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Let $R([0,1])$ be the unital commutative $C^*$-algebra of complex valued Riemann integrable functions on $[0,1]$ with pointwise operations and the supremum norm.

In the 1980 paper The Gelfand space of the Banach algebra of Riemann integrable functions by Jörg Blatter (MR0602719) the author states, that no satisfactory representation is known for the space of maximal ideals of $R([0,1])$.

For the smaller space of regulated functions on $[0,1]$ the maximal ideal space consists of the functionals $\beta_x,\delta_x, \gamma_x$ given by $\beta_x(f)=f(x)$, $\delta_x(f)=\lim_{y\to x^+}f(y)$ and $\gamma_x(f)=\lim_{y\to x^-}f(y)$ where $x\in [0,1]$, see e.g. The character space of the algebra of regulated functions by S. K. Berberian (MR0487932).

I wonder whether some progress has been made concerning the maximal ideal space of $R([0,1])$ since Blatter's paper. In fact I would be very thankful if someone could provide me with an example of a character on $R([0,1])$ which is not of the form $\beta_x$, $\delta_x$, $\gamma_x$ as above or some intuition why the determination of the maximal ideal space of $R([0,1])$ is so much harder than the corresponding problem for the space of regulated functions.

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    lvb, just a remark: your last sentence seems to imply that the characters $\beta_x$, $\delta_x$, $\gamma_x$ extend to $R([0,1])$ which I don't think they do. By the way, what's the topology on $[0,1]\sqcup [0,1]\sqcup [0,1]$ regarded as the promitive ideal space of the regulated functions?2011-06-26
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    In fact *every* maximal ideal of R is an extension of one of $\beta_x$, $\delta_x$, $\gamma_x$. It's just that they do not extend in a unique way. You can construct the maximal ideals via ultrafilters.2011-06-26
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    Thank you. You are right. I should have seen that. Do you know a reference where this construction is carried out?2011-06-27
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    lvb, if you want to make sure that George Lowther notices your comment, use @George Lowther.2011-06-27
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    lvb, I think it's an interesting question. A simple remark is that $R([0,1])$ is a non-separable Banach space. Similarly, the complex Banach space $l^\infty ([0,1])$ is also non-separable. Can one find a concrete maximal ideal of $l^\infty ([0,1])$ ? I don't know. (supplementary question) David Bernier2011-07-10
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    @David: As a new user with < 50 rep, you are unable to leave comments. However, as your post was really a comment rather than an actual answer to the question, I manually converted it into a comment. You will soon have enough rep to leave comments directly yourself.2011-07-12
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    May be his can help you. Consider algebra of sets $\Sigma=\{E\subset[0,1]: \lambda(\operatorname{int}(E))=\lambda(\operatorname{cl}(E))\}$. Then $L_\infty([0,1],\Sigma,\lambda|_\Sigma)$ is isomorphic to the space Riemann integrable functions with $\sup$-norm. (see exercise 363Yi in Measure theory D. H. Fremlin Vol 3)2013-12-09

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