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 $1<p\leq 2$? 



Reflexive subspaces of non-separable abstract $L_1$ spaces
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. 


