Are there spaces with non-integer Hausdorff dimension that are differentiable? I realize that fractal curves usually represent non-differentiable functions in some sense, but would it to be possible to have a space that has non-integer Hausdorff dimension, and still have a reasonable notion of differentiation?  Essentially I'm curious about the ability for a space with a (pseudo)Riemannian metric, to have non-integer Hausdorff dimension.  If so, can you please provide an example and/or construction?
   If not, why?
 A: I'm not sure if this is exactly what you are looking for, but if it is, I believe that the answer is basically "no". For an integer $k$, a Radon measure $\mu$ is said to be $k$-rectifiable if, for $\mu$-a.e. $x$, $\delta_{k,r,x} \mu$ converges weak-* as $r \to 0$ to $c_x \mathcal H^k |_\pi$ for some $k$-plane $\pi$ and some number $c_x > 0$. Here the notation $\nu|_S$ means $\nu|_S(E) = \nu(S \cap E)$, and $\delta_{k,r,x} \nu(E) = r^{-k} \nu(r(E-x) + x)$. It is a deep theorem of David Preiss that rectifiability is implied by the mere convergence of $r^{-k} \mu(B(x,r))$ to some number $c_x$ as $r \to 0$.
Note that rectifiability is basically the property that a measure has a tangent space in an extremely weak sense almost everywhere.
So we might give a definition of rectifiability as follows: a Radon measure $\mu$ is $s$-rectifiable if, for $\mu$-a.e. $x$, $\mu(B(x,r))r^{-s} \to c_x$ for some $c_x > 0$. By Preiss's theorem this is equivalent to the usual definition when $s$ is an integer. However, there is a theorem of Marstrand (far, far more accessible than Preiss's theorem) that the condition above implies either $\mu = 0$ or that $s$ is an integer.
These notes by de Lellis give a proof of Preiss's theorem (it takes over a hundred pages), and chapter 3 is dedicated to a proof of Marstrand's theorem (about 10 pages). Also, an alternative, easier and more elegant proof of Marstrand's theorem by Kirchheim-Preiss is available, exploiting remarkable facts about so-called uniformly distributed measures.
A: For standard differentiation on the real line, the answer is (unsurprisingly) no. The following corollary appears in the paper Sets of Fractional Dimensions (V): on Dimensional Numbers of Some Continuous Curves by A.S. Besicovitch and H.D. Ursell:
Corollary: The dimensional number of the curve $y=f(x)$, where $f(x)$ has finite derivative at all points is $1$.
A: So, fractal curves are usually non-differentiable, but that's within the space in which they are embedded. There's no reason why you couldn't have a perfectly reasonable notion of differentiation inside, say, a Koch curve. Just use the usual definition of the derivative, but allow your limits to only consider points within the curve. I'm not sure how useful that is, though.
On the other hand, as I understand the idea of a (pseudo) Riemannian metric, it does require that the space be a manifold; a manifold cannot have noninteger dimension under most reasonable notions of fractal dimension.
