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Let $X$ be a locally compact Hausdorff space, $\beta X$ its Stone-Cech compactification and $\Delta: X\to\beta X$ the inclusion map. Given a Borel probability measure $\mu^{\beta}$ over $\beta X$, is it possible to find a Borel probability measure $\mu$ over $X$ so that $L_1(X,\mathscr{B}(X),\mu)$ is isometrically isomorphic to $L_1(\beta X,\mathscr{B}(\beta X),\mu^{\beta})$.

I guess the answer is no (at least in such generality) and perhaps some conditions on the support of $\mu^{\beta}$ are needed in order to get the isomorphism. Probably counter-examples can be found when $\mu^{\beta}(\Delta X)<1$, but I am needing some help to construct them.

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  • $\begingroup$ I don't think this can work, as you suspected: if $X=\mathbb N$, then $L^1(X)$ is a weighted $\ell^1$ space whose unit ball has many extreme points, and it must be possible (easily, one would think) to come up with a measure on $\beta\mathbb N$ for which the unit ball of $L^1(\beta\mathbb N)$ has no extreme points. $\endgroup$
    – Christian Remling
    Commented Mar 31, 2017 at 0:42
  • $\begingroup$ Well, a dumb example where this works is when $\mu^\beta$ is atomic, since then you can just arbitrarily choose a countable subset of $X$ and give it the same weights. But it makes it hard to see why the question is interesting because $\mu$ and $\mu^\beta$ need not be related in any canonical way. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2017 at 5:03
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    $\begingroup$ For a cruder idea, if $X = \mathbb{N}$ then $L^1(\mathbb{N},\mu)$ is separable for any $\mu$, but $L^1(\beta \mathbb{N}, \mu^\beta)$ is typically not separable. For instance, suppose $\mu^\beta$ comes from a non-principal ultrafilter on $\mathbb{N}$. The equivalence relation on $2^{\mathbb{N}}$ induced by the ultrafilter has uncountably many equivalence classes. If $\{A_i\}$ are pairwise inequivalent subsets of $\mathbb{N}$, let $f_i$ be the continuous extension of $1_{A_i}$ to $\beta \mathbb{N}$. These functions are at pairwise distance 1 in $L^1(\mu^\beta)$. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2017 at 5:08
  • $\begingroup$ Hi @NateEldredge thanks for the comment as I was suspecting the answer in general should be no. Although I forgot to mention I was more interested in cases where $X$ is uncountable. $\endgroup$
    – Leandro
    Commented Mar 31, 2017 at 11:21
  • $\begingroup$ Well, I believe the same thing will happen when $X$ is $\mathbb{R}$ or any non-compact Polish space. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2017 at 14:04

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