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Similarly to the decomposition $L_2(\mathbb R^n) = L_2(\mathbb{R})^{\otimes n}$ as vector spaces (and even as Hilbert spaces) , do we have $bm(\mathbb{R^n}) = bm(\mathbb{R})^{\otimes n}$

where $bm(\mathbb{R})$ refers to the space of borel measures over the reals?

The reason I was thinking of this decomposition is because we have such a decomposition for the $\sigma$-algebras $\mathbf{B}(\mathbb{R^n})$ and the construction of product measures as product of marginales measures is similar to the wa the isomorphism between the $l_2$ spaces is constructed.

Thank you

Christian Remling
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Elesthor
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3 Answers3

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In its present form, your question is not well-posed (and this helps to explain the apparent discrepancy between the posted responses). In the case of infinite dimensional spaces, there are many possible definitions of the tensor product, even for Banach spaces, and the answer to your question will depend on the choice of a particular one. As pointed out in the first answer, there will be no such representation if you regard the space of measures as a Banach space in the usual way. However, there are suitable concepts of tensor products for which a representation of the type you desire is valid. In the simpler case of compact metric cases, e.g., products of compact intervals, the appropriate category is that of Waelbroeck spaces, in the non compact case you will require a more baroque one, that of CoSaks spaces. Unfortunately, the rather obscure nature of these constructions means that there is no explicit version to be found in the literature, to my knowledge, but it is quite easy to create one from known results in the Banach space situation.

Added as an edit: I have decided to edit this answer since there seems to be some confusion around. Firstly, I persist in my claim that the question cannot be answered as it stands since it specifies neither the appropriate structure on the space of measures nor the sense in which the tensor product is to be taken. However, I repeat that there are such specifications which allow a positive answer. They are, in my opinion, perfectly natural but, sadly, haven't found their way into the literature to my knowledge.

I will try to give brief sketch. It is more convenient to consider the case of a general (completely regular) space $X$. I will look firstly at the special cases where this is compact, resp., locally compact.

  1. If $X$ is compact, then the space of measures is, as the dual of a Banach space, a so-called Waelbroeck space. This concept was introduced by H. Buchwalter based on work by the name-giver. It is dicussed in some detail in "Categories of Banach spaces" by Cigler et al. There is a natural concept of a projective tensor product for such spaces (nota bene related to, but not identical to, the corresponding one for Banach spaces) and the result you require is then valid.

  2. the case of a locally space is similar but here you use the structure of the space of measures as the dual of $C_0(X)$. This corresponds to vague convergence.

  3. Finally, for the general case, it is appropriate to consider the space of bounded, Radon measures as the dual of $C^b(X)$, the bounded continuous functions with the so-called strict topology. This corresponds to weak convergence of measures. Again, there is a suitable category for which your question has a positive response.

Which of 2. or 3. are relevant depends on which type of convergence you are interested in---vague or weak.

report
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  • Naive question: for Waelbroeck spaces (which I think of as duals of Banach spaces, and morphisms being the weak-star continuous maps between these duals) does this tensor product linearize jointly continuous maps or separately continuous maps? (I have a very vague memory of this tensor product when I looked at the CLM book many years ago) – Yemon Choi Apr 21 '15 at 10:33
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    If $X$ is locally compact not compact, as far as I remember, the dual of the Banach space $\mathcal{C}^{(0)}(X)$ is the space of bounded measures (and not all $\mathcal{M}(X)$). Am I right ? – Duchamp Gérard H. E. Apr 21 '15 at 12:01
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    I had understood that the question was about bounded Borel measures. If this is not the case, then we are talking about the dual of the $LF$-space(in the sense of L. Schwartz) of continuous functions with compact supporst. However, the answer remains the same: yes, if one uses the correct (non standard) version of the tensor product (the fact mentioned by Christian Remling remains a barrier to the use of any of the standard tensor products of Banach spaces or even locally convex spaces) – report Apr 21 '15 at 12:16
  • .@report (for 2, cont'd) I read in the title "the space of all Borel measures" so as the concerned spaces are very particular ($\mathbb{R}^d$), I take it as concerning all measures (which includes the Lebesgues one). – Duchamp Gérard H. E. Apr 21 '15 at 12:37
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    Yes, mea culpa, I misread the question. I will leave my answer in the hope that it might be of interest to somebody and also in view of the fact that as indicated in my comment above the same general remark holds: there is a natural but nonstadard interpretation of the data (structure on the dual space, interpretation of the tensor product) for which the response is yes. – report Apr 21 '15 at 12:46
  • .@report I amended my (first draft of) answer in the direction of an attempt (not yet completely successful) to solve the general problem. It can turn out that the restriction to bounded measures be necessary. – Duchamp Gérard H. E. Apr 21 '15 at 13:08
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This won't work. You can't approximate the restriction $\rho$ of one-dimensional Hausdorff measure to $\{(x,x): 0\le x\le 1\}$, say, in total variation norm by a finite linear combination $\sum c_j \mu_j\otimes\nu_j$. No matter what you do ($\mu_j,\nu_j$ continuous, or with a point part), $\mu\otimes\nu$ and $\rho$ will always be mutually singular.

