Can an Osgood curve be almost everywhere differentiable? It is known that you can “reparametrize” Osgood curves to make them almost-everywhere smooth curves (simply compose one after the Cantor function). However doing this breaks injectivity, stopping them from being Osgood curves anymore.
Is it possible to use some other kind of trick or construction, to build an Osgood curve which is almost-everywhere differentiable (and remains injective)?
I know the answer is false if you replace almost-everywhere differentiable with differentiable (or if you only allow countably many non-differentiable points).
Bonus: If the answer turns out to be no, is there any comment about how differentiable an Osgood curve can be? Specifically ones where every sub-arc is also an Osgood curve?
 A: Take a set $K\subset\mathbb{R}^2$ that is homeomorphic to the Cantor set and has positive $2$-dimensional Lebesgue measure. Take a homeomorphism of the ternary Canor set onto $K$ and extend it as a piecewise $C^\infty$ curves in the complement of the ternary Cantor set. If you make it carefully, you will obtain an injective curve whose image has positive 2-dimensional Lebesgue measure and which is $C^\infty$ smooth in the complement of the set of measure zero - in the complement of the ternary Cantor set.
In higher dimensions $n\geq 2$, it is possible to construct an embedding $f:\mathbb{S}^n\to\mathbb{R}^{n+1}$ that it is $C^\infty$ smooth outside a compact set $C\subset\mathbb{S}^n$ of Hausdorff dimension zero and such that $f(\mathbb{S}^n)$ has positive $(n+1)$-dimensional Lebesgue measure.
In fact for any set $K\subset\mathbb{R}^{n+1}$ that is homeomorphic to the ternary Cantor set, one can construct $f$ as above so that $f(C)=K$. Since $K$ can have positive measure, the above result follows. That construction also includes various generalizations of the horned sphere when taking wild Cantor sets as $K$.
A construction of such an embedding with additional properties can be found in:
P. Hajłasz, X. Zhou, Sobolev homeomorphism on a sphere containing an arbitrary Cantor set in the image. Geom. Dedicata 184 (2016), 159-173.
arXiv
A: I have given this some thought and realised that the answer to my question is yes. In case anyone comes across this question and is interested, I will post my answer here. To begin with I will make the following claim (which is essentially the crux of the result, but I will postpone a proof in order to show how the claim solves my problem first). Here is the claim:

Given any non-decreasing function $\nu : (0,\infty) \rightarrow (0,\infty)$, there exists a function $F_\nu : [0,1] \rightarrow [0,1]$ satisfying the following properties:

*

*$F_\nu$ is continuous, bijective and strictly increasing

*For almost-all $x \in [0,1]$ we have that $\frac{F_\nu(x+h) - F_\nu(x)}{\nu(|h|)} \rightarrow 0$ as $h \rightarrow 0$

So assuming this claim is true for now, here is a proof to the main question. To start let $J : [0,1] \rightarrow \mathbb{R}^2$ be any Osgood curve. By the Heine–Cantor theorem, $J$ is uniformly continuous and so it has some modulus of continuity $\omega : (0,\infty) \rightarrow (0,\infty)$. Without loss of generality we may assume that $\omega$ is non-decreasing and bijective.
Define $\nu : (0,\infty) \rightarrow (0,\infty)$ via $\nu(x) := \omega^{-1}(x^2)$. Now use the above claim to construct $F_\nu$ and let $A$ be the set of points satifying the second property. Then for all $x \in A$ we have that (for small enough $h$):
$$||J(F_\nu(x+h)) - J(F_\nu(x))|| \leq \omega(|F_\nu(x+h) - F_\nu(x)|) \leq \omega(\nu(|h|)) = h^2$$
Therefore $J \circ F_\nu$ is an almost-everywhere differentiable reparameterization of $J$ which retains the properties of being an Osgood curve (namely being continuous and injective).

