Fubini's theorem for Riemann integrals?

Question

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!

Answer 1

Yes. Both Calculus on Manifolds and Analysis on Manifolds have proofs of this without any reference to measures. The "almost everywhere" conditions turn into something slightly different. This is from Analysis on Manifolds:

Let $Q=A \times B$, where $A$ is a rectangle in $\mathbb{R}^k$ and $B$ is a rectangle in $\mathbb{R}^n$. Let $f:Q\to\mathbb{R}$ be a bounded function; write $f$ in the form $f(x,y)$ for $x\in A$ and $y\in B$. For each $x\in A$, consider the lower and upper integrals $$ \underline\int_{y\in B} f(x,y) \quad\mathrm{and}\quad \overline\int_{y\in B} f(x,y). $$ If $f$ is integrable over $Q$, then these two functions of $x$ are integrable over $A$, and $$ \int_Q f = \int_{x\in A}\underline\int_{y\in B} f(x,y) = \int_{x\in A}\overline\int_{y\in B} f(x,y). $$

I'm not sure what happens to the other part that says under certain conditions, the integrability of $f_x(y) = f(x,y)$ and the existence of $\int |f_x|<\infty$ implies that $f$ is integrable.

Answer 2

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.