Prove that the piecewise function involving Cantor sets is measurable

Question

Note: $\Psi: \mathcal{Z} \to \mathcal{Z}$, where $\mathcal{Z}$ is the system of finite, disjoint, closed intervals in $[0,1]$

and $\Psi(\dot{\bigcup}_{j=1}^{J}[a_{j},b_{j}]):=\dot{\bigcup}_{j=1}^{J}[a_{j},\frac{2a_{j}+b_{j}}{3}]\cup[\frac{a_{j}+2b_{j}}{3},b_{j}]$ with $C_{n}:=\Psi^{n}([0,1])$

Let $(C_{n})_{n}$ be the sets such that $\bigcap_{n}C_{n}:=C$, where $C$ is the Cantor set. Note $C_{0}:=[0,1]$. Let $f: \mathbb R \to \mathbb R$, where

$f(x)$ = \begin{cases} \text{0,} &\quad\text{if x $\in \mathbb R - C_{0}$}\\ \text{k,} &\quad\text{if x $\in C_{k-1} - C_{k}, k \in \mathbb N$}\\ \text{$\infty$,} &\quad\text{if $x \in C$}\\ \end{cases}

prove that $f$ is measurable, and calculate the integral $\int_{\mathbb R}f(x)d\lambda(x)$.

My ideas: We have previously proven that a set

$E^{*}:=\{f:\mathbb R \to [0,\infty]: f-$measurable $\}=\{f:\mathbb R \to [0, \infty]: \exists (f_{n})_{n}$ monotone increasing measurable functions with $f_{n} \to f, n \to \infty \}$

so in attempting to find such a $f_{n}$, I defined

$f_{n}(x)$ = \begin{cases} \text{0,} &\quad\text{if x $\in \mathbb R - C_{0}$}\\ \text{$\frac{(n-1)k}{n}$,} &\quad\text{if x $\in C_{k-1} - C_{k}, k \in \mathbb N$}\\ \text{$n$,} &\quad\text{if $x \in C$}\\ \end{cases}

it is clear that $f_{n}$ is monotone increasing and $f_{n}$ is measurable as staircase functions $\forall n \in \mathbb N$ and $f_{n} \to f$. Have I shown $f$ is measurable?

Then, I am quite stumped on finding an appropriate integral to calculate $\int_{\mathbb R}f(x)d\lambda(x)$. My idea is:

$\int_{\mathbb R}f(x)d\lambda(x)= \int_{\mathbb R - C_{0}}f(x)d\mu(x)+\sum_{k \in \mathbb N}\int_{C_{k-1}-C_{k}}f(x)d\mu(x)+\int_{C}f(x)d\mu(x)$

and then I am unsure on how to progress. Any help is greatly appreciated.

Answer 1

The function $f$ measurable. Its range is $\mathbb N$, and for each $k\in\mathbb N$, $$ f^{-1}[\{k\}]=C_{k}\setminus C_{k-1}, $$ which is a measurable subset of $\mathbb R$, since it is the difference between two closed sets. In general, if $A\subset\mathbb R$ measurable, then $$ f^{-1}[A]=f^{-1}[A\cap \mathbb N]=\bigcup_{k\in A\cap\mathbb N}f^{-1}[\{k\}] $$ and hence $f^{-1}[A]$ is measurable, as a countable union of measurable sets.

Therefore $f$ is measurable.

Next $$ \int_{\mathbb R}f\,d\lambda=\sum_{k=1}^\infty k\,\lambda\big(\,f^{-1}[\{k\}]\big)= \sum_{k=1}^\infty k\,\lambda\big(C_k\setminus C_{k-1}\big)=\sum_{k=1}^\infty k\cdot \frac{2^{k-1}}{3^k}=\frac{1}{2}\sum_{k=1}^\infty k\cdot \left(\frac{2}{3}\right)^k=s $$ Observe now that $$ s-\frac{2}{3}s=\frac{1}{2}\sum_{k=1}^\infty k\cdot \left(\frac{2}{3}\right)^k-\frac{1}{2}\sum_{k=1}^\infty k\cdot \left(\frac{2}{3}\right)^{k+1} =\frac{1}{2}\sum_{k=1}^\infty k\cdot \left(\frac{2}{3}\right)^k-\frac{1}{2}\sum_{k=2}^\infty (k-1)\cdot \left(\frac{2}{3}\right)^{k}\\=\frac{1}{3}+\frac{1}{2}\sum_{k=2}^\infty \left(\frac{2}{3}\right)^{k}=\frac{1}{3}+\frac{1}{2}\cdot\frac{2^2}{3^2}\cdot \frac{1}{1-\frac{2}{3}}=1. $$ Hence $s=3$ and so is $\int_{\mathbb R}f\,d\lambda$.

Answer 2

There are several unclear places in the question. Trying to recover it, I assume that $\dot{\bigcup}$ denotes a union of mutually disjoint sets, $C_{n}:=\Psi^{n}([0,1])$ for each $n$, and $\lambda$ is the standard Lebesgue measure on $\Bbb R$. Since $f$ have to be a function from $\Bbb R$ to $\Bbb R$, it cannot attain a value $-\infty$, so I assume that $f(x)=-1$ for each $x\in C$. Since $\lambda(C)=0$, this should not affect measurability of $f$ and value of $\int_{\Bbb R} f d\lambda$.

Since range $\operatorname{ran} f$ of $f$ is countable and $f^{-1}(a)$ is Borel for each $a\in R$, $f^{-1}(A)$ is Borel (and, hence $\lambda$-measurable) for each (Borel) subset $A$ of $\Bbb R$.

By the definition,

$$\int_{\Bbb R} f d\lambda=\sum_{k\in\operatorname{ran} f\setminus\{0\}} k\lambda(f^{-1}(k))=\sum_{k=1}^\infty k\lambda(C_{k-1}\setminus C_{k})=\sum_{k=1}^\infty k(\lambda(C_{k-1})-\lambda(C_{k}))=$$ $$\sum_{k=1}^\infty k\left(\left(\frac 23\right)^{k-1}-\left(\frac 23\right)^{k}\right)=\sum_{k=1}^\infty\left (\frac 23\right)^{k-1}=\frac{1}{1-\frac 23}=3.$$

Answer 3

I think its pretty clear from first principles. A function is measurable if the pre images of measurable sets is measurable.

Note that we have that each $C_n$ is measurable because it is a countable union of intervals, and $C$ is measurable because it is the intersection of the $C_n$, a countable intersection of measurable sets. So, $f^{-1}(a)$ is measurable for $a \in \{\infty, 0, 1, 2, 3, \ldots\} = f(\mathbb{R})$. For any $A \subseteq \mathbb{R}$, $f^{-1}(A) = f^{-1}(A \cap f(\mathbb{R}))$, which is a (disjoint) union of a countable collection of measurable sets, and therefore measurable.

Hence, for measurable $A$, $f^{-1}(A)$ is measurable, so $f$ is measurable.

Ping me if you want some argument around the integral.