Let q1,q2,... be a fixed enumeration of rationals in [0,1]. Define f(x) so $f(q_n)=a_n $ for all a and 0 otherwise. Prove that it is Riemann integrable if the sequence an approaches 0, and justify will it be integrable if the sequence doesn’t approach 0.
Here I would know how to prove if there is a finite no of rationals by using epsilon lengths of rectangale, but unfortunately it is infinite.
$0$ is always a possible Riemann sum no matter what the partition, so the only hope the function has of being integrable is if the integral is $0$.
Therefore you need to show for each $\varepsilon$ there is a $\delta$ such that every Riemann sum for a partition that is finer than $\delta$ is absolutely less than $\varepsilon$. A natural strategy is to budget some of the $\varepsilon$ to contributions from infinitely many small $a_n$, and the rest to intervals that contain finitely many large function values.
So since we're assuming $a_n\to 0$ we can find $N$ such that $|a_n| < \varepsilon/2$ for $n\ge N$. This bounds the contributions for "small" $a_n$s to $\varepsilon/2$. Can you show that there must be a $\delta$ such that the first $N$ $a_n$s together cannot contribute more than $\varepsilon/2$ to the Riemann sum?
In the case that $a_n\not\to 0$, you can't say anything in general about integrability. To wit, $$ f(x) = \begin{cases} 1 & \text{when }x=1/k\text{ for some }k\in\mathbb N_+ \\ 0 & \text{otherwise} \end{cases} $$ is Riemann integrable, but $$ f(x) = \begin{cases} 1 & \text{when }x=(2m+1)/2k\text{ for some }m,k\in\mathbb N_+ \\ 0 & \text{otherwise} \end{cases} $$ is not.