Lebesgue Measure of the Graph of a Function Let $f:R^n \rightarrow R^m$ be any function. Will the graph of $f$ always have Lebesgue measure zero?
$(1)$ I could prove that this is true if $f$ is continuous.
$(2)$ I suspect it is true if $f$ is measurable, but I'm not sure. (My idea was to use Fubini's Theorem to integrate the indicator function of the graph, but I don't know if I'm using the Theorem properly).
If $(2)$ is incorrect, what would be a counterexample where the graph of $f$ has positive measure?
If $(2)$ is correct, can we prove the existence of a non-measurable function whose graph has positive outer measure?
 A: No function can have a graph with  positive measure or even positive inner measure, since every function graph has uncountably many disjoint vertical translations, which cover the plane. 
Meanwhile, using the axiom of choice, there is a function whose graph has positive outer measure. The construction is easiest to see if one assumes that the Continuum Hypothesis is true, so let me assume that.
To begin, note first that there are only continuum many open sets in
the plane, since every such set is determined by a
countable union of basic open balls with rational center
and rational radius. Next, it follows that the number of
$G_\delta$ sets is also continuum, since any such set is
determined by a countable sequence of open sets, and
$(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$.
Thus, we may enumerate the $G_\delta$ sets in the plane as
$A_\alpha$ for $\alpha\lt \aleph_1$ (using CH). Build a
function $f:\mathbb{R}\to\mathbb{R}$ by transfinite
induction. At any stage $\alpha\lt \aleph_1$, we have
the approximation $f_\alpha$ to $f$, and we assume that it
has been defined on only $\alpha$ many points. Given
$f_\alpha$, consider the $G_\delta$ set $A_\alpha$. If we
can extend $f_\alpha$ to a function $f_{\alpha+1}$ by
defining it on one more point $x$, so that
$(x,f_{\alpha+1}(x))$ is outside $A_\alpha$, then do so.
Otherwise, $A_\alpha$ contains the complement of
countably many vertical lines in the plane, and thus has
full measure.
After this construction, extend the resulting function if
necessary to a total function $f:\mathbb{R}\to\mathbb{R}$.
It now follows that the graph of $f$ is not contained in
any $G_\delta$ set with less than full measure. Thus, the
graph has full outer measure.
Now, finally, the same construction works without CH, once you realize that any $G_\delta$ set containing the complement of fewer than continuum many vertical lines has full measure.
