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Let

$\displaystyle f(x)=\begin{cases} \frac{1}{n}, & \text{if }x=\frac{m}{n},m,n\in\mathbb{N}\text{ and m and n has no common divisor} \\ 0, & \text{otherwise} \end{cases}$

Show $f\in R[0,1]$.

How to find a partition $P_\epsilon$ such that $U(P_\epsilon,f)-L(P_\epsilon,f)<\epsilon$ for all $\epsilon>0$?

$L(P_\epsilon,f)=0$, so only need to consider $U(P_\epsilon,f)$.

Gamma Function
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user130916
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  • $f$ is continuous at irrationals and discontinuous at rationals. You know that ? – OBDA Mar 20 '14 at 08:30
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    Some related questions: http://math.stackexchange.com/questions/106488/upper-riemann-sum-of-the-thomae-function, http://math.stackexchange.com/questions/419042/help-on-showing-a-function-is-riemann-integrable or http://math.stackexchange.com/questions/17275/is-thomaes-function-riemann-integrable (Probably you can find a few more questions about integrability of the same function.) – Martin Sleziak Mar 20 '14 at 08:47

3 Answers3

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Hint. A bounded function $f:[a,b]\to\mathbb R$ is Riemann integrable iff for every $\varepsilon>0$, here exists a partition $P$ of $[a,b]$, such that $$ U(f,P)-L(f,P)<\varepsilon. $$ In the case of this function $L(f,P)=0$, for all $P$ of $[0,1]$. Given an $\varepsilon>0$, we simply need to find a partition $P$ of $[0,1]$ for which $U(f,P)<\varepsilon$.

Clearly, there are only finitely many points $\{x_k\}$ in $[0,1]$, where $$ f(x_k)>\frac{\varepsilon}{2}. $$ Say $N=N(\varepsilon)$ such point. Simply take a partition, where all these $x_k$'s lie in very small subintervals $[t_{k_1},t_{k_2}]$, i.e., $$ t_{k_1}=x_k-\delta<x_k<x_k+\delta=t_{k_2}, $$ with $\delta<\varepsilon/2N$.

Then check that for this partition $U(f,P)<\varepsilon$.

Yiorgos S. Smyrlis
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You can avoid using any partition. Show that $f$ is continuous at the irrationals and discontinuous at all the rationals. Then $f$ is a function with countable number of discontinuities and hence is Riemann Integrable see this Proof that a function with a countable set of discontinuities is Riemann integrable without the notion of measure.

Community
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wanderer
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  • 'Show that f is continuous at the irrationals and discontinuous at all the rationals.' This can be done using simple counting arguments – user136920 Apr 17 '14 at 23:41
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Take the intervals between the points in $\frac{1}{n!}ℤ ∩ [0..1]$. Any rational between $\frac{k}{n!}$ and $\frac{k+1}{n!}$ for any $k ∈ ℤ$ must have denominator greater than $n$ (you can expand any fraction with denominater $n$ or less to a fraction with denominator $n!$, so any such fraction is already in $\frac{1}{n!}ℤ$). This means that in each open interval between points in $\frac{1}{n!}ℤ∩[0..1]$, $\sup f$ is less than $1/n$.

k.stm
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