Are any two Cantor sets ; "Fat" and "Standard" Diffeomorphic to each Other? All:
I know any two Cantor sets; "fat" , and "Standard"(middle-third) are homeomorphic to each other. Still, are they diffeomorphic to each other? I think yes, since they are both $0$-dimensional manifolds (###), and any two $0$-dimensional manifolds are diffeomorphic to each other. Still, I saw an argument somewhere where the claim is that the two are not diffeomorphic. 
The argument is along the lines that, for $C$ the characteristic function of the standard Cantor set integrates to $0$ , since $C$ has (Lebesgue) measure zero, but , if $g$ where a diffeomorphism into a fat Cantor set $C'$, then: $ f(g(x))$ is the indicator function for $C'$, so its integral is positive.
And (my apologies, I don't remember the Tex for integral and I don't have enough points to look at someone else's edit ; if someone could please let me know )
By the chain rule, the change-of-variable $\int_0^1 f(g(x))g'(x)dx$ should equal $\int_a^b f(x)dx$ but $g'(x)>0$ and $f(g(x))>0$ . So the change-of-variable is
contradicted by the assumption of the existence of the diffeomorphism $g$ between $C$ and $C'$.
Is this right? 
(###)EDIT: I realized after posting --simultaneously with "Lost in Math"* , that the Cantor sets {C} are not 0-dimensional manifolds (for one thing, C has no isolated points). The problem then becomes, as someone posted in the comments, one of deciding if there is a differentiable map $f:[0,1]\rightarrow [0,1]$ taking C to C' with a differentiable inverse.


*

*I mean, who isn't, right?

 A: I'm pretty sure Hausdorff dimension is a diffeomorphism invariant.  Hausdorff measure of course is not.  The basic idea is that if you have a ball of radius $r$ and a diffeomorphism the image of the ball of radius $r$ contains a ball of radius $Mr$ where $M$ is the maximum of the norm of $(f^{-1})'$.  Also, the image of the ball is contained in a ball of radius $Nr$, where $N$ is the maximum of the norm of $f'$.   Basically you just have to worry about how diffeomorphisms distort the radius of balls (up to inclusion).  Diffeomorphisms do so in a tame fashion, provided they're at least $C^1$.  
So although all Cantor sets are homeomorphic, up to diffeomorphism you have at least the Hausdorff dimension that separates them -- I think likely there are many more invariants but I haven't given it much thought. 
More generally speaking, given a homeomorphism between two metric spaces $f : X \to Y$ which is bi-lipschitz, 
$$d(f(x),f(y)) \leq Md(x,y)$$
and 
$$d(f^{-1}(x),f^{-1}(y)) \leq Nd(x,y)$$
where $N,M > 0$, the Hausdorff dimension of $A \subset X$ is equal to the Hausdorff dimension of $f(A)$. 
A diffeomorphism has the property that it's bi-lipschitz with $M = max ||f'||$ and $N = max ||(f^{-1})'||$. 
A: I think I found an answer to my question, coinciding with the idea in Ryan's last paragraph: absolute continuity takes sets of measure zero to sets of measure zero. A diffeomorphism defined on [0,1] is Lipshitz continuous, since it has a bounded first derivative (by continuity of f' and compactness of [0,1]), so that it is absolutely continuous, so that it would take sets of measure zero to sets of measure zero.
