An assumption of Fubini's theorem is that: $$\displaystyle{\iint_{R} |f(x,y)| \thinspace \mathrm{d} A < \infty}$$ If this assumption is not met, and the integral of the absolute value of the function is infinite, does that imply that the function is discontinuous at some point in the region R?
If the assumption is met, is that the same thing as saying that f(x,y) is continuous over R?
Edit: This question is based on Fubini's theorem for rectangular regions.
Yes, provided that $R$ is closed (I assume this will be the case since this looks like a Calculus III question, and we tend to use closed regions for that course). You can use the fact that if $R$ is a closed and bounded region, then if $f(x,y)$ is continuous, $|f(x,y)|$ must obtain a maximum somewhere. Call it $M$. Then $$-M\cdot A(R) \leq \iint |f(x,y)|dA \leq M\cdot A(R)$$ where $A(R)$ is the area of the rectangular region.
If you want a function that does not work on an open region, consider $$f(x,y) = \frac{1}{x}$$ on the region $R = (0,1)\times (0,1)$. Then $f(x,y)$ is continuous on $R$, yet the integral is infinite.