Can passing to a length metric increase Hausdorff measure? For concreteness, let's say that $(X,d)$ is a metric space homeomorphic to $\mathbb{R}^2$ whose Hausdorff 2-measure $\mathcal{H}_d^2$ is locally finite.
We can pass from $(X,d)$ to the length metric, denoted by $\overline{d}$, defined by the infimum of the length of rectifiable curves joining two points in $X$. The inequality $d \leq \bar{d}$ holds in general, so $\mathcal{H}_d^2(E) \leq \mathcal{H}_\bar{d}^2(E)$ for every Borel set $E$. My question is whether this inequality can be strict.
This question does need at least one refinement, however, since a negative answer can arise in an uninteresting way. Namely, you might have an uncountable collection of points in $X$ that are not accessible by a rectifiable curve. For such a point $x \in X$, we have $\bar{d}(x,y) = \infty$ for all $x \neq y$. Taking the definition of Hausdorff measure literally, in this situation the Hausdorff 2-measure of an uncountable set of inaccessible points is infinite. Thus the set of inaccessible points should be ignored. Let $F \subset X$ be the set of inaccessible points. The question is now: does $\mathcal{H}_d^2(E\setminus F) = \mathcal{H}_{\bar{d}}^2(E\setminus F)$ hold for every Borel set $E \subset X$?
 A: It's often the case that a question is easier to answer on my own after I've posted it to mathoverflow.
The answer seems to be "yes", that Hausdorff measure and even dimension can be increased. It might be some work to write down the details carefully (which I intend to do), but here is the idea: Pick a suitable Cantor set in $\mathbb{R}^3$. Build a surface in $\mathbb{R}^3$ consisting of branching narrow tubes whose fingers are the Cantor set. By controlling the length of each tube carefully, we can increase the distance between points in the Cantor set in a similarly controlled way, thus increasing the Hausdorff dimension of this set. On the other hand, we can make the tubes short enough that each point is connected by a rectifiable curve. More on this type of construction can be found e.g. in P. Mattila, Geometry of sets and measures in Euclidean spaces, p. 65., and I'm familiar with it via the paper K. Rajala, Uniformization of two-dimensional metric surfaces. 
An interesting thing for me is that if the Cantor set has dimension less than one, the surface constructed is 2-rectifiable. 2-rectifiability is perhaps the next strongest condition on a surface after the property being a Lipschitz image of $\mathbb{R}^2$, under which assumption the question here should have a negative answer.
