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Give an example of a function $f:[0,1] \rightarrow \mathbb{R}$ such that...

(a) $f$ is bounded, but not Riemann integrable on $[0,1]$. $ f(x) := \begin{cases} 2x & \text{if $x$ is rational}\\ x & \text{if $x$ is irrational.} \end{cases} $ (b) $f$ is Riemann integrable on $[0,1]$ but not monotone.

$f(x) := 2$

(c) $f$ is Riemann integrable on $[0,1]$ but neither continuous nor monotone.

$f(x) := \begin{cases} 0 & \text{if $x$ is $0$}\\ 2 & \text{otherwise.} \end{cases} $


Is this correct? Thanks!

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    If $f$ is Riemann integrable and you change the value in finitely many points, the result will be again Riemann integrable. If you known this fact, you should be able to find some examples for (b) and (c).2012-06-22

2 Answers 2

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In fact there are at least an uncountable number of elements in each of the three classes above.

  1. For $\alpha>0$, the class of functions $f_\alpha(x) := \begin{cases} \alpha & \text{if $x$ is rational}\\ 0 & \text{if $x$ is irrational.} \end{cases}$ satisfy (a)
  2. For $\beta>0{}$, the class of functions $f_\beta(x)=\beta x(1-x){}$ satisfy (b)
  3. For $\gamma>0$, the class of functions $f_\gamma(x)=\gamma x(1-x){}$ for 0 and $f_\gamma(x)=-5{}$ satisfy (c)

The domain of definition of each function above is (of course) understood to be $[0,1]$

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    @leo thanks, I have fixed it now :)2012-04-19
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(a) is absolutely correct, (b): see comments - a constant function is usually considered monotone, but not strictly so, (c): correct. Of course, you can just your example from (c) for (b)...

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    To find a function that is not monotone anywhere but integrable, consider the Weierstrass function.2012-06-22