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

Question

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 $[0,1]$.

Answer 1

What you did is Ok. You could simplify by considering $\epsilon$ instead of $\epsilon/2$ which is not necessary here.

Answer 2

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.