If $f:\mathbb{R^n} \rightarrow \mathbb{R}$ is an integrable function, then its graph has content equal zero. Show that the converse isn't true via counterexample.
First of all, by graph of $f$ I mean the set of the points $(x,f(x))$.
I tried showing the converse isn't true, but couldn't finish, even though I think my idea will work.
My attempt: The Dirichlet function $f:[0,1] \rightarrow \mathbb{R}$, $f(x) = 1, x \in \mathbb{Q}; f(x) = 0, x \in \mathbb{R-Q}$. The Dirichlet function is nowhere continuous, since the irrational numbers and the rational numbers are both dense in every interval $[0,1]$. The supremum of $f$ is $1$ and the infimum is $0$, therefore it is not Riemann integrable. The graph of $f$ will be countable points in the plane over the line $y=1$ and uncountable points of the plane over the line $y=0$, so if I take neighborhood balls of each point and unite them, for a given $\epsilon$, this cover would show me the set has null content. But my problem is how to construct such finite cover, since I have uncountable many points...
Can someone help me? And also the $\Rightarrow$ part. Thanks.
EDIT: I'm not anymore sure that $f$ isn't Riemann integrable. The Lebesgue criterion for integration says $f$ is in fact integrable, so my argument about supremum and infimum isn't right. Why isn't it right?
There are two ideas here that we should clarify:
Content (or "Jordan content" or "Jordan 'measure'") is defined only for bounded subsets $A$ of $\mathbb R^n$, as $\int_A 1$, where this is the Riemann integral. (Similarly, we can define the Lebesgue measure of a set $A \subseteq \mathbb R^n$ to be $\int_{\mathbb R^n} \mathbb{1}_A$, where this is the Lebesgue integral.)
Lebesgue measure $0$: $A \subseteq \mathbb R^n$ has Lebesgue measure $0$ iff for all $\varepsilon > 0$, there is a countable collection $B_i$ of boxes such that $A \subseteq \bigcup_i B_i$ and $\sum_i |B_i| < \varepsilon$, where $|B_i|$ denotes the usual measure of a box.
The theorem isn't quite clear as stated. The Riemann integral is only well-defined for bounded subsets of $\mathbb R^n$ (hence the restriction on the definition of Jordan content) due to various issues that come up with uniquely defining improper integrals.
However, you could fix the theorem to require the domain of $f$ to be a bounded subset of $\mathbb R^n$.
In that case, your example of the Dirichlet function $f$ works. To see that its graph has Jordan content $0$, note that it is a subsetq of $\big( [0, 1] \times (-a, a) \big) \cup \big( [0,1] \times (1-a, 1+a) \big)$ for all $a > 0$; taking $a = \varepsilon/8$, we see that the graph of $f$ lies inside two rectangles with total Jordan content $\varepsilon / 2$, so it has content $0$.