An abstract $L_1$ space is a Banach lattice $E$ such that $\|x+y\|=\|x\|+\|y\|$ for disjoint $x,y\in E$. The space $L_1[0,1]$ is a separable example that contains subspaces isomorphic to $L_p[0,1]$ for $1<p\leq 2$. The space of Borel measures $M[0,1]$ is a non-separable example, but each reflexive subspace of $M[0,1]$ is separable because $M[0,1]$ is isomorphic to the dual space of the separable space $C[0,1]$.

Given an uncountable set $I$, we endow $[0,1]^I$ with the product measure associated to the Lebesgue measure on $[0,1]$, and $L_1([0,1]^I)$ is a non-separable abstract $L_1$ space. Does it contain subspaces isomorphic to $L_p([0,1]^I)$ for $1<p\leq 2$?

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    $\begingroup$ $\ell_p(I)$ is easy, because $L_1[0,1]^I$ obviously contains an IID family of size $|I|$ of $p$-stable random variables. If the classical approach for embedding $L_p[0,1]$ into $L_1[0,1]$ does not obviously generalize, try the approach that Maurey, Schechtman, and I used in our Memoirs. $\endgroup$ – Bill Johnson Nov 27 '14 at 17:37

I guess so. Here's an outline which reduces everything to the separable case.

Let $p\in (1,2]$. Then of course $L_p[0,1]^I$ is finitely representable in $L_1$. Take an ultrafilter $U$ (on some ridiculously large index set) such that $L_p[0,1]^I$ is a subspace of $L_1^U$. By Kakutani's theorem, $L_1^U = L_1(\mu)$ for some silly measure. By Maharam's theorem, we have

$$L_1(\mu) = \big(\bigoplus_{\lambda\in \Lambda} L_1 [0,1]^{\kappa_\lambda}\big)_{\ell_1(\Lambda)}$$

for some indexed family $(\kappa_\lambda)_{\lambda\in \Lambda}$ of cardinal numbers. It is not difficult to prove that there exists $\lambda\in \Lambda$ such that $L_p[0,1]^I$ embeds into $L_1[0,1]^{\kappa_\lambda}$.

As $L_p[0,1]^I$ and $L_1[0,1]^I$ have the same density, the sublattice of $L_1[0,1]^{\kappa_\lambda}$ generated by the copy of $L_p[0,1]^I$ should do the job.

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  • $\begingroup$ Nice, Tomek. I missed that even after using Maharam's theorem recently. $\endgroup$ – Bill Johnson Nov 28 '14 at 22:43

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