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According to the comments of Nate Eldredge I did revise the question. In particular I change "$C^{*}$ algebras" to "Banach algebras".

Is there a reference who introduce the following measure on Banach algebras and investigate it from the view point of Banach algebras or from the view point of ergodic theory?:

Let $A$ be a Banach algebra. We define an outer measure $ m^{*}$ on $A$ as follows: For $B \subset A$ we put $$m^{*}(B)=\mu ^{*} \left(\bigcup_{b\in B} \mathrm{sp}(b)\right)$$ where $\mu^{*}$ is the standard outer measure of $\mathbb{C} \simeq \mathbb{R}^2$. Then the Caratheodory methods gives us a measure space structure on $A$ with measure $m$. This measure is an invariant measure for both scalar translation and inner automorphism of algebras.

Does this construction gives us nothing new when $A$ is the matrix algebra? In particular is it true to say that, for the matrix algebra, a subset is measurable in this spectral sense if and only if it is mea surable in the classical sense?

As the final question: Is there a Banach algebra $A$ for which this measure construction does not lead to triviality: that is $A\neq \mathbb{C}$ would have subsets with positive finite measure?

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    $\begingroup$ It definitely doesn't make the Borel sets measurable. As you suggest, let $X$ be two points, so $A = \mathbb{C}^2$ and the spectrum of $(z,w)$ is $\{z,w\}$. Let $C \subset \mathbb{C}$ be your favorite Borel set of finite positive measure, let $v \notin C$, and consider the sets $B_1 = \{(u,v) : u \in C\}$, $B_2 = \{(v,u) : u \in C\}$. Then $B_1, B_2$ are certainly Borel, and $B_1 \cap B_2 = \emptyset$. But we have $\bigcup_{b \in B_1} \operatorname{sp}(b) = C \cup \{v\}$ and the same for $B_2$, as well as $B_1 \cup B_2$. [...] $\endgroup$
    – Nate Eldredge
    Commented Jun 29, 2016 at 6:44
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    $\begingroup$ So $m^*(B_1) = m^*(B_2) = m^*(B_1 \cup B_2) = \mu(C)$ and thus $m^*$ is not countably additive on these sets, so they are not measurable. $\endgroup$
    – Nate Eldredge
    Commented Jun 29, 2016 at 6:44
  • $\begingroup$ @NateEldredge Thank you for your very interesting comments. What about the unit Ball in $M_{n}(\mathbb{C})$?Is it easy to compute the integral of trace on this ball, if it is measurable? $\endgroup$
    – Ali Taghavi
    Commented Jun 29, 2016 at 7:12
  • $\begingroup$ @NateEldredge According to your very helpful comment I realize that the unit ball is not measurable since $\begin{matrix} \lambda&k\\0&\lambda \end{matrix}$ is disjoint from the unit ball where $\lambda$ varies in the unit disc and $k$ is a very large constant. Your comment perhaps shows that we do not get interesting measure theory on C* algebras because of exsistence of nilpotent elements in non commutative case, so I revise the question with replacing C* algebras by Banach algebras.Thanks again for your comments. $\endgroup$
    – Ali Taghavi
    Commented Jun 29, 2016 at 11:28
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    $\begingroup$ I'm voting to close this question as off-topic because it is just a fishing expedition; it does not seem that sufficient thought was put into the question beforehand $\endgroup$
    – Yemon Choi
    Commented Jun 29, 2016 at 18:58

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