Characteristic Function

**manjohn12** · March 15th 2009, 01:48 PM

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

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

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

Is this correct?

**ThePerfectHacker** · March 15th 2009, 03:56 PM

Originally Posted by manjohn12

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

Prove $f$ 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 $[x_{j-1},x_j]$ (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 $\Sigma_{j=1}^n (1)(x_{j-1}-x_j) = b-a$ . If for any partition we choose only the rational points then the Riemann sum has value $\Sigma_{j=1}^n (0)(x_{j-1} - x_j) = 0$ . The problem is that $b-a$ and $0$ cannot be made arbitrary close to one another with a fine enough parition (since it is independent of $||P||$ ).

**manjohn12** · March 15th 2009, 03:58 PM

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 $[x_{j-1},x_j]$ (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 $\Sigma_{j=1}^n (1)(x_{j-1}-x_j) = b-a$ . If for any partition we choose only the rational points then the Riemann sum has value $\Sigma_{j=1}^n (0)(x_{j-1} - x_j) = 0$ . The problem is that $b-a$ and $0$ cannot be made arbitrary close to one another with a fine enough parition (since it is independent of $||P||$ ).

See here. Yes that is the standard way of proving it.

**manjohn12** · March 15th 2009, 07:11 PM

So would the contradiction be correct?

**manjohn12** · March 23rd 2009, 12:21 AM

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?

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