# Let $f(x)=1$ if $x=0$ and $f(x)=0$ if $x>0$. Show that $f$ is integrable.

Let $$f(x)=1$$ if $$x=0$$ and $$f(x)=0$$ if $$x>0$$. Show that $$f$$ is Riemann integrable on $$[0,1]$$. I think for everyone this question is really basic, but I'm just training myself on proving the integrability of functions. Certainly, there are easiests ways to prove the statement, but I would like to prove it considering upper and lower Darboux sums. So, I would just like to know if my approach is correct, please x)

By definition, $$f$$ is integrable iff. $$\exists$$ a subdivision $$\sigma$$: $$\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon \ \forall \epsilon>0$$.

First, I remark that for all subdivisions, the lower Darboux sum is equal to $$0$$. So, we have to show that $$\overline{S}_{\sigma}(f)<\epsilon \ \forall \epsilon>0$$.

Then, consider $$0<\epsilon<2$$, the intervals $$[0,\epsilon/2]$$ and $$(\epsilon/2,1]$$ and two first terms of partition $$x_0=0,x_1=\epsilon/2...$$ We don't consider the rest of partition, as the upper Darboux sum is trivially $$0$$ whatever the partition on the interval $$(\epsilon/2,1]$$.

Thus, $$\overline{S}_{\sigma}(f)=\sum_{i=0}^{0}1\cdot (\epsilon/2-0)+0<\epsilon$$.

We conclude then that $$f$$ is Riemann integrable on $$[0,1]$$.

What you did is Ok. You could simplify by considering $$\epsilon$$ instead of $$\epsilon/2$$ which is not necessary here.
There are a few problems with your approach: when you write that you have to show that$$\overline{S_\sigma}<\varepsilon\ \forall\varepsilon>0,\tag1$$you don't tell was which partition $$\sigma$$ is, but then you acto as if it was a concrete partition. For instance, how do you know that $$\frac\varepsilon2$$ belongs to the partition. Besides, what you need to prove is that, for each $$\varepsilon>0$$, there is some partition $$\sigma$$ such that $$\overline{S_\sigma}<\varepsilon$$, which is not the same thing as $$(1)$$. And it is easy to do: just take $$\sigma=\left\{0,\frac\varepsilon2,1\right\}$$, if $$\frac\varepsilon2<1$$; otherwise, $$\sigma=\{0,1\}$$ will do.
• But in the beginning I wrote that: $\exists$ partition $\sigma$... And then I just explicited 2 first terms of the partition, as the other terms it could be watever we want as it will always be equal to $0$ (in the upper Darboux sum) – Daniil Feb 25 at 18:04
• Where did you write “$\exists$ partition $\sigma$”? – José Carlos Santos Feb 25 at 18:09
• As I wrote in my answer, you are not supposed to prove that there is an upper Darboux sum smaller than $\varepsilon$ $\forall\varepsilon>0$. It's the other way around: for all $\varepsilon>0$, you are supposed to prove that there is an upper Darboux sum smaller than $\varepsilon$. And if $\varepsilon>1$, I just take $\sigma=\{0,1\}$. Otherwise, I take $\left\{0,\frac\varepsilon2,1\right\}$. – José Carlos Santos Feb 25 at 18:56
• No. It's the partition that depends on $\varepsilon$. And this is the third time that I tell you this. – José Carlos Santos Feb 25 at 21:04