Let $g:[a,b]\to\mathbb R$ be defined by
$g(x) =
\begin{cases}
3, \text{ if x is rational}\\
2, \text{ if x is irrational}
\end{cases}$
Show that $g(x)$ is not integrable.
—
So far I have said that for any dissection $D=(x_0,…,x_n)$ we have
$\inf f[x_{i-1}, x_i]=2$, and
$\sup f[x_{i-1}, x_i]=3$
as we know that every interval will contain both rationals and irrationals.
From here, if I had a defined interval eg. [0,1] I would simply find the upper and lower sums but I am not sure what to do in the case of our arbitrary interval [a,b] as ultimately I think we need to show
$\sup${$\underline{s}(D):D\in D$} < $\inf${$\overline{s}(D):D\in D$}
Where would I go from here?
Best Answer
The information that you have given regarding $\sup$ and $\inf$ of $f$ over an interval suffices.
Indeed, a function is Riemann integrable, if its upper Riemann sum coincides with its lower Riemann sum.
However, let us take any interval $[a,b]$ over which we are checking Riemann integrability, and any partition $a = t_0 < t_1 < ... < t_n = b$. Then, the lower Riemann sum is $$\sum_{i=0}^{n-1} (t_{i+1} - t_i)[\inf_{[t_i,t_{i+1}]} f] = 2 \sum_{i=0}^{n-1} (t_{i+1} - t_i) = 2(b-a)$$
Regardless of the partition.
Similarly, the upper Riemann sum is $3(b-a)$ regardless of the partition. Even if I make the partitions finer and finer, the lower and upper sums don't come closer to each other because they are fixed non-equal numbers.
Consequently, the function is not R.I. over any $[a,b]$, $b > a$.