Characteristic Function

• Mar 15th 2009, 01:48 PM
manjohn12
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?
• Mar 15th 2009, 03:56 PM
ThePerfectHacker
Quote:

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 $\displaystyle ||P||$).
• Mar 15th 2009, 03:58 PM
manjohn12
Quote:

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.
• Mar 15th 2009, 07:11 PM
manjohn12
So would the contradiction be correct?
• Mar 23rd 2009, 12:21 AM
manjohn12
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?