Consider the function h defined by h(x) := x+1 for x an element of [0,1] rational, and h(x) := 0 for x an element of [0,1] irrational. Show that h is not Riemann integrable.
The hint in the back of the book says let Pdot_n be the partition of [0,1] into n equal subintervals with t_1= 1/n and Qdot_n be the same subintervals tagged by irrational points.
So it seemed obvious that |S(f;Pdot)-S(f;Qdot| >= 1 and we're done. However I got to thinking, how is this valid? As in, how can the tag t_1=1/n possibly tag all n subintervals?
for example n=8, the tags are 1,1/2,1/3,…,1/8. Then wouldn't some subintervals have no tags, while others have multiple? Any help is appreciated.
Best Answer
With $n = 8$ the tags would be $1/8$, $2/8$, $3/8$, and so on all the way up to $8/8$.