Christian Remling
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My answer is more a plan than a real answer. It aims to not close too early by a brutal "no-go" algebraic statement. Of course, if you ask for finite (i.e. algebraic) tensor product this is not true.

If you are thinking of completed tensor products, then, I think yes. Of course, in order to have the completed tensor product one has to prove the following :

1) The natural map $bm(X)\otimes bm(Y)\rightarrow bm(X\times Y)$ is into ($X,Y$ are, in fact - and more generally - two topological spaces) and

2) For some convergence, $bm(X\times Y)$ is complete (at least for $X\times Y=\mathbb(R)^d$) and the image of $bm(X)\otimes bm(Y)$ is dense in it.

... which I thought was standard but now needs some elaboration. Let me add, after the comment of Yemon below, that the plan for tensor product completion I describe above seems not in general to give the projective tensor product of Banach/locally convex spaces.

Note 1 We know that $bm_+(X)$ (cone of positive measures) is complete for the vague topology. From this, the hole space $bm(X)$ is complete for the quasi-strong topology defined by the seminorms $p_f(\mu)=|\mu|(f)$ for $f$ a positive test function and $|\mu|=\mu^++\mu^-$. What I do not yet know is about the density of $bm(X)\otimes bm(Y)$ within $bm(X\times Y)$ for the quasi-strong topology.

Note 2 If you restrict your problem to the spaces of bounded measures $bm_b(Z)$, you can equip the space of measures with the ultrastrong topology (Boubaki's terminology Integration Ch III paragraph 1 exercise 15) given by the (Banach) norm. For this norm $bm_b(X\times Y)$ is complete.

Duchamp Gérard H. E.
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  • Yes, it was in the sense on completed product. Thank you for the answer. – Elesthor Apr 10 '15 at 12:19
  • If needed, we can elaborate a bit. I am at your disposition – Duchamp Gérard H. E. Apr 10 '15 at 14:55
  • That would be very nice of you, thanks! Or a reference if you have. – Elesthor Apr 14 '15 at 13:06
  • .@Elesthor As a reference, you can have a look at thm 417C p89 of D.H.Fremlin : Measure theory, Topological Measure Spaces. Once you know the (1) is true (embedding) and that the space of all bm is complete (easy, this is a weak topology), then comes naturally the completed tensor product ! – Duchamp Gérard H. E. Apr 16 '15 at 20:28
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    Just to clarify for anyone else reading: the tensor product completion Gerard describes above is not in general the same as the projective tensor product of Banach spaces – Yemon Choi Apr 19 '15 at 18:17
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    There are two things that are wrong with the above answer. Firstly it is not true that the space of measures is complete for vague convergence. Secondly, one would require denseness in some norm to finish the argument, not that in vague convergence. I repeat that there is a notion of tensor product for which the result is true, but it is not one in the category of Banach spaces. – report Apr 20 '15 at 06:40
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    .@report you are right. I tried to repair (using the cone of positive measure and changing the seminorms), but not in general so far. So I amend my answer which becomes only partial. – Duchamp Gérard H. E. Apr 21 '15 at 08:17
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    .@report I made precise a set of semi-norms which would fit the plan, see the note added in the answer. Thanks again for your comment(s) – Duchamp Gérard H. E. Apr 21 '15 at 09:37