What is the Onsager-Machlup function for the Cantor measure on $\mathbb R$? Given a Borel measure $\mu$ on metric space $(X,d)$, if the limit
$$\lim_{\epsilon\to 0}\frac{\mu(B(z_1,\epsilon))}{\mu(B(z_2,\epsilon))}=\exp\left(F(z_2)-F(z_1)\right)$$
exists (including when $F(z)=\infty$), then $F$ is called the Onsager-Machlup function associated to $\mu$. The OM function is a generalization of a density that exists even when a density does not. For example, on non locally compact spaces.
It is well known and easy to see that if $\mu$ is a measure on $\mathbb R^n$ and has a density wrt either Lebesgue or counting measure, then the OM function is just the negative log of the density. However, what if $n=1$ and the measure $\mu$ is the Cantor measure?
 A: Partial answer (with a gap).
I expect the limit
$$
\lim_{\epsilon\to 0}\frac{\mu(B(z_1,\epsilon))}{\mu(B(z_2,\epsilon))}
\tag1$$
fails to exist for $\mu\otimes\mu$-almost all pairs $(z_1,z_2) \in C \times C$.
Reference I will follow here:
Falconer, Kenneth, Techniques in fractal geometry, Chichester: John Wiley & Sons. xvii, 256 p. (1997). ZBL0869.28003.
Let $C$ be the middle-thirds Cantor set.  Write $s = (\log 2)/(\log 3)$ for the fractal dimension of $C$.  Let $\mu$ be the Cantor measure on $C$.  Then $\mu$ is the restriction of Hausdorff measure $\mathcal{H}^s$ to $C$.

 For most self-similar fractals, these two measures are merely within a constant factor of each other (Lemma 6.4(c)); but in this case Hausdorff's original paper proves equality.  This is one of the few fractals where we can compute the exact value of the Hausdorff measure.

The upper and lower $s$-dimensional densities for $C$ are defined by
$$
\overline{D}^s(z) := \limsup_{r\to 0^+}\frac{\mu(B(z,r))}{(2r)^s} ,
\\
\underline{D}^s(z) := \liminf_{r\to 0^+}\frac{\mu(B(z,r))}{(2r)^s} .
$$
From Proposition 6.5: For a self-similar set like $C$ we have two constants $\overline{d}, \underline{d}$ with $0 < \underline{d} \le \overline{d} < +\infty$ such that
$$
\overline{D}^s(z) = \overline{d},\quad \underline{D}^s(z) = \underline{d}
$$
for $\mu$-almost all $z \in C$.
Corollary 9.8: For cases like $C$ where $s$ is not an integer, we have strict inequality
$\underline{d} < \overline{d}$.

 If I recall correctly, $\overline{d} = 2^{-s}$ and $\underline{d} = 4^{-s}$; but the exact values are not needed here.

Now for $z_1,z_2 \in C$ and $r>0$
$$
\frac{\mu(B(z_1,r))}{\mu(B(z_2,r))} =
\frac{\mu(B(z_1,r))/(2r)^s}{\mu(B(z_2,r))/(2r)^s}
\tag2$$
Now as $r \to 0^+$, the numerator has limit points ranging from $\underline{d}$ to $\overline{d}$ and so does the denominator.  For typical $z_1,z_2$, I expect (here is where there is a gap...) the numerator and denominator limits are not synchronized, so that $(2)$ has limsup equal to $\overline{d}/\underline{d}$ and liminf equal to $\underline{d}/\overline{d}$.  This means that limit $(1)$ does not exist.
I say this for "typical" $z_1, z_2$.  For some pairs, say where $z_1$ and $z_2$ differ in only finitely many places in base $3$, I expect the numerator and denominator would be synchronized, and then we get limit $1$.  The collection of synchronized pairs $(z_1,z_2)$ has $\mu\otimes\mu$ measure $0$.

What about the gap mentioned in the proof?  Perhaps we can do this from the ergodic properties of $\mu$ seen in Lemma 6.4.  Perhaps also use of the Renewal Theorem as in Chapter 7.
