If $f:[0,1]\to \mathbb{R}$ st $f(x)=x$ if $x\in \mathbb{Q}$ and $f(x)=0$ if $x\in \mathbb{Q}^c$, show that f is not R- Integrable

Question

If $f:[0,1]\to \mathbb{R}$ such that $f(x)=x$ if $x\in \mathbb{Q}$ and $f(x)=0$ if $x\in \mathbb{Q}^c$, then using Riemann sum show that f is not Riemann Integrable over $[0,1]$.

Riemann sum: If $f:[a,b]\to \mathbb{R}$ be a function and $(P,T)$ is a tagged partition such that $P=\{x_0, x_1, ..., x_n\}$ where $a=x_0<x_1<...<x_n=b$ and $T=\{t_0, t_2, ...t_{n-1}\}$ for $t_i\in [x_i, x_{i+1}]$ then the Riemann sum is $S(P,T,f)=\sum_{i=0}^{n-1}f(t_i)(x_{i+1}-x_i)$.

Using it how to show this? How to take $P$ and $T$ suitably to show this? Please help.

Answer 1

Let $(P,T)$ be a tagged partition such that $$ P={x_0,x_1,...,x_n}$$ where $$ a=x_0<x_1<...<x_n=b$$ and $$T={t_0,t_2,...t_{n−1}}$$ for $$t_i∈[x_i,x_{i+1}]$$ then the Riemann sum is $$S(P,T,f)=∑_{i=0}^{n−1}f(t_i)(x_{i+1}−x_i)$$

You can find a rational number $q_i$ and an irrational number $r_i$ in $ (x_{i-1}, x_i)$.

If we pick $t_i=q_i$, the rational points, and evaluate the Riemann Sum, we get a positive number which approaches to $1/2$ as $n \to \infty $.

On the hand if you pick $t_i=r_i$, irrational numbers, the Riemann Sum is $0$.

The limits do not match. Thus the Riemann integral does not exist.

Answer 2

The following definition is the one you have used (I believe):

$f$ is Riemann integrable if there exists an $L$ such that for every $\epsilon>0$, there exists some $\delta>0$, for every partition $P=\{0=x_{0}<\cdots<x_{n}=1\}$ on $[0,1]$ with $\max_{0\leq i\leq n-1}(x_{i+1}-x_{i})<\delta$ and $T=\{t_{0},....,t_{n-1}\}$ with $t_{i}\in[x_{i},x_{i+1}]$ for $i=0,...,n-1$, then $|S(P,T,f)-L|<\epsilon$.

Back to the question, if it were Riemann integrable with integral $L$, using the symbols in the above definition, and using the density of irrational numbers, by choosing all $t_{i}$ to be irrational numbers, then $S(P,T,f)=0$: \begin{align*} S(P,T,f)=\sum_{i=0}^{n-1}f(t_{i})(x_{i+1}-x_{i})=\sum_{i=0}^{n-1}0\cdot(x_{i+1}-x_{i})=0, \end{align*} and hence $|L|<\epsilon$, this is true for all $\epsilon>0$, so $L=0$.

Now choose $\epsilon$ to be a specific number, say, $1/4$, and by the above definition again, by density of rational numbers, we choose the partition $P$ to be $\{0,1/n,2/n,...,1\}$ such that $1/n<\delta$ and $T=\{1/n,...,1\}$, then $|S(P,T,f)-L|=|S(P,T,f)-0|<1/4$ but \begin{align*} |S(P,T,f)|&=\sum_{i=0}^{n-1}f(t_{i})(x_{i+1}-x_{i})\\ &=\sum_{i=0}^{n-1}f\left(\dfrac{i+1}{n}\right)\left(\dfrac{i+1}{n}-\dfrac{i}{n}\right)\\ &=\sum_{i=0}^{n-1}\dfrac{1}{n}f\left(\dfrac{i+1}{n}\right)\\ &=\sum_{i=0}^{n-1}\dfrac{1}{n}\cdot\dfrac{i+1}{n}\\ &=\dfrac{1}{n^{2}}\sum_{i=0}^{n-1}(i+1)\\ &=\dfrac{1}{n^{2}}\cdot\dfrac{n(n+1)}{2}\\ &=\dfrac{1}{2}\left(1+\dfrac{2}{n}\right)\\ &>\dfrac{1}{2}, \end{align*} a contradiction.