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Suppose we are given an uncountable set $\Gamma$ and a measure space $(\Omega, \mathcal{F}, \mu)$. I would like to know when the Banach space $\ell_1(\Gamma)$ embeds into $L_1(\mu)$. Of course, this is impossible when $\mu$ is a probability measure (or more generally, when $\mu$ is $\sigma$-finite). On the other hand, this is not automatic when we drop the $\sigma$-finiteness as is easily seen by considering the restriction of the Lebesgue measure to the trivial $\sigma$-field $\{\varnothing, \mathbb{R}\}$. My qustion is: what are the sufficient conditions for this phenomenon? In particular,

Is it true that if $\mu$ is localizable but non-$\sigma$-finite then $\ell_1(\omega_1)$ embeds into $L_1(\mu)$?

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    $\begingroup$ Haskell Rosenthal studied this kind of problem in the late 1960s and maybe early 1970s. They should be easy to find on MathSciNet. $\endgroup$ Commented Nov 3, 2013 at 16:05

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If $(X,\Sigma, \mu)$ is semi-finite (it has no infinite atoms) and non-$\sigma$-finite, by transfinite induction there is a family $\{E_\alpha\}_{\alpha \in \omega_1}$ of disjoint measurable sets of positive finite measure (indeed, having found $E_\beta$ for all $\beta < \alpha$, the complement of the countable union $\cup_{\beta < \alpha}E_\beta$, is measurable and has infinite measure, so by assumption it contains a measurable subset of positive finite measure).

Then the closed subspace of $L_1( X,\Sigma, \mu)$ generated by the normalized characteristic functions of the $E_\alpha$ is isometric to $\ell_1(\omega_1)$.

Note that we do not need $\mu$ to be localizable, as we do not need the uncountable union $\cup_{\alpha < \omega_1}E_\alpha$ to be measurable.

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  • $\begingroup$ And if $\mu$ is localizable, $\ell_1(\omega_1)$ embeds isometrically as a direct summand. $\endgroup$ Commented Nov 5, 2013 at 21:58

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