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!
 A: (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). 
