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



Is the space of all borel measures on $\mathbb R^n$ isomorphic to the tensor product of spaces of borel measures on $\mathbb R$?
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. 


