## Cantor Function

Recall the definition of the Cantor set: Let

be the middle third of the interval= [ 1/3, 2/3]

*[0, 1]*. Let

be the middle thirds of the intervals remaining after deleting from [0, 1]. Let= [ 1/9, 2/9], = [ 7/9, 8/9 ]

be the middle thirds of the intervals remaining after deleting , , and from= [2/27, 3/27], = [7/27, 8/27],

= [19/27, 20/27], = [ 25/27, 26,27]

*[0, 1]*. Continue in this fashion so that at the

*n*-th stage we have the intervals

Then the complement of the union of all these intervals is the Cantor set without the endpoints. Now define the following function, ..., , ...,

Then, for example, we have thatiftis in the interval

if 1/3 t 2/3

Then *F(t)* is defined everywhere in *[0, 1]* except at the Cantor
set minus the end points *0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, ...*
If *t* is a number where *F* is not defined, then there exists an
increasing sequence *{ x _{n} }* of
these endpoints converging to

*t*, and a decreasing sequence

*{ x*of these endpoints converging to

_{n}' }*t*. Since

*F*is defined at those endpoints x

_{n}and x

_{n}', we define

F(t) = F(x_{n}) = F(x_{n}')

Now we have defined completely the Cantor function. It has the following properties:

*F*is defined everywhere in the interval*[0, 1]**F*is not constant*F*is increasing*F*is continuous in the interval*[0, 1]**F*is differentiable in the interval*[0, 1]**F'*is zero at every interior point of the intervals

*F'*is zero at points of total length 1 in the interval

*[0, 1]*, yet

*F*it is not constant.

### Proof

Some of these properties are obvious, and some require more thought.
In particular, why does the above limit of endpoints exist ? That
is a crucial point, because we used this limit to extend the function
to the whole interval *[0, 1]*. For details and hints about the
Cantor Function, please consult Kolmogorov and Fomin, p 334 ff.