Riemann integral of cantor set $1/5^{n}$ Form a sequence $(En)_{n\ge0}$ of subsets of $R$ as follows:
$E_{0} = [0, 1]$;
$E_{1}$ is obtained from $E_{0}$ by removing the open middle interval of length $1/5$, i.e. $E_{1} = [0,2/5]∪[3/5, 1]$;
$E_2$ is obtained by removing the open middle intervals of length $1/25$ from each of the $2$ intervals
forming $E_{1}$, i.e. $E_{2} = [0,9/50 ] ∪ [11/50 ,2/5] ∪ [3/5,39/50 ] ∪ [41/50 , 1]$;
In general, $E_{n}$ is obtained by removing the open middle intervals of length $1/5^{n}$
from each of the $ 2^{n−1}$
intervals forming $E_{n−1}$.
Let
$E =\bigcap_{n=1}^{\infty} E_n$.
I did prove that :
$\int \chi_{E} d\mu =\mu(E)=1/2$
How can i calculate the Riemann integral of the characteristic function of $E$?
First I tried to use the fact the $\chi_{E}$ is a step function , then it's Riemann integrable , since its Lebesgue integrable ,the Riemann integral and the Lebesgue integral are equal .
Can some one advice me with a more convincing way ?
Like using the fact that 
$$\lim_{n \to \infty } \chi_{E_{n}} = \chi_{E}$$
and
 $$\chi_{E}<\chi_{E_{n}} , \forall n\in N$$
 A: Notice that each upper Riemann sum can be realized as a Lebesgue integral of a suitable step function $\geq f$ and an analogous statement holds for lower Riemann sums. So it follows that
$$\underline{\int} f(x) \, dx \leq \int f \, d\mu \leq \overline{\int} f(x) \, dx$$
where $\int f \, d\mu$ is the Lebesgue integral of $f$. Now since
$$\int \chi_E \, d\mu \leq \overline{\int_{0}^{1}} \chi_E(x) \,dx \leq \int_{0}^{1} \chi_{E_n}(x) \, dx = \mu(E_n) \xrightarrow[n\to\infty]{} \mu(E), $$
by the squeezing lemma we obtain the value of the upper Riemann integral of $\chi_E$.
For the other direction, the crucial observation is that $[0, 1]\setminus E$ is dense in $[0, 1]$. Thus for each non-degenerate interval $I \subseteq [0,1]$ we have $\inf_{x\in I} \chi_E(x) = 0$. So each lower Riemann sum is exactly zero and therefore $\underline{\int_{0}^{1}} \chi_E(x) \, dx = 0$.

Another way of showing that $\chi_E$ is not Riemann integrable is to realize that the set of discontinuity of $\chi_E$ is exactly $E$, which has non-zero Lebesgue measure.
