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.
Best Answer
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 $\int_{\mathbb R}f\,d\lambda$.