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Let X be a locally compact (Hausdorff) topological space, K the vector space (over $\mathbb{C}$) of continuous functions with compact support, defined in X and with complex values. A measure $\mu$ on X is defined in the linear theory as an element of the dual space of K, where K is equipped with a topology having some complicated definition (using a direct limit), but more simply $\mu$ is a linear mapping from K to $\mathbb{C}$ such that, for every compact subset C of X, the restriction of $\mu$ to the subspace $K_C$ of K consisting in those functions that have their support contained in C, is continuous when $K_C$ is equipped with the topology of uniform convergence. Looks still complicated, but appears as natural because this condition is automatically satisfied by any linear mapping K->$\mathbb{C}$ that is positive, i.e. takes positive real values on positive real functions (positive meaning $\ge 0$) - and real measures are differences of two positive ones, complex measures are equal to $\mu_1+i\mu_2$ ($i$ the imaginary unit) where $\mu_1$ and $\mu_2$ are real measures.

Using such a measure, one defines - with respect to $\mu$ - a (pseudo-) semi-norm for all functions X->$\mathbb{C}$ (pseudo because it takes the value $\infty$ for some functions) turning the vector space of these functions into a topological space said to have the topology of convergence in the mean. The functions of the closure $\bar K$ of K in this space are by definition the $\mu$-integrable functions. And $\mu$ can be extended by continuity to $\bar K$ and the value at an element $f$ of this will be (defined as) its integral $\int f(x) d\mu (x)$. When $f$ has only the values $0$ and $1$, the set $A = f^{-1}(1)$ is said to be $\mu$-integrable if $f$ is, and its integral is written $\mu(A)$ and called the measure of $A$.

Let's consider the case when $\mu$ is bounded, which means that the whole space X is $\mu$-integrable. Then one shows that we get in a classical sense a complex measure space X, the $\mu$-integrable subsets of X being the measurable ones and the values of the classical measure on X being as stated.

Is the mapping sending any bounded $\mu$ to the classical measure obtained from it restricted to the set of the borelian subsets of X (which are measurable for any $\mu$) injective? Proof?

Jean-Claude Arbaut
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Ulysse Keller
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    BTW: Clearly - in case the answer is yes - any proof is probably rather long ... so I don't expect details here. An external link or reference to a paper article / book leading to a result from which one can easily infer what I'm looking for or at least a weak form of it ... will do – Ulysse Keller Dec 04 '22 at 09:56
  • I saw just yesterday that my question got an edit, and trying to find out how, I eventually discovered the added reference (-) request. Being relatively new to this website, I am not sure what is the "normal" reaction to such a request. I guess that I'll have to make some (more) comments which have to contain references related to my original post. Also I am not sure FOR WHAT aspect of my post references are requested. So I'll give several comments of the type indicated – Ulysse Keller Dec 07 '22 at 11:42
  • BTW the user J.-C. Arbaut is from France and lives in a place not far from my country Switzerland ... so I decided to put similar informations into my profile. My "About (me)" shows my background - also for the question above – Ulysse Keller Dec 07 '22 at 16:59
  • By the classical measure theory I mean what you can read nearly everywhere when a measure theory is presented. So no need for an explicit reference. Often "measure" means positive (i.e. nonegative) measure. Usually they say signed measures cannot take both +00 and -00 as values, and complex measures must be bounded ... which seems to be due to the fact that measures are supposed to be defined on a sigma-algebra of subsets so the whole space is measurable. This shocked me when I read Halmos' measure theory and I preferred Bourbaki's theory – Ulysse Keller Dec 07 '22 at 17:20
  • By the linear measure theory I mean what you find in N.Bourbaki's book Intégration originally 3 volumes at edition Hermann (Paris) which I have (paper) and studied mainly 1st volume, chapt. 3 & 4. The present version is edited by Springer (Berlin / Heidelberg) and they have English translations. Probably including this. Not all of the content is needed for my question and simplification is possible concerning continuity as mentioned in my orig. post – Ulysse Keller Dec 07 '22 at 19:21
  • My mention of Laurent Schwartz, mainly known as the "inventor" of the distributions, is due to his support of Bourbaki's measure theory to which his distributions are quite analogous. In his book (2nd ed. of 1957) on page 17, one reads: "Le théorème de Riesz a pris une importance de plus en plus grande. Aujourd'hui, il est devenu indispensable de définir une mesure mu comme forme linéaire continue ..." (In case translation is needed, I leave it to others.) According to Wikipedia (fr) L. Schwartz has been a member of Bourbaki, I guess he contributed to Bourbaki's measure theory – Ulysse Keller Dec 07 '22 at 19:55
  • Bourbaki's measure theory names measures what is called in the classical theory Radon measures. So it's not quite correct to say (what I read somewhere) that Bourbaki defines differently R. measures. Bourbaki defines measures only as linear functionals in the context of locally compact spaces - except in chapter 9 which I have not yet seen; so no abstract measures so far; in my opinion these are really interesting only when positive (because of the drawbacks for signed / complex m.) – Ulysse Keller Dec 07 '22 at 20:10
  • If the refer. requ. concerns specifically my question, I can't serve. It isn't anything like an exercise I found. My motive is: I am currently studying chapter 12 in W. Rudin's Functional Analysis on bounded operators on a Hilbert space ... Like everywhere in this book where measure theory is used, it's the classical version. For my purposes, I adapted (to linear theory) the definition of a resolution of the identity (his § 12.17) ... getting a little problem: if I replace the set function E_x,y in the def. by a linear funct., this one might a priori not be unique. Thus my question. – Ulysse Keller Dec 07 '22 at 20:35
  • In fact, my original post asked itself for a reference - a direct answer would be too long for this place! So I could have put myself the tag reference-request: Indeed, looking at the selection of posts with this tag, I found one with no other tag. I am nearly sure that the tag was activated by the original poster of that one. Now I finally found - in the recent edition of the Bourbaki text - a result that implies a yes to my quest.: it is stronger because 1) the hyp. concerns the restriction to compact sets only: compact => closed => borel in H'dorff space and the measure need not be bounded – Ulysse Keller Dec 07 '22 at 20:51

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I finally found an answer in Bourbaki's Intégration newest edition at Springer (Berlin / Heidelberg) - I looked there for the first time, trying to answer the reference request I received. It does more than to answer my question, it seems to practically complete my program of a bridge between the two versions of measure theory. The pages 159-169 of the book containing chapters 1 to 4 (French original) do this (parts 9 to 11 of chapter 4, §4); for my question, parts 9 & 10 of said § suffice (ending on page 163 with corollary 3 of prop.19 which implies a yes to my question). Although I have access to the book online from my home PC via the library of ETHZ (tech. high school of Zurich) I doubt that the following URL will be usable by most readers of this: here

Ulysse Keller
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