**Problem Statement** Show that the Dirichlet function \( f \) defined on \([0, 1]\) by \[ f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} \] is not Riemann integrable on \([0, 1]\). **Explanation** The Dirichlet function is an example of a function that is defined differently for rational and irrational numbers. It is 1 for any rational \( x \) and 0 for any irrational \( x \). This function serves as a classic example in analysis to illustrate properties related to integrability and discontinuity, especially in the context of Riemann integration. To show that it is not Riemann integrable, consider that for a function to be Riemann integrable, it must be continuous almost everywhere on the interval of integration. The Dirichlet function is discontinuous at every point in \([0, 1]\), as between any two rationals there exist irrationals and vice versa. Hence, it fails to meet the criteria for Riemann integrability.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**Problem Statement**

Show that the Dirichlet function \( f \) defined on \([0, 1]\) by 

\[ 
f(x) = 
\begin{cases} 
1 & \text{if } x \text{ is rational} \\
0 & \text{if } x \text{ is irrational}
\end{cases} 
\]

is not Riemann integrable on \([0, 1]\). 

**Explanation**

The Dirichlet function is an example of a function that is defined differently for rational and irrational numbers. It is 1 for any rational \( x \) and 0 for any irrational \( x \). This function serves as a classic example in analysis to illustrate properties related to integrability and discontinuity, especially in the context of Riemann integration. 

To show that it is not Riemann integrable, consider that for a function to be Riemann integrable, it must be continuous almost everywhere on the interval of integration. The Dirichlet function is discontinuous at every point in \([0, 1]\), as between any two rationals there exist irrationals and vice versa. Hence, it fails to meet the criteria for Riemann integrability.
Transcribed Image Text:**Problem Statement** Show that the Dirichlet function \( f \) defined on \([0, 1]\) by \[ f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} \] is not Riemann integrable on \([0, 1]\). **Explanation** The Dirichlet function is an example of a function that is defined differently for rational and irrational numbers. It is 1 for any rational \( x \) and 0 for any irrational \( x \). This function serves as a classic example in analysis to illustrate properties related to integrability and discontinuity, especially in the context of Riemann integration. To show that it is not Riemann integrable, consider that for a function to be Riemann integrable, it must be continuous almost everywhere on the interval of integration. The Dirichlet function is discontinuous at every point in \([0, 1]\), as between any two rationals there exist irrationals and vice versa. Hence, it fails to meet the criteria for Riemann integrability.
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