1. ## Characteristic Function

Define $\displaystyle f$ as follows: $\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \text{is rational} \\ 1 \ \ \ \ \text{if} \ x \ \text{is irrational} \end{cases}$

Prove $\displaystyle f$ is not Riemann integrable using the "addition" laws.

Proof. So we can break up $\displaystyle f$ as follows: $\displaystyle f(x) = f_{R}(x) + f_{I}(x)$. These are the rational and irrational parts of $\displaystyle f$ respectively. Suppose for contradiction that $\displaystyle f$ is Riemann integrable on some given interval $\displaystyle [a,b]$. Then $\displaystyle f_{R}(x)$ and $\displaystyle f_{I}(x)$ are Riemann integrable. But they are not. Contradiction. $\displaystyle \diamond$

Is this correct?

2. Originally Posted by manjohn12
Define $\displaystyle f$ as follows: $\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \text{is rational} \\ 1 \ \ \ \ \text{if} \ x \ \text{is irrational} \end{cases}$

Prove $\displaystyle f$ is not Riemann integrable using the "addition" laws.
I am not really sure what you mean by "addition laws". It happens to be really easy to prove this by using the definition of the Riemann integral. For any subinterval $\displaystyle [x_{j-1},x_j]$ (in the partition) there exists a rational and an irrational point. If for any partition we choose only the irrational points then the Riemann sum has value $\displaystyle \Sigma_{j=1}^n (1)(x_{j-1}-x_j) = b-a$. If for any partition we choose only the rational points then the Riemann sum has value $\displaystyle \Sigma_{j=1}^n (0)(x_{j-1} - x_j) = 0$. The problem is that $\displaystyle b-a$ and $\displaystyle 0$ cannot be made arbitrary close to one another with a fine enough parition (since it is independent of $\displaystyle ||P||$).

3. Originally Posted by ThePerfectHacker
I am not really sure what you mean by "addition laws". It happens to be really easy to prove this by using the definition of the Riemann integral. For any subinterval $\displaystyle [x_{j-1},x_j]$ (in the partition) there exists a rational and an irrational point. If for any partition we choose only the irrational points then the Riemann sum has value $\displaystyle \Sigma_{j=1}^n (1)(x_{j-1}-x_j) = b-a$. If for any partition we choose only the rational points then the Riemann sum has value $\displaystyle \Sigma_{j=1}^n (0)(x_{j-1} - x_j) = 0$. The problem is that $\displaystyle b-a$ and $\displaystyle 0$ cannot be made arbitrary close to one another with a fine enough parition (since it is independent of $\displaystyle ||P||$).
See here. Yes that is the standard way of proving it.

4. So would the contradiction be correct?

5. Instead of the traditional way of doing it, would breaking it up into a "rational part" and an "irrational part" be the best way to proceed?