I saw this example on internet as I was studying for my exam. I am a little bit confused as why the supremum is 1 and infimum is 0? Also at the end it has a conclusion that f is not integrable in [2,3], is it because L and U have different answers? Please help me to clarify this. Thank you.
We'll see this in words, and then translate it into mathematical language. Then the supremum/infimum thing will fall out.
We have a dissection of $[2,3]$.This will give the intervals $(x_{i-1},x_i)$.
$\sup_{(x_{i-1},x_i)} f(x)$ is the largest value that $f$ takes in this interval. However, this interval contains at least one irrational, call that irrational $r$ , then $f(r) =1$, where $r \in (x_{i-1},x_i)$. Now, the maximum value $f$ takes anywhere is $1$ (from definition), so the supremum of $f$ over every interval is $1$, since in every interval, $f$ takes the value $1$ at least once.
Similarly, $\inf_{(x_{i-1},x_i)} f(x)$ is the smallest value that $f$ takes in this interval. However, this interval contains at least one rational, call that rational $s$ , then $f(s) =0$, where $s \in (x_{i-1},x_i)$. Now, the minimum value $f$ takes anywhere is $0$ (from definition), so the infimum of $f$ over every interval is $0$, since in every interval, $f$ takes the value $0$ at least once.
Now, we have from the definition of $\mathcal U(f,D)$ (for all dissections $D$):
$$ \mathcal U (f,D) = \sum_{i=0}^{n-1} (x_{i} - x_{i-1}) \left(\sup_{(x_{i-1},x_i)} f(x)\right) = \sum_{i=0}^{n-1} (x_{i} - x_{i-1}) \times 1 = 3-2 = 1 $$
Now, we have from the definition of $\mathcal L(f,D)$ (once again, for all dissections $D$): $$ \mathcal L(f,D) = \sum_{i=0}^{n-1} (x_{i} - x_{i-1}) \left(\inf_{(x_{i-1},x_i)} f(x)\right) = \inf_{i=0}^{n-1} (x_{i} - x_{i-1}) \times 0 = 0 $$
This is essentially what is happening in the proof.