Hi I am confused about the following question. I am trying to understand the conditions for when the rationals intersect the closed interval. I know that $\mathbb{Q}$ is a subset of $\mathbb{R}$ but don't fully grasp how the function is set up.
Let $f \colon[a,b] \to \mathbb{R}$ be defined by $$ f(x) = \begin{cases} 1,& x \in \mathbb{Q} \cap [a,b] \\ -1,& x \in \mathbb{Q}^c \cap [a,b]\end{cases}$$ Determine whether $f$ is Riemann integrable on $[a,b]$.
The interval $[a,b]$ is the set of all real numbers which are greater than or equal ot $a$ and less than or equal to $b$
Most of these real numbers are irrational and the other ones are rational.
One very important fact is that both rationals and irrationals are dense in the interval $[a,b]$ meaning that no matter how small a non-empty open interval you choose inside $[a,b]$ there are both rationals and irrationals in it.
The way your function is defined for every closed interval you get the least upper bound of $1$ and the greatest lower bound of $-1$
You can take it from here.