When does the diagonal have zero measure? Let $\mu$ be a Radon measure on a space $X$ and consider the product measure $\mu \otimes \mu$ on $X \times X$. Is there some necessary and/or sufficient condition on $\mu$ to guarantee that the diagonal $\Delta \subset X \times X$ has zero $\mu \otimes \mu$-measure?
 A: Let us assume that $\mu$ is sigma-finite. In this case, a variant of the Fubini-Tonelli theorem applies (see e.g. Folland, Real Analysis, Theorems 7.26, 7.27).
Hence ($M(x) := \{y \in X \mid (x,y) \in M\}$ denotes the section of $M$ at height $x$):
$$
(\mu \times \mu)(\Delta) = \int_X \mu(\Delta (x)) \,d \mu(x) = \int_X \mu(\{x\}) \, d\mu(x).
$$
Thus, if $\mu(\{x\}) = 0$ for all $x \in X$, we get $(\mu \times \mu)(\Delta) = 0$. As @Crotsul noted, this is also a necessary condition.

Further comments: The diagonal $\Delta$ is closed in $X \times X$. Thus, it is measurable with respect to the Borel sigma-algebra on $X \times X$. Nevertheless, if $X$ is not second countable, it can happen that $\Delta$ is not measurable with respect to the product sigma algebra on $X \times X$. Thus, you should add which interpretation of $\mu \times \mu$ you are using (usual product measure or Radon product). Of course, if $X$ is second countable, these are the same.
Finally, I don't know what happens if $\mu$ is not $\sigma$-finite. But as long as $X$ is second countable, it will also be $\sigma$-compact and hence $\sigma$-finite, so that the possible problem vanishes.
