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(Context: I have been using this resource from UC Davis to help me out with Riemann integrability.)

So, I more or less get how Riemann integrability works when we specify $x$ over an interval, say $x>0$. But what if $x$ is a function of $n$? I came across this problem:

Prove that $f(x)$ is Riemann Integrable over $[0,1]$ and that $\int_{0}^{1}f(x) = 0$ given $$f(x)=\left\{\begin{matrix} 1 & x=\frac{1}{n} \\ 0 & \text{otherwise} \end{matrix}\right.$$

I'm not sure how to partition this (because of the $n$) to show this. Anyone have any insight?

quid
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kuseno
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  • What is $n$ defined as? Any integer? – Element118 Nov 29 '15 at 03:58
  • $n$ here is not meant to be understood as a variable with a definite value, it's just an arbitrary natural number. Thus $f(1)=1,f(1/2)=1,f(1/3)=1,\dots$ but all other values get mapped to zero. – Ian Nov 29 '15 at 04:00
  • Possible duplicate of http://math.stackexchange.com/questions/1158070/proving-riemann-integrability-of-a-function-with-countably-many-discontinuities – Element118 Nov 29 '15 at 04:03
  • It's...not quite a full duplicate. The question there is more general, and the proof I gave there omits a step that is probably important for this question. – Ian Nov 29 '15 at 04:07

1 Answers1

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The idea here is that all but finitely many of the points in $\{ 1/n : n \in \mathbb{N} \}$, where the discontinuities in $f$ are, are very close to $0$. So you can work this way. Let $\varepsilon >0$. Choose $N \in \mathbb{N}$ such that $1/(N+1)<\varepsilon/2$. Then make the first two points of the partition be $0$ and $1/(N+1)$.

Now you just have $N$ discontinuities to work with. Make your partition so that they are each surrounded by intervals of length at most $\varepsilon/2N$. Now you can show that the lower sum of this partition is $0$ while the upper sum is some positive number which is less than $\varepsilon$.

Ian
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  • (Forgive me if this seems dumb, I'm still learning the topic.)

    So, what exactly is this $1/(N+1)$? In my readings, I've only seen epsilon notation brought up when "minimizing" the difference between t he upper and lower sum. Since you yourself said that we will eventually show how the lower sum of the partition is 0, what is it?

     – kuseno Nov 29 '15 at 04:19
  • Please ignore that previous comment. I got it! Thanks for the help. – kuseno Nov 29 '15 at 04:32