A function $f$ is defined on $[0,1]$ by $$f(x) = \begin{cases} \sin x, & x\ \text{is rational},\\ x, & x\ \text{is irrational} \end{cases}$$
Find $\underline{\int}_0^{\pi/2}f(x) \, dx, \overline{\int}_0^{\pi/2} f(x) \, dx$ and hence show that $f$ is not integrable on $[0, \pi/2]$.
It may be look like $f(x) = \sin x $, if $x$ is rational and $f(x) = x$, if $x$ is irrational.
But here, f is defined on $[0,1]$ and have to check integrability on $[0, \pi/2]$ where $[0,1]\subset [0, \pi/2]$.
Q1: Then partition would be taken on $[0,1]$ or $[0, \pi/2]$.
Q2: Also as f is not defined on $[1, \pi/2]$, is it realistic to check integrability on $[0, \pi/2]$.
I don't know. Please help.
Consider $f(x) - \sin(x)$. Is this integrable or not? Note that $x-\sin(x) >0$ except when $x=0$. It should be quite clear that $f(x) - \sin(x)$ isn't Riemann integrable. Think back to the proof as to why $\mathbb{1}_\mathbb{Q}$ isn't integrable: within any $\epsilon$-wide interval, the oscillation of the function is always $1$ no matter how small $\epsilon$ is.
Now for our function $f(x) -\sin(x)$: The oscillation over any $\epsilon$-wide interval will always be equal to $x+\epsilon - \sin(x)$ due to the density of rationals (infinitely many rationals in every interval), which does not grow small as $\epsilon$ grows small. Therefore, $f(x)-sin(x)$ isn't Riemann integrable.
The sum of two integrable functions is integrable. $\sin(x)$ is clearly integrable. This implies $f(x)$ is not integrable.
(This proof can be done easily without subtracting $\sin(x)$, but I chose to do this because it brings the problem closer to a classic example.)
As to your concern about $f$ being defined on $[0, 1]$ yet you're being asked to check integrability over $[0, \frac{\pi}{2}]$, it is possible that $f$ vanishes outside $[0, 1]$, in which case the proof above still makes sense. If $f$ truly is not defined everywhere on $\mathbb{R}$, and it is not explicitly stated that $f$ can go to zero wherever it is not defined, then you cannot integrate it over a region where it's not defined—what would that even mean?
That's a minor concern, however. It's also possible that they meant $f: [0, \frac{\pi}{2}] \mapsto [0, 1]$, which is perfectly true and then your problem is perfectly solvable.