I am trying to prove that the function on $[0,1]$ defined by $$f(x)=\begin{cases} 1 & \text{if $x=\frac{1}{n}$ , $n \in \mathbb{N}$}\\ 0 &\text{else} \end{cases}$$

Is Riemann integrable on this interval. I would like to use squeeze theorem but I am having trouble choosing an appropriate upper bound.

Let a lower bound, integrable function be $\alpha(x)=0$

I need an integrable function, $\omega$, dependent on $\epsilon$ such that

$\alpha(x) \le f(x) \le \omega(x)$ all $x \in [0,1]$

with $\int_{0}^{1} \omega-\alpha \lt \epsilon$ for all $\epsilon \gt 0$

But I dont know I can use the definition of $f$ to choose a suitable $\omega$

I could also use some sort of arguments involving changing things only at finite points, the Archemdian property, etc

Overall, I am just really confused. I have been working on it for hours but I cant make any progress. I really need some advice

Can anyone offer some help?



Let's choose a sequence of functions $f_N$, which is $1$ in little $\delta$ intervals around those $\frac{1}{n}$ with $n \leq N$, and is $1$ for $x \in [0, \frac{1}{N}]$. Then you have exactly $N$ little $\delta$ balls and a little remainder.

Note that the integral of $f_N$ is always at least as large as the integral of $f$, since it's at least as large everywhere. We can over-estimate the integral of $f_N$ by $N \cdot (2\delta) + \frac{1}{N} = 2N\delta + \frac{1}{N}$.

Now, given an $\epsilon$, you can show the integral of $f$ is less than $\epsilon$ by comparing with $f_N$ with $N > \frac{2}{\epsilon}$ and $\delta < \frac{\epsilon}{4N}$. The first inequality guarantees that $\frac{1}{N} < \frac{\epsilon}{2}$, and the second guarantees that $2N\delta < \frac{\epsilon}{2}$. Together, along with our naive upper bound for the integral of $f_N$ from above, this guarantees that the integral of $f$ is at most $\epsilon$.

As this can be done for any $\epsilon > 0$, and the integral of $f$ is bounded below by $0$, you can conclude that the overall integral of $f$ is $0$. $\diamondsuit$

  • $\begingroup$ I used a diffirent method but similar , I will post it later but I will accept this as it is a good method that could be helpful in the future $\endgroup$
    – Quality
    Feb 2 '16 at 6:03

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