What is the motivation for defining the cantor function $c(x)$ to be The value of $x$ in base $3$, with digits replaced and interpreted in base $2$?
It seems very arbitrary. I can't understand why.
Edit: I'm mainly interested in why taking middle thirds out of the cantor set is equivalent to expelling all members with a 1 in their base 3 example, so long as the 1 is not followed by zeros.
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$\begingroup$That was not the original definition, but it is a quick-and-easy way of defining the same function. The idea was to form a function whose derivative is $0$ almost everywhere, but still is not constant. To do this, the well-known Cantor set, which has measure $0$, is exploited.
I believe the original definition was recursive, following these lines: $f_0(x) = x$ on [0,1]. Then at each stage $n$, $f_n$ is a "slope-step" function, defined piecewise on a finite partition of [0,1]. On each interval of the partition, $f_n$ is either constant, or else is an increasing line segment. At each partition point the value of $f_n$ from below and above are the same, so $f_n$ is continuous, and also $f_n(0) = 0$ and $f_n(1) = 1$. $f_{n+1}$ is defined as follows:
- On each interval where $f_n$ is constant, $f_{n+1} = f_n$.
On any interval $[a, b]$ of the partition where $f_n(b) > f_n(a)$, let $h = {f_n(a) + f_n(b) \over 2}$, $f_{n+1}$ is defined on $[a,b]$ by
- rising from $f_n(a)$ to $h$ on $[a, a+ (b-a)/3]$,
- staying constant at $h$ on $[a + (b-a)/3, a + 2(b-a)/3]$,
- and rising from $h$ to $f_n(b)$ on $[a+2(b-a)/3, b]$.
The Cantor-Lesbegue function is $\lim\limits_{n \to \infty} f_n$
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