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 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.
Best Answer
Let me work with general Banach spaces (good references are Vector Measures by Diestel and Uhl, or Introduction to Tensor Products of Banach Spaces 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 we find that $M(X) \hat\otimes_\epsilon M(Y)$ is the compact operators from $C(X) \rightarrow M(Y)$.
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.
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.
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, $F$ 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).