For $0\leq d\leq 1$, let $\eta_d$ be the $d$-dimensional Hausdorff measure on $\mathbb{R}$ (for some normalization); recall that it is translation-invariant.
Motivation for what follows: Up to recently, I was convinced that if $f$ and $g$ are in $L^1(\mathbb{R},\eta_d)$ (that is, Borel functions $\mathbb{R}\to\mathbb{R}$, integrable w.r.t. $\eta_d$, modulo equality $\eta_d$-a.e) one could define their convolution as usual by $(f*g)(x) = \int_{\mathbb{R}} f(y)\, g(x-y)\, \mathrm{d}\eta_d(x)$ and that this would satisfy $\|f*g\|_1 \leq \|f\|_1\,\|g\|_1$ as expected (where $\|f\|_1 := \int_{\mathbb{R}} f(t)\, \mathrm{d}\eta_d(t)$). Upon thinking about this question, I realized that this is not the case: the usual proof relies on translation-invariance and Fubini, but Fubini itself relies on $\sigma$-finiteness which is not true of $\eta_d$; for the Haar measure on a locally compact abelian group, which is not necessarily $\sigma$-finite, one can get away with "regularity" instead (even though things appear to be confused), but now I'm struggling to understand what works or does not work for Hausdorff measures. So let me ask about a concrete case:
Let $d = \log 2/\log 3$ and let $\eta = \eta_d$ be the $d$-dimensional Hausdorff measure on $\mathbb{R}$, normalized so that the usual Cantor middle-third set $C$ has $\eta(C)=1$. Let $f = \mathbf{1}_C$ be the indicator function of $C$ and $\nu = f \eta$ (that is, $E \mapsto \int_E f\, \mathrm{d}\eta$) be the "uniform" probability measure on $C$. We define the convoluted function $g := f\mathbin{*_\eta}f = f*\nu$ as $$x \mapsto \int_{\mathbb{R}} f(y)\, f(x-y)\, \mathrm{d}\eta(y) = \int_{C} f(x-y)\,\mathrm{d}\nu(y)$$ and we define the convoluted measure $\xi := \nu*\nu$ as $$E \mapsto (\nu\times\nu)(\{(y,z)\in\mathbb{R}^2 : y+z\in E\})$$
To put it more simply, $g(x)$ is the probability that, if $y$ is chosen uniformly in the Cantor set, $x-y$ also falls in the Cantor set; whereas $\xi(E)$ is the probability that two given reals $y,z$ both chosen uniformly and independently in the Cantor set have a sum in $E$.
Naïvely one would like to think that $\int_E g\, \mathrm{d}\eta = \xi(E)$ (for $E$ a Borel set) with the following "proof": $$\begin{aligned}\int_E g\, \mathrm{d}\eta &= \int_E\left(\int_{\mathbb{R}} f(x-y)\, \mathrm{d}\nu(y)\right) \mathrm{d}\eta(x)\\ &= \int_{E\times\mathbb{R}} f(x-y)\, \mathrm{d}\nu(y)\, \mathrm{d}\eta(x)\\ &= \int_{\{(y,z)\in\mathbb{R}^2 : y+z\in E\}} f(z)\, \mathrm{d}\nu(y)\, \mathrm{d}\eta(z)\\ &= \int_{\{(y,z)\in\mathbb{R}^2 : y+z\in E\}} \mathrm{d}\nu(y)\, \mathrm{d}\nu(z)\\ &= \xi(E)\end{aligned}$$ but as explained above, this proof does not work. However, both $g$ and $\xi$ are still well-defined objects.
Questions:
How can we describe $g = f\mathbin{*_\eta}f$ and $\xi = \nu * \nu$ more concretely?
Specifically, what is the support of $g$?
What is the exact relation between $g$ and $\xi$?
Specifically, is $\xi$ absolutely continuous w.r.t. $\eta$? (I believe it isn't.)
Bonus question: I am aware that the Fourier-Stieltjes transform $\hat\xi$ of $\xi$ is $u\mapsto e^{-2i\pi u} \prod_{j=1}^{\infty} \cos^2(2\pi u/3^j)$ (the square of that of $\nu$). But what is the Fourier transform of $g$?