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.
Hausdorff outer measure is defined for all sets, and then we use the definition of Caratheodory to restrict it to a subalgebra of "measurable" sets to get the Hausdorff measure. In $\mathbb R^n$, the $n$-dimensional Hausdorff outer measure is the same (up to a constant factor) as $n$-dimensional Lebesgue outer measure, so they have the same measurable sets as well.
Another counterexample to 2. In interval $[n,n+1]\;$ take a subset with Hausdorff dimension $1-1/n\;$. The union of these sets has Hausdorff dimension 1 but Lebesgue measure zero.
So to prove your fact about Lipschitz functions, you cannot do it by considering only sets with Hausdorff dimension ${}\lt 1$, you have to consider Lebesgue measure itself, and use its definition. Try it, it's not too hard!
Best Answer
If $\mu^{*}$ is an outer measure, then we define the measurable sets in terms of $\mu^{*}$; the measure is then defined to be the restriction of the outer measure to the measurable sets.
To be more explicit: if you have a $\sigma$-algebra and an outer measure $\mu^{*}$ on the algebra, then we say that a set $E$ is $\mu^{*}$-measurable if and only if for every $A$ in the $\sigma$-algebra, $$\mu^{*}(A) = \mu^{*}(A\cap E) + \mu^{*}(A\cap E').$$
As Halmos says in his book Measure Theory,
Once you have the definiiton of $\mu^{*}$-measurable, then let $S$ be the set of all measurable sets, and you define the measure $\mu$ on $S$ by $\mu(E) = \mu^{*}(E)$ for all $E\in S$.
In particular, this holds for the Lebesgue measure: if $E$ is Lebesgue measurable, then the Lebesgue measure of $E$ is equal to the outer measure of $E$, because the Lebesgue measure of $E$ is defined to be the outer measure of $E$. This holds for any Lebesgue measure, not just for measure $0$.
The point of the theorem you state earlier is that having outer measure zero implies that the set is measurable; that's the nontrivial part of the statement (not that the Lebesgue measure of the set will then be zero).