Let $X, Y$ be locally compact Hausdorff spaces with nonnegative regular measures $\mu, \lambda$. By definition (in the book I'm reading) a regular measure is a Borel measure for which every Borel set can be approximated from below by compact sets, or from above by open sets. In the appendix in Rudin's book Fourier Analysis on Groups, he makes a couple of statements which I have not found anywhere else. I just wanted to confirm that they are correct. First there is:
There is a unique regular measure $\mu \times \lambda$ on $X \times Y$ for which $\mu \times \lambda(A \times B) = \mu(A) \lambda(B)$ for $A \subseteq X, B \subseteq Y$ Borel.
This statement seems fishy to me, because I know product measures are in general not uniquely determined by their values on products of Borel sets. Some things can happen when $X$ or $Y$ is not $\sigma$-finite. Next there is:
("Fubini's theorem") If $f \geq 0$ is a Borel function on $X \times Y$, then $$ \int\limits f d(\mu \times \lambda) = \int\limits_X \int\limits_Y f(x,y)d\lambda(y) d\mu(x) = \int\limits_Y \int\limits_X f(x,y)d \mu(x) d \lambda(y) $$
The reason I am concerned with this is that there is a very similar result stated in Hewitt and Ross, Abstract Harmonic Analysis, Theorem 13.9, which makes the same claim for $f \geq 0$ Borel (although they allow $f$ to take on the value infinity, Rudin clearly does not). This theorem gives the same result, but requires the addition hypothesis that $f$ vanishes off of a countable union of $(\mu \times \lambda)$-measurable sets which each have finite measure.
Would any experts on the subtleties of regular measures be willing to explain why or confirm that Rudin's statements are correct? One thing I am wondering is whether Rudin is tacitly assuming $X,Y$ and $X \times Y$ have an additional property (which is true of locally compact Hausdorff abelian groups, which is what his book is mainly concerned with), namely that each space is a disjoint union of $\sigma$-compact subspaces $X_{\alpha}$, such that every $\sigma$-compact subset of $X$ is contained in a countable union of the $X_{\alpha}$.
Best Answer
One question is what you mean by "every Borel set can be approximated by compact sets from inside and by open sets from outside". The usual definition I take (adopted by Folland, who also provides essentially the example mentions below) is that $\mu(M) = \inf_{U \subset M}\mu(U)$ (with $U$ open) should hold for all Borel sets $M$, whereas $\mu(M)=\sup{K \subset M}\mu(K)$ is only required for open $M$.
The standard counterexample to things like this is the following: Let $X = \Bbb{R}_d$ be the group of reals (with the usual addition), but with the discrete topology and let $Y=\Bbb{R}$.
Now, take $A\times B=\Bbb{R}_d \times \{0\}$. Note that the relevant sets are closed, hence Borel. By your requirement, $\mu\times\lambda (A \times B)= \mu(A)\lambda(B) =\infty \cdot 0=0$, where $\mu,\lambda$ are the Haar measures. Note that $\infty \cdot 0=0$ is the only sensible choice for produce measures (consider $A\times \emptyset$).
But for any open set $U \supset A \times B$, we have $U \supset \{x\}\times (0,\epsilon_x)$ for some $\epsilon_x >0$. Since $\Bbb{R}$ is uncountable, there is some $n$ and infinitely many $x_1, x_2, \dots$ with $\epsilon_{x_\ell} \geq 1/n$ for all $\ell$. Together with your assumptions on the product measure, this easily yields $\mu \times \lambda (U) = \infty$. Thus, the product is not outer regular.
Since a product measure as described by Rudin does not exist in general, talking about the Fubini problem is of little use. Anyway, the above discussion shows that a regular version of the product measure would yield $\mu\times\lambda (A\times B)=\infty$, so that your Fubuni formula fails when applied to the indicator of this set.
Finally, a very nice discussion of these topics can be found in Folland's books "Real Analysis" and "Abstract Harmonic Analysis". There he constructs a Radon product of two Radon measures and explains the difference to the usual product measure.