How to integrate a function which is unbounded on a set of volume zero I am trying to compute the iterated integral,
$$\int_0^1 \int_{\sqrt{y}}^1 e^{y/x}\ dx dy$$
by changing the order of integration using Fubini's theorem. In order to do this, I need to first show that the corresponding double integral, 
$$\int_\Omega e^{y/x}\ dx dy$$
exists. 
If I take any sub-rectangle of a partition of $\Omega$ containing the origin, no matter how small, then the supremum of $f$ on this sub-rectangle doesn't exist. Then it seems like the upper sum of $f$ doesn't exist either.
May I please have a small hint about how to prove that $f$ is integrable?
In general, it seems like if I have a function which is unbounded or even undefined on a set $S$ of volume zero, I should be able to integrate the function and say that the integral is equal to whatever it would be if I set the function values on $S$ to $0$ (or some constant.) The book I'm using only defines the Riemann integral for bounded functions, so I don't know how to handle cases like these. 
Thanks!
 A: The integrand should be $f(x,y)=\chi_{\sqrt{y}\leq x\leq 1}e^{y/x}$. On the little square $[0,\delta]\times[0,\delta]$, for all $x$ such that $\sqrt{y}\leq x$, then $f(x,y)\leq e^{y/\sqrt{y}}=e^{\sqrt{y}}$ which is bounded as $0\leq y\leq\delta$. For any $x$ such that $x<\sqrt{y}$, then $f(x,y)=0$, which is good.
Alternate Way:
Consider the domain $\Omega=\{(x,y):\sqrt{y}\leq x\leq 1, 0\leq y \leq 1\}$, the only trouble point from this domain for $f(x,y)=e^{y/x}$ is exactly the point $(0,0)$. You have to define $f(0,0)$ for any other value. For any subrectangle $[0,\delta]\times[0,\delta]$ which intersects $\Omega$, still, except for $(0,0)$, for any other points from this domain, one still has $\sqrt{y}\leq x$ and hence $f(x,y)\leq e^{\sqrt{y}}$ which is bounded on $([0,\delta]\times[0,\delta])\cap\Omega$.
To prove for integrability, one goes with Lebesgue integral, and you just need to find a nonnegative bounded function, because $\Omega$ is a bounded domain, while $e^{\sqrt{y}}$ serves as a candidate.
