Fubini's theorem for Riemann integrals? The integrals in Fubini's theorem are all Lebesgue integrals. I was wondering if there is a theorem with conclusions similar to Fubini's but only involving Riemann integrals? Thanks and regards!

To see the difficulties of Fubini with Riemann integrals, study two functions $f$ and $g$ on the rectangle $[0,1]\times[0,1]$ defined by:
(1) $\forall$integer $i\ge0$, $\forall$odd integer $j\in[0,2^i]$,
$\forall$integer $k\ge0$, $\forall$odd integer $\ell\in[0,2^k]$,
define $f(j/2^i,\ell/2^k)=\delta_{ik}$ (here, $\delta_{ik}$ is the Kronecker delta, equal to one if $i=k$ and $0$ if not) and $g(j/2^i,\ell/2^k)=1/2^i$;
and
(2) $\forall x,y\in[0,1]$, if either $x$ or $y$ is not a dyadic rational, define $f(x,y)=0$ and $g(x,y)=0$.
Then both iterated Riemann integrals of $f$ are zero, i.e., $\int_0^1\int_0^1 f(x,y)\,dx\,dy=\int_0^1\int_0^1 f(x,y)\,dy\,dx=0$. However, the Riemann integral, over $[0,1]\times[0,1]$, of $f$ does not exist.
Also, the Riemann integral, over $[0,1]\times[0,1]$, of $g$ is zero. However, $\forall$dyadic rational $x\in[0,1]$, the Riemann integral $\int_0^1 g(x,y)\,dy$ does not exist. Consequently, $\int_0^1\int_0^1 g(x,y)\,dy\,dx$ does not exist, in the Riemann sense.
