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 $L\_1(\mu)$? 
> 
> 
> 



When does $\ell_1(\Gamma)$ embed into $L_1(\mu)$?
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. 


