I have learned that the common ways to define distributions (measures) are by either density probability functions or their integral, the cdf.
I was wondering for fun if an alternate mechanism I am imagining could also be usable or if it has already been studied and is in use.
This would be similar to the density idea, but we assume a $uniform$ on a curve $f(x)$ and project it straight (normalizing as appropriate) on the real line to induce a suitable measure.
This would differ from the density mechanism because the density, intuitively, is a bit like projecting equispaced points with respect to the y axis, while here the equal segments are taken on the curve $f(x)$. (Furthermore, in general, the same projection, I guess, could be obtained with infinitely many different curves.)
I would like to ask if this idea makes any sense and if it is something that has already been explored.
First, you can't assign every curve $y=f(x)$ a uniform distribution, e.g. the real line itself.
Thus we need more conditions. Let's assume $f$ is nonnegative and integrable (with nonzero integral), then $\tilde f(x)=f(x)\mathbin/\int f(x)\, \mathrm dx$ is a density of a continuous random variable $X$ on $\mathbb R$.
We can assign a uniform distribution on this curve $y=f(x)$ by means of c.d.f $F(x)$ of $X$, since $F(X)\sim\mathrm{Uniform}(0,1)$. In this way, the distribution of the projection defined by $(x,f(x))\mapsto x$ is just the distribution of $X$, so we have defined a distribution over the curve $y=f(x)$ whose p.d.f is $\tilde f(x)$ while assigning a uniform distribution on it.
So, it can be seen that defining a distribution on a curve is nothing new because of the fact that the c.d.f follows a uniform distribution.
EDIT: I think I realize what the "uniform distribution" means in this problem, i.e., the probability of an arc is proportional to the arc length.
Now suppose we have a smooth curve $y=f(x)$ where $a\leqslant x\leqslant b$, and the arc length from $a$ to $t$ is $s(t)$. For a continuous random variable $X$ with range $[a,b]$ defined by this curve, the distribution of it is given by $F(t)=s(t)/s(b)$ where $F(t)$ is the c.d.f of $X$.
Conversely, if we have $F(t)$, then $f(t)$ can by solved from $$ s(t)=\int_a^t \sqrt{1+f'(x)^2}\, \mathrm dx $$ by taking derivative.
However, I don't think the distribution defined in this way is practical. If you know $F(t)$ then simply taking derivative you will get the density but you have to solve the integral to obtain the curve $y=f(x)$. Integration is more difficult than differntiation.