I need an example of this, and my thoughts were to try this:
let $\{ r_1, r_2, r_3, …, \}$ be an enumeration of all the rationals in $[0,1]$
define $g_n(x) = 1$ if $x=r_n$ or $0$ otherwise.
I think this works, the sequence is pointwise convergent to Drichlet's function
Best Answer
Yes, you seem to be correct (as far as Riemann integrability is concerned), though your definition isn't completely clear, I understand which functions you refer to.
Each $g_n$ is discontinuous at $r_j, 1\leq j \leq n$, since there exists a sequence of irrationals $i_m$ converging to each $r_j$, but $g_n(i_m) =1 \not \to g_n(r_j) = 0$. However, $g_n$ is continuous at all other points, since given any such point, you can find a neighbourhood of it that avoids $r_j, 1 \leq j \leq n$. So it is locally constant, hence continuous.
$g_n$ is integrable,since you can consider the partition as follows : take points $a_j,b_j$ around each $r_j$, for $1 \leq j \leq n$, such that $[a_j,b_j]$ are disjoint. Now, use these points to create a partition, and then bring $a_j,b_j$ closer to $r_j$, the upper Riemann sum will then converge to zero, while the lower one will always be zero. That means the Riemann integral is zero.
However, the pointwise limit $g$, which is the characteristic function of rationals in $[0,1]$, isn't Riemann integrable, for this reason: suppose you take a partition of $[0,1]$. The upper Riemann sums involve supremums to be taken over each interval of the partition, while the lower Riemann sums involve infimums to be taken over the interval. However, over every interval, $g$ has supremum $1$ and infimum $0$. Hence, regardless of which partition you take, the upper Riemann sum will always be $1$ and the lower Riemann sum will be $0$, so the Riemann integral can't exist.