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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 spammed 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 any bounded function on $X$, the function that maps $\delta_{x_i}$ 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 !

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 !

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 spammed 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 any bounded function on $X$, the function that maps $\delta_{x_i}$ 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 and your second question simply don't make sense.

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 spammed 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 any bounded function on $X$, the function that maps $\delta_{x_i}$ 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 !

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 spammed 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 any bounded function on $X$, the function that maps $\delta_{x_i}$ 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 mean that $\psi_{|X}$ can be basically any bounded function on $[0,1]$. In particular it has no reason to be measurable and your second question simply don't make sense.

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 spammed 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 any bounded function on $X$, the function that maps $\delta_{x_i}$ 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 and your second question simply don't make sense.