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.
So, essentially what's going on (as far as I was taught when I learned the subject) is this: we have the real line $\mathbb{R}$ and we have sets which we "know" what the measure of them ought to be: intervals. For convenience of the general theory, authors usually use half open intervals $(a,b]$ since they form an algebra $\mathcal{A}\subset \mathcal{P}(\mathbb{R})$. These also generate $\mathcal{B}_{\mathbb{R}}$. Now, we use the theory of premeasures: we define a premeasure $\mu_0$ on this algebra by defining $\mu_0(\emptyset)=0$ and $$\mu_0(\bigcup_1^{\infty} (a_i,b_i])=\sum_1^{\infty}\mu_0((a_i,b_i])=\sum_1^{\infty}(b_i-a_i)$$(actually, in general we can take any right continuous function here, defining $\mu_0((a,b])=f(b)-f(a)$ for $f$ right continuous, but it is most convenient to just use the identity).
Then, you use the general theory to show that you can derive an outer measure $m^{\ast}$ from this premeasure, and use Caratheodory's theorem to construct the full measure $m$ and the $\sigma$-algebra $\mathcal{L}$. We know that $\mathcal{B}_{\mathbb{R}}\subset \mathcal{L}$ since $\mathcal{A}$, the algebra of half intervals, generates $\mathcal{B}_{\mathbb{R}}$. However, it is not obvious that $\mathcal{B}_{\mathbb{R}}\not=\mathcal{L}$, but in fact this is true.
The Lebesgue measure space is in fact $(\mathbb{R},\mathcal{L},m)$, although you can often restrict to just the Borel sets.
Best Answer
Here's an alternate idea:
Use Tonelli's theorem to show that for any measurable function $f \geq 0$: if we define the sets $$ \Gamma = \{(x,t): 0 \leq t < f(x)\}\\ \Gamma' = \{(x,t): 0 \leq t \leq f(x)\} $$ Then $(\mu\times \lambda)(\Gamma) = (\mu\times \lambda)(\Gamma') = \int_{X}f(x)\,dx$
Extend this to include general measurable $f$. Conclude that your set, $\Gamma' - \Gamma$, has measure zero.