Can be iteratedly integrate w.r.t. all direction $\Rightarrow$ integrable? Let $f:[0,1]\times[0,1]\to\Bbb R$ be bounded. Suppose $\int_0^1(\int_0^1 f(x,y)dy)dx$ and $\int_0^1(\int_0^1 f(x,y)dx)dy$ are all exist and equal. Then is $f$ Riemann integrable on $[0,1]\times [0,1]$? I tried to use Fubini's theorem, but the condition of that theroem doesn't suit in this case. Can I avoid digging into $\epsilon-\delta$ proofs?
 A: The answer seems to be no, but it depends possibly upon how you define
the objects in question.
I presume here that definitions are as follows:
For every $x\in [0,1]$: $F_x = \int_0^1 f(x,y)dy$ is defined as a Riemann integral, i.e. that the upper and lower integrals (w.r.t. step-functions on finite partitions into intervals) are equal and similarly for $F_y=\int_0^1 f(x,y)dx$.
For the double integral upper and lower integrals w.r.t finite partitions into rectangles should be equal.
In that case, let  $\Omega\subset [0,1]\times[0,1]$ be a countable dense subset of points $(x_n,y_n)$ with the property that for every $x\in [0,1]$, $y\in [0,1]$,
the sets $\{n: x_n=x\}$ and $\{n: y_n=y\}$ have cardinality 0 or 1. 
Let $f$ be the indicator function on $\Omega$. Then you check that $F_x=F_y=0$ for every $x,y$ (since e.g. $y\mapsto f(x,y)$ is zero everywhere except for at most one point) but that $f$ is not Riemann integrable. The upper integral equals one (you can not avoid the dense set) and the lower is zero.
