Determine if function is Riemann integrable on interval

Question

I am supposed to determine if function given by:

$$\varphi \left ( x \right )=\left\{\begin{matrix} \ 1 , \left ( \exists m,n\in\mathbb{N} \right )x=\frac{m}{2^{n}}\\ \\0, else \end{matrix}\right.$$

is Riemann integrable on interval $\left [ 0,1 \right ]$.

What I did so far:

I found out partition, $P_{n}=\left ( \frac{1}{n} \right )$, and I know that to any $\varepsilon > 0$ we choose $n\in\mathbb{N}$, so that $\frac{1}{n}< \frac{\epsilon }{2}$. Then $U\left ( Pn,\varphi \right )=\frac{1}{n}\varphi \left ( t_{i} \right )$, where $t_{i}=\frac{m}{2^{n}}$,

but I do not know how to continue, or what I have to do with the second part of the function.

Can anyone help me?

Answer 1

The lower sums $L(\phi,P)$ are clearly zero.

Also note that the set $\{\frac{m}{2^n}:m\in \Bbb{Z},n \in \Bbb{N}\}$ is dense on the reals and thus on $[0,1]$(and on $[0,1]$ you need only that $m \in \Bbb{N}$)

So if you take a partition $P$ you will always have $U(\phi,P) \geq 1$ because the supremum of the function on each subinterval will be $1$.

Thus $U(\phi,P)-L(\phi,P) \geq 1$ so the function cannot be Riemman integrable on $[0,1]$

Answer 2

For a somewhat different approach, if $p\in [0,1]$ is irrational, then the density of the rationals in $[0,1]$ implies that $f$ is not continuous at $p.$ And since the irrational numbers in $[0,1]$ have positive measure, $f$ is not Riemann integrable.