Since $C[0,1]^{\*}$ is an abstract $L$-space, $C[0,1]^{\*}$ is order isometric to $L\_{1}(\mu)$ for some measure $\mu$. 


My question is: the measure $\mu$ can be choosen to be a finite positive measure?


Thank you!



The dual space of $C[0,1]$
In an $L^1(\mu)$ space, we can tell when two elements are disjoint (that is, have disjoint support):
$f\_1, f\_2$ are disjoint if and only if $\|f\_1\pm f\_2\| = \|f\_1\|+\|f\_2\|$.


I claim that if $L\_1(\mu)$ is isometric to $C[0,1]^\*$, then $\mu$ is not sigma-finite. To do this, it suffices to exhibit an uncountable pairwise disjoint family in $C[0,1]^\*$. Here is the easiest one:
$$
\varepsilon\_t(f) := f(t)
$$
defines the "point evaluations" $\varepsilon\_t$, one for each $t \in [0,1]$. So now all we have to do is observe that
$$
\|\varepsilon\_t \pm \varepsilon\_s\| = \|\varepsilon\_t\| + \|\varepsilon\_s\|,
\qquad s \ne t $$
where $\|\cdot\|$ is the dual space norm.


