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Question: Let $f:[0,1] \to \Bbb R$ be a function s.t. $ \begin{cases} 1 & x=\frac 1n \\ 0 & \text{otherwise}\end{cases}$

prove that $f$ is integrable and that $\int _0^1 f(x)dx=0$.

What we did:

  • Let $\varepsilon>0$ and we shall pick a $\delta>0$ s.t. for each partitioning $\Pi$ s.t $\lambda(\Pi)<\delta$ , the oscillation $\omega(f,\Pi)<\varepsilon$.

  • We pick $\delta=\frac \varepsilon2$. For each partitioning $\Pi$ it's partioning the function to inervals s.t $0=x_0<x_1<...<x_m=1$

  • We mark $M_i=\max(f)_{x_i,x_{i+1}}$, $m_i=\min(f)_{x_i,x_{i+1}}$ Now $\omega(f,\Pi)\le \sum_{i=0}^m(M_i-m_i)(x_{i+1}-x_i) =$ $ (M_0-m_0)(x_1-x_0)+\sum_{i=1}^m(M_i-m_i)(x_{i+1}-x_i)\le$ $1\delta +\frac \varepsilon2=\varepsilon$ (the first part is true because each interval is smaller than $\delta$ by definition and the second part is true because this function in the interval $[x_1,1]$ identifies with $f=0 $ for all but a finite number of points and therefore it's oscillation is smaller than $\frac \varepsilon2$.

Therefore from the definition of Riemann's integral $I=0$.

Is this proof correct??

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  • $\begingroup$ What is your definition of integrabilty? For each $\varepsilon >0$ there exists a partition $\Pi$ such that $U(f,\Pi)-L(f,\Pi)<\varepsilon$? $\endgroup$
    – Pedro
    Commented Nov 2, 2013 at 14:22
  • $\begingroup$ Yes, we use the definition: For each $\varepsilon>0$ $\exists \delta>0$ that for any partitioning $\Pi$ that $\lambda(\Pi)<\delta$ , the oscillation $\omega(f,\Pi)<\varepsilon$. and that $\omega(f,\Pi) = \sum_{i=1}^n(M_i-m_i)(x_{i+1}-x_i)$. $\endgroup$
    – Splash
    Commented Nov 2, 2013 at 14:53

2 Answers 2

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A function $f:\ [a,b]\to{\mathbb R}$ is Riemann integrable if it passes the following simple test: For any given $\epsilon>0$ we can find a partition $\Pi$ of $[a,b]$ such that $$\omega(f,\Pi)<\epsilon\ .\tag{1}$$ Your integrability condition seems to be stronger: You require the existence of a $\delta>0$ such that $(1)$ holds for all partitions $\Pi$ with a mesh $<\delta$. But one can show that the simple test quoted at the beginning already guarantees the existence of such a $\delta$.

Nevertheless I shall show that your $f$ is integrable according to your definition.

Let a positive $\epsilon\ll 1$ be given. There is a positive integer $N$ with ${\displaystyle{1\over N}<{\epsilon\over4}}$. I claim that $$\delta:={\epsilon\over 4N}$$ suffices.

Proof. Assume $\Pi$ is an arbitrary partition with mesh $<\delta$.

All points ${1\over n}$ with $n>N$ lie in the interior of the interval $\bigl[0,{1\over N}\bigr]$. This interval is covered by Intervals of the partition having a total length $$\ell<{1\over N}+\delta<{\epsilon\over4}+{\epsilon\over4N}\leq{\epsilon\over2}\ .$$

Each point ${1\over n}$ with $1\leq n\leq N$ belongs to at most two intervals of the partition. These bad intervals have a total length $$\ell'< 2N\delta={\epsilon\over2}\ .$$

It follows that the total length $\ell+\ell'$ of the bad intervals is $<\epsilon$. For these intervals we have $M_i-m_i\leq1$; for all other intervals of the partition we have $M_i-m_i=0$. Putting it all together we see that $(1)$ holds for $\Pi$.

Considering your own work, in the last bulleted part you didn't take care of the discontinuities away from $0$ individually, but argued that they can be dealt with "somehow". At any rate the partition you worked with has $M_k-m_k=1$ for many intervals, and it is unclear why their total contribution should be less than ${\epsilon\over2}$.

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  • $\begingroup$ One of the best answers I got on this site. Thanks very much. Very detailed and accurate. $\endgroup$
    – jreing
    Commented Nov 5, 2013 at 19:14
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Note that, a function is Riemann integrable if

i) it is bounded,

2) it has a countable number of discontiuity ( or it is discontinuous on a set of measure zero).

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    $\begingroup$ We didn't learn about a set of measure zero. We can only use the fact the finite number of discontinuity does not change the integral value. $1/n$ is not finite. $\endgroup$
    – Splash
    Commented Nov 2, 2013 at 14:31
  • $\begingroup$ @Splash: So, that's why I wtote the two possibilities $\endgroup$ Commented Nov 2, 2013 at 19:38
  • $\begingroup$ I was looking again but all we taught was that if a function is Riemann Inegrable -> it is bounded, but not vice versa. Tho I found something else: if $f$ is bounded and it is Riemann Inegrable for every [a,x] while a<x<b so it's Riemann Inegrable on [a,b]. Is that what you meant? $\endgroup$
    – Splash
    Commented Nov 3, 2013 at 19:25

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