Riemann integrability of a function $f:[0,1]\to\mathbb{R}$

Question

I was trying to show whether or not the function:

$f: [0,1 ] \rightarrow \mathbb{R}$

$f(x)= \frac {1}{n}$ for $x = \frac {1}{n}$ $(n \in \mathbb{N})$

and

$f(x) = 1$ if the condition isn't satisfied

was integrable, but I'm having trouble figuring out the upper and lower sums. If anyone could point me in the right direction I would appreciate it.

Answer 1

You may also be aware that a function is Riemann integrable if it has only countably many discontinuities; here the only discontinuities occur at $\frac{1}{n}$ for $n\in\mathbb{N}$, thus it is certainly Riemann integrable and $f(x)=1$ a.e., thus $\int_0^1f(x)\:dx=1$.

Answer 2

Any upper sum must have the value $1$, since the maximum value of $f$ on any non-degenerate subinterval of $[0,1]$ is $1$. So, you need to show that there are lower sums that take values as close to $1$ as you wish.

Here is an, informal, outline of how to do this:

Note that the leftmost subinterval, $I$, in a partition contains all but finitely many numbers of the form $1/n$, $n\in \Bbb N$. When constructing your partition, take $I$ to have small width.

The finitely many points of the form $1/n$, $n\in \Bbb N$ that are not in $I$ can be "put into" subintervals so that the sum of the lengths of these subintervals is small (for each of these points, choose a subinterval containing it of small width).

In this manner, The contribution to the lower sum from all these subintervals can be made as small as you like (it's bounded by the sum of the lengths of the subintervals).

On the remaining subintervals, $f$ has the constant value $1$; so the contribution to the lower sum from these subintervals is the sum of their lengths. But the sum of their lengths is close to $1$.

Of course, to make this formal, you need to set $\epsilon>0$ and find a partition so that the corresponding lower sum $L$ satisfies $L\ge 1-\epsilon$.

I hope this helps.