Integrable function has a graph that's a null set?

I'm asked in an exercise to prove every integrable real function $f : R \rightarrow R$ on an interval $[a,b]$ has a graph that's a null set. I would appreciate some assistance with this!

More information: Integrable here means Darboux/Riemann integrable. This is not a measure theory question and I am unfamiliar with measure theory results and theorems.

We have already proven that if f is continuous, then its graph will be a null set. I was trying to generalize the ideas from that proof to this proof but I haven't succeeded yet... I really have no idea how to approach this.

Thanks a lot!

-
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Arturo Magidin May 23 '12 at 5:39
What happens if you try to relate the graph to the sums which define the integral? – Mark Bennet May 23 '12 at 5:53
Mark: If the function is positive or negative for the most part, then that would immediately prove the question. But what if the function is more chaotic? – daro May 23 '12 at 5:54
@daro: Do you know the characterization of integrable functions in terms of the upper and lower Riemann sums? Consider what the difference between the upper and lower Riemann sums represents in terms of areas and the graph. – Arturo Magidin May 23 '12 at 6:17
@ArturoMagidin: If the function is say, positive, the difference would, in terms of area, represent a set of rectangles containing the graph of f with an area as small as we like. The thing is, f is not necessarily continuous so we can't divide it into positive and negative parts..... hmm. – daro May 23 '12 at 6:20

(Note: I'm assuming below that $f$ is bounded; Riemann integrability is usually only considered for bounded functions, so this is not an unreasonable assumption.)
Given a function $f$, an interval $[a,b]$, and a partition $P$ of $[a,b]$, let us denote by $\overline{S}(f,P)$ the upper Riemann sum of $f$ on $[a,b]$ relative to the partition $P$; that is, we consider $$\overline{S}(f,P) = \sum_{i=1}^{n}s_{i}\Delta_i$$ where $a=x_0\lt x_1\lt\cdots\lt x_n=b$ is the partition, $s_i$ is supremum of the values of $f(x)$ on $[x_{i-1},x_i]$, and $\Delta_i=x_i-x_{i-1}$. Similarly, let $\underline{S}(f,P)$ be the lower Riemann sum of $f$ on $[a,b]$ relative to the partition $P$, that is $$\underline{S}(f,P) = \sum_{i=1}^n m_i\Delta_i$$ where $m_i$ is the infimum of the values of $f$ on $[x_{i-1},x_i]$.
$f$ is integrable on $[a,b]$ if and only if it is bounded, and for any sequence of partitions $P_n$ of $[a,b]$ such that the mesh size $\lVert P_n\rVert\to 0$ as $n\to\infty$, we have $$\lim_{n\to\infty}\underline{S}(f,P_n) = \lim_{n\to\infty}\overline{S}(f,P_n),$$ or equivalently, if $$\lim_{n\to\infty}\Bigl( \overline{S}(f,P_n) - \underline{S}(f,P_n)\Bigr) = 0.$$
Now think about what $\overline{S}(f,P_n) - \underline{S}(f,P_n)$ represents. At any given subinterval of the partition $[x_i,x_{i+1}]$, we are taking $\bigl(s_i - m_i\bigr)\Delta_i$. Now, $\Delta_i$ is the length of the interval; $s_i$ is the supremum of the values the function takes, and $m_i$ is the infimum value the function takes. That means that the graph of the function on this interval is contained in the rectangle $[x_i,x_{i+1}]\times[m_i,s_i]$, which has area exactly $\bigl(s_i-m_i\bigr)\Delta_i$. That is, the difference between $\overline{S}(f,P_n)$ and $\underline{S}(f,P_n)$ can be interpreted as the sum of the areas of a collection of rectangles that contains the graph of $y=f(x)$.
Can you take it from here? (Note that the above does not depend in any way on whether $f$ is positive, negative, chaotic, continuous, or not; just on the fact that it is bounded and integrable).