For a compact topological space $X$, denote by $\mathcal{M}(X)$ the Banach space of finite signed Borel (Radon) measures on $X$ with the total variation norm. This is canonically isometric to the dual space of the space of continuous functions $C(X)$. 


If $X$ and $Y$ are compact spaces, then we have the isometric isomorphism
$$\mathcal{M}(X) \hat{\otimes}\_\pi \mathcal{M}(Y) \cong \mathcal{M}(X \times Y),$$
where on the left side, we have the completed projective tensor product (*Side question: Does anybody know a reference for this statement? I only found the corresponding statement for the subspaces $L^1$, although the proof is very similar*). 
**[EDIT: this claim is incorrect in general -- see below]**


Now of course, one can also complete with respect to the injective tensor product, where the norm is given by
$$\varepsilon(z) = \sup\_{\xi, \eta} \bigl\{ (\xi \otimes \eta)(z) \mid \xi \in \mathcal{M}(X)^\prime, \eta \in \mathcal{M}(X)^\prime, \|\xi\| = \|\eta\| = 1\bigr\}$$
for $z \in \mathcal{M}(X) \otimes \mathcal{M}(Y)$ (the algebraic tensor product). However, since the measure space are dual spaces, there is also a third alternative, namely
$$\tilde{\varepsilon}(z) = \sup\_{f, g} \bigl\{ z(f \otimes g) \mid f \in C(X), g \in C(Y), \|f\| = \|g\| = 1\bigr\}.$$
Notice that (at least formally), the latter a much smaller norm, since die dual space of $\mathcal{M}(X)$ are huge. It seems to me that here we have the isomorphism
$$\mathcal{M}(X)\hat{\otimes}\_{\tilde{\varepsilon}} \mathcal{M}(Y) \cong B\bigl(C(X) \times C(Y)\bigr),$$
with the space of bounded bilinear maps on $C(X) \times C(Y)$. Is this correct?


In any case, we have the inclusions
$$\mathcal{M}(X) \hat{\otimes}\_\pi \mathcal{M}(Y) \subseteq \mathcal{M}(X)\hat{\otimes}\_{{\varepsilon}} \mathcal{M}(Y) \subseteq \mathcal{M}(X)\hat{\otimes}\_{\tilde{\varepsilon}} \mathcal{M}(Y)$$


**Now the questions:** 


* Is there a characterization of the middle space?
* Are there "good" (i.e. somewhat natural examples) that show that this inclusion is in fact strict?




---


**Edit:** As Yemon Choi notices, I was possibly wrong by claiming that $\mathcal{M}(X) \hat{\otimes}\_\pi \mathcal{M}(Y) \cong \mathcal{M}(X \times Y)$; in fact, it is only clear that it is an isometrically embedded subspace. Furthermore, by the comment of Matthew Daws, we have $\varepsilon = \tilde{\varepsilon}$. 


**Edit2:** Thanks to the answers below, we obtain the following picture: First, we have
$$ \mathcal{M}(X) \hat{\otimes}\_\varepsilon \mathcal{M}(Y) \cong \mathcal{M}(X) \hat{\otimes}\_{\tilde{\varepsilon}} \mathcal{M}(Y),$$
using [the fact](https://en.wikipedia.org/wiki/Goldstine_theorem) that the inclusion of the closed unit ball of $C(X)$ into the closed unit ball of $\mathcal{M}^\prime(X)$ is dense with respect to the weak-$\*$-topology on $\mathcal{M}^\prime(X)$. From this follows that the norms $\varepsilon(z)$ and $\tilde{\varepsilon}(z)$ coincide.


Now due to the great answer by Matthew Daws below, the diagonal measure $\mu \in \mathcal{M}([0, 1]^2$ is an example of an element which is not contained in $\mathcal{M}([0, 1]) \hat{\otimes}\_\varepsilon \mathcal{M}([0, 1])$, hence neither in $\mathcal{M}([0, 1]) \hat{\otimes}\_\pi \mathcal{M}([0, 1])$.


In fact studying some literature tells me that $\mathcal{M}(X \times Y)$ can be identified with the subset of *integral forms* of $\mathrm{Bil}(C(X) \times C(Y))$. However, it is not yet clear to me yet whether every element of $\mathcal{M}(X) \hat{\otimes}\_\varepsilon \mathcal{M}(Y)$ is actually integral, i.e. contained in $\mathcal{M}(X, Y)$, and whether the inclusion
$$\mathrm{Bil}(C(X) \times C(Y)) \supseteq \mathcal{M}(X \times Y)$$
is really a proper inclusion.



Tensor product of measure spaces
Let me work with general Banach spaces (good references are [Vector Measures](http://www.ams.org/mathscinet-getitem?mr=0453964) by Diestel and Uhl, or [Introduction to Tensor Products of Banach Spaces](http://www.ams.org/mathscinet-getitem?mr=1888309) by Ryan).


