I have two questions on the second dual of $C[0,1]$:


R. D. Mauldin ([[1](http://www.math.unt.edu/~mauldin/papers/no5.pdf)]) proved that: For a given bounded linear functional 
$T: C[0,1]^\*\to \mathbb{C}$ there is a bounded function $\psi$ defined on $B$, the set of all Borel subsets of $[0,1]$, with $T(\mu)=\int \psi d\mu$.


I guess the following two items. Could you please let me know your ideas about?


1) Let $\psi$ be a bounded function on $B$ which is also in $C[0,1]^{\*\*}$. The restriction $\psi\_{|\_{X}}$ forms a bounded Borel measurable function. 


2) We denote $V\_{\infty}[0,1]$, by the set of all bounded Borel measurable functions on $[0,1]$. Then $\psi\to \psi\_{|\_{X}}$ defines a \* homomorphism from 
 $C[0,1]^{\*\*}$ onto $V\_{\infty}[0,1]$. It implies that $V\_{\infty}[0,1]=\frac{C[0,1]^{\*\*}}{J}$ where $J$ is a closed ideal. 



On the second dual of $C[0,1]$
It is easy to see that $\int \psi d\delta\_t$ where $\delta\_t$ is the Dirac mass at $t$ is $\psi(\{t\})$


So you are starting from a $T \in C([0,1])^{\*\*}$ and you are attaching to it the function $t \mapsto T(\delta\_t)$. This has absolutely nothing to do with Mauldin's result, so I'm not sure why you are mentioning it.


Now let $V \subset C([0,1])^\*$ be the vector subspace spanned by dirac masses.


It is easy to see that $v = \sum a\_i \delta\_{x\_i}$ is a finite linear combination of (distinct) Dirac masses, then $\Vert v \Vert = \sum |a\_i|$.


Hence, pick $f$ any bounded function on $X$, the linear map that send $\delta\_{x\_i}$ to $f(x\_i)$ is continuous on $V$, of norm $\max |f(x)|$.


By the Hahn-Banach theorem it extend into an element of $C([0,1])^{\*\*}$ of the same norm. 


So this means that $\psi\_{|X}$ can be basically any bounded function on $[0,1]$. In particular it has no reason to be measurable.


For you second question, if you replace $V$ by the set of all bounded functions on $[0,1]$ then the answer is yes:


$V$ is a von neumann algebra and there is a morphism $C[0,1] \rightarrow V$ so this can be extended into a normal morphism $C[0,1]^{\*\*} \rightarrow V$. this extension indeed corresponds to the map described above hence is a surjective normal morphism of von Neuman algebra, hence $V$ is a quotient of $C[0,1]^{\*\*}$, by a two sideed ideal, but even by a projection !


