The dual space of $C[0,1]$

Question

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!

Answer 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.