < \infty$.

Denote by $L_p (\mu)_+$ the positive cone of $L_p (\mu)$. One can easily see that $L_p (\mu)_+$ is complete as a closed subset of Banach space (lattice) $L_p (\mu)$.

I would like to eradicate the element $0$ from the positive cone $L_p(\mu)_+$, it is clear that the set $L_p(\mu)_+ \backslash \{0\}$ is not complete now.

My Question is:

May I ask that what kind of technique I could use to ensure that even if I eradicate the zero element from the positive cone, I still could get a complete set corresponding to $L_p(\mu)_+ \backslash \{0\}$?

I've been thinking this problem for several days. I think establishing a new quotient space with respect to $L_p(\mu)_+ \backslash \{0\}$ might work, but I'm not sure does it work? and how to implement it?

Or could you please provide any idea or technique about how to reach the completeness of a space associated with $L_p(\mu)_+ \backslash \{0\}$? Any idea are most welcome!

Tomasz Kania
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Paradiesvogel
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1 Answers1

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A subset of a complete metric space $X$ is completely metrisable if and only if it is $G_\delta$. (In other words, $G_\delta$ sets are exactly those sets $A$ for which you may change the restriction of the metric in $X$ to a $A$ with an equivalent complete metric.)

Certainly the set you get after removing one point is $G_\delta$ so you may apply this in your situation.

Tomasz Kania
  • 16,361
  • Could you please explain more about your answer? Did you mean that $L_p (\mu)+ \backslash {0}$ is $G{\delta}$? If so, would you mind proving this briefly? – Paradiesvogel May 10 '17 at 23:28