# $f:B\subset \mathbb{R}^m\longrightarrow \mathbb{R}$ is integrable if and only if its graph has zero volume

Let $f:B\longrightarrow \mathbb{R}$ a bounded function in an m-rectangle $B\subset \mathbb{R}^m$
Prove that $f$ is integrable if and only if its graph has zero volume.
If the function $f$ is integrable then (Upper sum) - (Lower sum) $<\epsilon$ generates rectangles covering the graph of $f$, where the sum of their volumes is (Upper sum) - (Lower sum) <$\epsilon$ then $m-measure(B)=0$ then $vol(B)=0$, but I have problems in the necessary condition. – felipeuni Oct 24 '12 at 22:15
Try to take the Riemann sums, or even the lower and upper sums of the function in some partition of $B$. To say that a function is integrable is to say that these sums converge as the norm of the partition tends to 0. From there you can build a countable collection of sets whose volume sum less then $a$ for all $a > 0$. The converse is similar. If you can bound the graph of f in some countable collection of sets whose volumes sum less then a, then the function has to be integral again by that boundedness of the upper and lower sums. – Henrique Tyrrell Oct 24 '12 at 22:19