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.

projectivetensor product linearises bilinear maps, so I think (unless I misunderstand) that $B(C(X)\times C(Y))$ is $M(X) \hat\otimes_\pi M(Y)$. $\endgroup$ – Matthew Daws Jun 1 '17 at 14:25