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

R. D. Mauldin ([1]) 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.

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  • $\begingroup$ can you explain what is this integral ? (integrating a function defined on the set of all borel set with respect to a measure ?) Also what do you call $\psi_{|X}$ ? I think the answer to 2. is immediate and has nothing to do with Mauldin result: $V_{\infty}$ is a von neuman algebra in which $C([0,1])$ is weakly dense so it is the quotient of the enveloping algebra (which is isomorphic to the second dual) $\endgroup$
    – Simon Henry
    Commented Apr 8, 2016 at 9:18
  • $\begingroup$ Also, please note that Mauldin's result holds under the Continuum Hypothesis. $\endgroup$
    – Tomasz Kania
    Commented Apr 8, 2016 at 10:14
  • $\begingroup$ @Simon Henry, Mauldin's paper says that: If $\psi$ and $\mu$ are real valued functions on $B$ then "the number $\omega$ is the integral of $\psi$ with respect to $\mu$" means that if $\epsilon>0$, then there is a subdivision $D$ of $[0,1]$ such that if $D'$ refines $D$, then $$|\sum_{E\in D'} \psi(E)\mu(E)-\omega |\leq \epsilon$$. $\endgroup$
    – ABB
    Commented Apr 8, 2016 at 10:29
  • $\begingroup$ Yes, Mauldin's result holds under the Continuum Hypothesis. Thanks for your comment. $\endgroup$
    – ABB
    Commented Apr 8, 2016 at 10:32
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    $\begingroup$ I mean't that you should edit your question so that what you have explained to me in the comment appears in the question. In its present form there is no way to understand what you mean by $\psi_{|X}$ as there is no mention of "$X$" in your question, and there is several way to restrict function from $B$ to $[0,1]$. Mentioning that it requires the continuum hypothesis and what is this integral would also be a good things. $\endgroup$
    – Simon Henry
    Commented Apr 8, 2016 at 11:11

1 Answer 1

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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 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 !

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  • $\begingroup$ Dear Simon Henry, It was really nice. Thanks. $\endgroup$
    – ABB
    Commented Apr 8, 2016 at 11:27
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    $\begingroup$ Surely you mean "spanned" and not "spammed" ;) $\endgroup$
    – Loïc Teyssier
    Commented Apr 8, 2016 at 12:40

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