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Let $f: [0,1] \rightarrow \mathbb{ R}$ be determined by:

$f(x) = 0 $when $x$ is irrational

$\frac{(-1)^p}{q}$ when $x$ is rational (and in reduced form $\frac{p}{q}$)

Show that $f$ is Riemann integrable on $ [0,1]$and that the integral equals $0$.

Any help would be much appreciated! Thank you!

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    Question as posted is not clear. You are missing information related to $p$ and $q$. – 2017-02-01
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    This definition is not correct. What is $p,q$? – 2017-02-01
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    $p, q$ is probably the reduced representation of any rational number $\frac{p}{q}$. – 2017-02-01
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    Of course, it should be so, but OP should take care about such things and write preciesely. – 2017-02-01

2 Answers 2

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Let $\epsilon >0$. Pick some $N$ such that $\frac{1}{N} < \frac{\epsilon}{2}$.

Show first that there are only finitely many fractions $\frac{p_1}{q_1},..., \frac{p_M}{q_M}$ such that $q_M

Now, pick any partition $0=x_0

Then $$ \left| \sum_{i}f(t_i)\cdot(x_{i+1} - x_i) \right| \leq \sum_{i}\left| f(t_i)\cdot(x_{i+1} - x_i) \right| \\ =\sum_{|f(t_i)|<\frac{\epsilon}{2}}\left| f(t_i)\cdot(x_{i+1} - x_i) \right| +\sum_{|f(t_i)|\geq\frac{\epsilon}{2}}\left| f(t_i)\cdot(x_{i+1} - x_i) \right| \\ \leq\sum_{|f(t_i)|<\frac{\epsilon}{2}}\frac{\epsilon}{2}\left| (x_{i+1} - x_i) \right| +\sum_{|f(t_i)|\geq\frac{\epsilon}{2}}\left| (x_{i+1} - x_i) \right| \\ \frac{\epsilon}{2} +\sum_{|f(t_i)|\geq\frac{\epsilon}{2}} \| P \|\leq \frac{\epsilon}{2}+M \| P\| $$

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    How are there only finitely many fractions meeting that condition when the rationals are dense? Maybe I am missing something in how you are defining that? – 2017-02-01
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    For example, consider the number of fractions of form $x/q$ that is reduced and lies between $[0,1]$, with denominator $q$. Then there are at most $q$ of them, namely $1/q, \ldots, q/q$ (of course, usually the numerator has common factor with $q$). – 2017-02-01
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    @user369580 The rationals are dense in $[0,1]$ but only two fractions have denominator 1. Only three fractions have denominator 2. and so on. This doesn't contradict the denseness, the denseness comes from the fact that the denominator can be arbitrarily large. – 2017-02-01
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Well the function is continous everywhere except for rational points. Thus the set of discontinuities is of (Lebesgue) measure $0$ so $f$ is Riemann integrable.

Now you can calculate the integral directly from the definition. Pick any partition $0=x_0

$$\sum_{i=1}^k f(t_i)\cdot(x_{i+1} - x_i)=\sum_{i=1}^k0\cdot(x_{i+1}-x_i)=0$$

Now you can go to the limit with this kind of partitions (remember that irrationals are dense) and since that limit has to be the Riemann integral (since the function is integrable) then the integral is $0$.

Note that values of $f(x)$ for rational $x$ doesn't matter at all. The integral will always be $0$ as long as the set of discontinuities of $f$ is of measure $0$ and $f=0$ elsewhere.

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    Unfortunately, we have not learned anything about measure/Lebesgue measure yet so can't use that to show the function is integrable. I had something like the rest of your proof to show that there exists a lower step function that sums to 0 based on the density of the irrationals (and that all lower step functions will be less than or equal to zero) but the upper step function is giving me problems. – 2017-02-01
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    @user369580 Oh man, doing it directly from the definition is so calculation intensive. I'll think about it later, but I can't promise anything. – 2017-02-01
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    Thanks! We have been working on a couple proofs involving showing the sequence of the difference of the upper sum and lower sums is less than or equal to epsilon but I can't get that to fit this one given the infinite number of partition points it would create no matter what (at least the way I have been thinking about it). – 2017-02-01