Cantor Function

Cantor ternary Function

if ${\mathcal {C}}$ is the Cantor set on [0,1], then the Cantor function c : [0,1] → [0,1] can be defined as

$c(x)={\begin{cases}\sum _{n=1}^{\infty }{\frac {a_{n}}{2^{n}}},&x=\sum _{n=1}^{\infty }{\frac {2a_{n}}{3^{n}}}\in {\mathcal {C}}\ \mathrm {for} \ a_{n}\in \{0,1\};\\\sup _{y\leq x,\,y\in {\mathcal {C}}}c(y),&x\in [0,1]\setminus {\mathcal {C}}.\\\end{cases}}$

Properties of Cantor Functions

• Cantor Function is continuous everywhere, zero derivative almost everywhere.
• lack of absolute continuity.
• Monotonicity
• Its value goes from 0 to 1 as its argument reaches from 0 to 1.

Cantor Function Alternative

The Cantor Function can be constructed iteratively using homework construction.

Measurable function with pre-image of Lebesgue measurable set not Lebesgue measurable

Define $f(x)={\begin{cases}\sum _{n=1}^{\infty }{\frac {2b_{n}}{3^{n}}},&x=\sum _{n=1}^{\infty }{\frac {b_{n}}{2^{n}}}\ \mathrm {for} \ b_{n}\in \{0,1\}\\0\ \mathrm {otherwise} \\\end{cases}}$

Then it can be shown $f(x)$ is the pointwise limit of simple functions. However f takes values in the cantor set on the set of non terminating decimals. We can find a non measurable set $F$ such that $E:=f(F)$ is a null set and thus lebesgue measurable. Therefore $f^{-1}(E)$ fails to be Lebesgue measurable despite E being measurable.

This is analogous to the construction in HW5 2d) to 2 g) and is useful for motivating the definition of the Lebesgue measurable functions to be $({\mathcal {L}},{\mathcal {B}}_{\overline {\mathbb {R} }})$ measurable