This leaves proving the initial claim. To begin, define the following set:
$$U_q := \bigcup_{k=0}^{q-1} \bigcup_{j=0}^{3^k - 1} (\frac{3j+1}{3^{k+1}} + \frac{1}{2} \frac{1}{3^q},\frac{3j+2}{3^{k+1}} - \frac{1}{2} \frac{1}{3^q})$$
This is a collection of points in $[0,1]$ which happen to all be a distance of at least $\frac{1}{2} \frac{1}{3^q}$ away from the Cantor set. The measure of $U_q$ is given by:
$$\sum_{k=0}^{q-1}2^k(\frac{1}{3^{k+1}} - \frac{1}{3^q}) = 1 + (\frac{1}{3})^q  - 2 (\frac{2}{3})^q$$
Define $C_{\frac{p}{3^n}} := \frac{p}{3^n} + \frac{1}{3^n} C$ (where $C$ is the Cantor set) and $U_{q, \frac{p}{3^n}} := \frac{p}{3^n} + \frac{1}{3^n} U_q$ to be rescaled and translated copies of the Cantor set and $U_q$ respectively.
If we take $U_q^* := \bigcap_{n=0}^{\infty} \bigcup_{p=0}^{3^n-1} U_{q(n+1), \frac{p}{3^n}}$, we find that it must have measure at least:
$$1 - \sum_{n=0}^{\infty} [2 (\frac{2}{3})^{q(n+1)} - (\frac{1}{3})^{q(n+1)}]$$
which is a quantity that tends to $1$ as $q \rightarrow \infty$.

Now let $\phi : [0,1] \rightarrow [0,1]$ be the Cantor function, and let $\phi_\frac{p}{3^n}$ be a transformation of it so that the entire staircase sits in the interval $[\frac{p}{3^n},\frac{p+1}{3^n}]$ (and define the output to be $0$ and $1$ on the left and right of this interval respectively).
What we have done so far allows us to make the following observation:

For all $x \in U_q^*$, we have that $\phi_\frac{p}{3^n}$ is constant along the interval $(x-\delta,x+\delta)$, provided that $\delta < \frac{1}{2} \frac{1}{3^{q(n+1)}} \frac{1}{3^n}$.

Define $A_n := \frac{1}{n} \nu(\frac{1}{2} \frac{1}{3^{n(n+1)}} \frac{1}{3^n})$, and $f_\nu : [0,1] \rightarrow \mathbb{R}$ via $f_\nu := \sum_{n=0}^{\infty} \sum_{p=0}^{3^n - 1} \frac{1}{6^n} A_{n+1} \phi_\frac{p}{3^n}$. The continuity of this function follows from the Weierstrass M-test, and strictly increasing follows from every interval containing at least one entire Cantor staircase (with non-zero height).
Now we have the following:

If $x \in U_q^*$:
then as long as $m \geq q$, we find that for all $h$ with $\frac{1}{2} \frac{1}{3^{(m+1)(m+2)}} \frac{1}{3^{m+1}} \leq |h| \leq \frac{1}{2} \frac{1}{3^{m(m+1)}} \frac{1}{3^m}$ we have:
$|f_\nu(x+h) - f_\nu(x)| \leq \sum_{n=m}^{\infty} \sum_{p=0}^{3^n - 1} \frac{1}{6^n} A_{n+1} = \sum_{n=m}^{\infty} \frac{1}{2^n} A_{n+1} \leq A_{m+1} \leq \frac{1}{m+1} \nu(|h|)$
Hence $\frac{f_\nu(x+h) - f_\nu(x)}{\nu(|h|)} \rightarrow 0$ as $h \rightarrow 0$.

Hence $\frac{f_\nu(x+h) - f_\nu(x)}{\nu(|h|)} \rightarrow 0$ as $h \rightarrow 0$, for all $x \in \bigcup_{q=1}^\infty U_q^*$. But this union has a measure zero compliment, which essentially gives us the claim.
The only thing missing is that the claim states that the image of the function should be $[0,1]$. To fix this we finally define $F_\nu := \frac{1}{f_\nu(1)} f_\nu$.
