Is this unbounded function in multiple variables Riemann integrable? I wondering if the following function $f:[-1,1]^2\to\mathbb{R}$:
$
    f(x,y)= 
\begin{cases}
    0,& x\not=0\\
    \frac{1}{y}, & x=0
\end{cases}
$
is Riemann integrable? I think not, because clearly the upper and lower sums do not converge. But on the other hand I've heard the the behaviour on a jordan-0 set doesn't matter?
More generally, can unbounded functions in multiple variables ever be Riemann integrable?
 A: You write "clearly the upper and lower sums do not converge". This is true, but I can turn it into an improper Riemann integral by removing a circle of radius $\delta$ about $(0,0)$. Then the function is bounded above by $1/\delta$, so is Riemann-integrable.
Next, I can show that the improper integral converges to zero:
I can cover the rest of the line $x=0$ with boxes with widths as small as I want, and arrange it for the upper and lower sums of this partition to tend to zero. For example, take the box
$$ [\delta,1 ) \times (-\epsilon/2,\epsilon/2), $$
which has area $(1-\delta)\epsilon<\epsilon$. The largest value of the function in this interval is $1/\delta$, so the contribution of this box to the Riemann sum is $\epsilon/\delta$, which for any $\delta>0$ can be made as close to $0$ as we like by choosing $\epsilon$ small enough. Hence the upper sums have limit inferior equal to $0$, by repeating the same argument for the other half of $x=0$ and noting that the function is $0$ everywhere else. Therefore the integral is zero for any $\delta>0$, and hence the answer as an improper integral is zero.

To answer the second part of your question, unbounded functions are never Riemann-integrable, but may exist as improper integrals in a similar way to the above. For example, since the area measure in polar coordinates is $r\, dr \, d\theta$, the function $(x^2+y^2)^{-1/2}=1/r$ has an improper Riemann integral on the disc $r<1$, equal to $2\pi \int_0^1 dr = 2\pi$
