# Working on a generalized Cantor set, with Lebesgue measure, and a certain inequality.

Let's consider the interval $[0,1]$ in the same way that we constructed the Cantor set, we can use the same idea, but instead of removing in the step $n$ middle open intervals of length $\frac{1}{3^n}$ we remove of length $\frac{1}{5^n}$. Call each step of this construction $C_n$ i.e

$C_0 = [0,1]$

$C_1= C_0-\left(\frac{1}{2}-\frac{1}{10},\frac{1}{2}+\frac{1}{10}\right)=\left[0,\frac{2}{5}\right]\cup \left[\frac{3}{5},1\right]$

And then to construct $C_2$ consider the middle point of $\left[0,\frac{2}{5}\right]$ which is $\frac{2}{10}$ , and remove from that interval the open interval $\left(\frac{3}{20}-\frac{1}{2\cdot 5^2},\frac{3}{20}+\frac{1}{2\cdot 5^2}\right)$ , the same with the interval $\left[\frac{7}{10},1\right]$ .

(It's the same as the usual Cantor construction) So we define $C = \bigcap C_n$. Let's define the function $f_n$ as the characteristic function of the set $C_n$ , and $f$ as the characteristic function of the set $C$.

Prove that $$\mathop {\sup }\limits_{m \geqslant n} \int\limits_0^1 {\left| {f_n \left( x \right) - f_m \left( x \right)} \right|\,\mathrm dm} \leqslant \frac{1} {3}\left( {\frac{2} {5}} \right)^n$$ and then prove that under any modification of the function $f$ on a set of measure zero, is not Riemann integrable.

I'm very scared of this problem, I don't know how to prove the inequality please help me, I'm a little dizzy . Thanks for the EDIT

Let $\{I_{k,j}:j=1,\ldots,2^k\}$ be the set of open intervals we are throwing away on $k$-th step of our construction. Note that $m(I_{k,j})=5^{-k-1}$ for all $k\in\mathbb{N}$ and $j=1,\ldots,2^k$.
Let $$C_0:=[0,1]\qquad\qquad C_{k+1}:=C_k\setminus\bigcup\limits_{j=1}^{2^k}I_{k,j}$$ Now we proceed to the proof of inequality. Define $f_n=\chi_{C_n}$, then for $m\geq n$ we have $$\int_0^1|f_m(x)-f_n(x)|dm= \int_0^1|\chi_{C_m}(x)-\chi_{C_n}(x)|dm= \int_0^1\chi_{C_n\setminus C_m}(x)dm= m(C_n\setminus C_m)= m\left(\bigcup\limits_{k=n}^{m-1}\bigcup\limits_{j=1}^{2^k} I_{k,j}\right)= \sum\limits_{k=n}^{m-1}\sum\limits_{j=1}^{2^k}m(I_{k,j})= \sum\limits_{k=n}^{m-1}\sum\limits_{j=1}^{2^k}5^{-k-1}= \sum\limits_{k=n}^{m-1}2^k5^{-k-1}$$ Hence $$\sup\limits_{m\geq n}\int_0^1|f_m(x)-f_n(x)|dm= \sup\limits_{m\geq n}\sum\limits_{k=n}^{m-1}2^k5^{-k-1}= \sum\limits_{k=n}^\infty 2^k5^{-k-1}=\frac{2^n 5^{-n-1}}{1-2\cdot 5^{-1}}=\frac{1}{3}\left(\frac{2}{5}\right)^n$$
We proceed to the second part. Define $$C:=\bigcap\limits_{k=1}^\infty C_k$$ Its measure equals $$m(C)= 1-m(C_0\setminus C)= 1-m\left(\bigcup\limits_{k=0}^\infty\bigcup\limits_{j=1}^{2^k}I_{k,j}\right)= 1-\sum\limits_{k=0}^\infty \sum\limits_{j=0}^{2^k} m(I_{k,j})=\\ 1-\sum\limits_{k=0}^\infty \sum\limits_{j=0}^{2^k} 5^{-k-1}= 1-\sum\limits_{k=0}^\infty 2^k 5^{-k-1}= 1-\frac{5^{-1}}{1-2/5}=\frac{2}{3}$$ Now note that $x\in C$ iff its representation in a fivefold number system have no digit $2$ at any place. Hence for a given $x\in C$ you can change some digit to $2$ to leave the set $C$. Moreover you can change digit with very big place number to get a number outside $C$ but very close to $x$. Thus for each $x\in C$ and each $\varepsilon>0$ there exist $\hat{x}\notin C$. This is equivalent to discontinuity of $f$ in point $x\in C$. So $f$ is discontinuous at any point of $C$.
Since $m(C)>0$ by Lebesgue critirion $f\notin\mathcal{R}([0,1])$, because set of discontinuities of $f$ have positive measure. Let $\hat{f}$ be a function equivalent to $f$. Assume $\hat{f}$ is Riemann interable, then it is bounded. Hence the function $\delta=\hat{f}-f$ is bounded. Since $\hat{f}$ is equivalent to $f$, then $\delta$ is non zero only on the set of zero measure. By Lebesgue criterion $\delta\in\mathcal{R}([0,1])$. Then $f=\hat{f}-\delta\in$$\mathcal{R}([0,1])$ as difference of two Riemann integrable functions. Contradiction, so $\hat{f}\notin\mathcal{R}([0,1])$