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


*

*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.

*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.

*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.
A: 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.
A: 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.   