For Banach spaces $E,F$ the bounded bilinear maps on $E\times F$ can be identified with the dual space of $E \hat\otimes\_\pi F$ (this is a special case of the universal property of the norm $\pi$) which in turn can be identified with $B(E,F^\*)$ in the obvious way. The injective tensor norm agrees with the norm induced by embedding $E^\*\otimes F^\*$ into $B(E,F^\*)$; the closure is the *approximable operators*. As $M(X)$ has the (metric) [approximation property](https://en.wikipedia.org/wiki/Approximation_property) we find that $M(X) \hat\otimes\_\epsilon M(Y)$ is the *compact* operators from $C(X) \rightarrow M(Y)$.



> 
> An example where $M(X) \hat\otimes\_\epsilon M(Y)$ is not the bounded bilinear maps on $C(X) \times C(Y)$.
> 
> 
> 


Let $X=Y=[0,1]$ with Lebesgue measure and consider $T:C(X) \rightarrow L^1(X) \subseteq M(X)$. By considering e.g. the functions $f\_n(t) = \exp(2\pi itn)$ we see that $T$ is not compact.



> 
> Furthermore, we **do not** have that $M(X\times Y) \subseteq M(X) \hat\otimes\_\epsilon M(Y) = K(C(X),M(Y))$.
> 
> 
> 


Again let $X=Y=[0,1]$ and let $\mu\in M(X\times Y)$ be
$$ \langle \mu, f \rangle = \int\_{[0,1]} f(t,t) \ dt \qquad (f\in C([0,1]^2) $$
so $\mu(E\times F) = |E\cap F|$ the Lesbegue measure of $E\cap F$, for $E,F\subseteq [0,1]$ measurable. This *does* induce a bounded linear map $T:C([0,1]) \rightarrow M([0,1])$ by
$$ \langle T(f), g \rangle = \langle \mu, f\otimes g \rangle
= \int\_{[0,1]} f(t) g(t) \ dt \qquad (f,g \in C([0,1])). $$
But this is just the same $T$ as I considered above: the inclusion of $C([0,1])$ into $L^1([0,1]) \subseteq M([0,1])$; this is not compact.



> 
> This $\mu \in M(X\times Y)$ is also an example of an element not in $M(X) \hat\otimes\_\pi M(Y)$.
> 
> 
> 


I don't know an elementary way to see this; but the theory is interesting. Again see the references at the start. A *vector measure* $\lambda$ on a measure space $X$ is a countably additive map $\lambda:X\rightarrow E$, for a Banach space $E$. $\lambda$ is regular if $\varphi\circ\lambda\in M(X)$ is regular for each $\varphi\in E^\*$. Every *weakly compact* operator $T:C(X)\rightarrow E$ has the form
$$ T(f) = \int\_X f \ d\lambda \qquad (f\in C(X)), $$
for a regular vector measure $\lambda$.


Let $\lambda\_0$ be a finite positive measure on $X$. Then $\lambda$ is *absolutely continuous* with respect to $\lambda\_0$ exactly when $\lambda\_0(E)=0 \implies \lambda(E)=0$. Say that $\lambda$ has the *Radon-Nikodym* property if, whenever $\lambda$ is absolutely continuous with respect to $\lambda\_0$, then $\lambda = \int F \ d\lambda\_0$ for some *Bochner integrable* $F:X\rightarrow E$. (The point is that not all vector measures have this property, in contrast to the scale case). An operator $T:C(X)\rightarrow E$ comes from some member of $M(X) \hat\otimes\_\pi E$ (that is, is a *nuclear* operator) if and only if the representing vector measure has the Radon-Nikodym property.


Again with $X=Y=[0,1]$ with $\mu$ and $T$ as before, the associated vector measure $\lambda:[0,1]\rightarrow M([0,1])$ is
$$ \lambda(E) : F \mapsto |E\cap F| $$
for measurable $E,F\subseteq [0,1]$. This is absolutely continuous with respect to Lebesgue measure. If $F:[0,1]\rightarrow M([0,1])$ is a Bochner integrable function inducing $\lambda$ then
$$ \langle \lambda(E), f \rangle = \int\_E f(t) \ dt =
\int\_E \langle F(s), f \rangle \ ds \qquad (f\in C([0,1])). $$
Thus $F(s) = \delta\_s$ the point-mass at $s$, for all $s\in [0,1]$. However, by the [Pettis Measurability Theorem](https://en.wikipedia.org/wiki/Weakly_measurable_function), $F$ should be *essentially separably valued* but clearly this is not the case, giving the required contradiction.


My *guess* would be that you can boost this example up to show that $M(X)\hat\otimes\_\pi M(Y)$ is strictly contained in $M(X\times Y)$ unless one of $X$ or $Y$ is discrete (in which case we do have equality).


