A Question about My Riemann Integral Notes

[The Student]

My notes say, if f(x) = {1 if x ∈ rationals, 0 if x ∈ irrationals, then the Riemann integral does not exist on [a,b] because L(P,f) = 0 and U(P,f) = b - a for every partition P of [a,b].

First of all, wouldn't L(P,f) = U(P,f) since they would have the same "height"/function of 1 or 0 to multiply each partition by? Secondly, why would U(P,f) = b - a? If there is no "width" for when f(x) = 1, how is U(P,f) anything but 0?

 

[pka]

My notes say, if f(x) = {1 if x ∈ rationals, 0 if x ∈ irrationals, then the Riemann integral does not exist on [a,b] because L(P,f) = 0 and U(P,f) = b - a for every partition P of [a,b].

First of all, wouldn't L(P,f) = U(P,f) since they would have the same "height"/function of 1 or 0 to multiply each partition by? Secondly, why would U(P,f) = b - a? If there is no "width" for when f(x) = 1, how is U(P,f) anything but 0?

Any upper sum is \(\displaystyle b-a\). In each sub-interval the maximum of the function is one and that max is multiplied by the length of the interval. Then added all together to equal \(\displaystyle b-a\).

Any lower sum is \(\displaystyle 0\). In each sub-interval the minimum of the function is zero and that min is multiplied by the length of the interval. Then added all together to equal \(\displaystyle 0\).

 

[The Student]

Any upper sum is \(\displaystyle b-a\). In each sub-interval the maximum of the function is one and that max is multiplied by the length of the interval. Then added all together to equal \(\displaystyle b-a\).

Any lower sum is \(\displaystyle 0\). In each sub-interval the minimum of the function is zero and that min is multiplied by the length of the interval. Then added all together to equal \(\displaystyle 0\).

What if the partition is rational and irrational sub-intervals? Or is that even possible?

 

[pka]

What if the partition is rational and irrational sub-intervals? Or is that even possible?

Do you have any idea how upper and lower sums are formed?

 

[The Student]

Do you have any idea how upper and lower sums are formed?

Yes, I have a pretty good but basic understanding of how to form sums. But the question that I posted is right before the notes for learning Reimann sums.

You had actually helped me understand the question and the answer. But a little while later, I wondered what would happen if each partition was a rational number, which would seem to make every sub-interval contain irrational numbers. Then this would seem to make every area equal zero for upper or lower sums. Does that make any sense?

 

[pka]

I wondered what would happen if each partition was a rational number, which would seem to make every sub-interval contain irrational numbers. Then this would seem to make every area equal zero for upper or lower sums. Does that make any sense?

No it does not.If \(\displaystyle [p,q]\) is an interval in the partition then whether \(\displaystyle p \text{ or }q\) is rational or irrational has nothing to do with the sum. Only the evaluation point matters. In upper and lower sums we take the greatest and least value of the function any where in the interval. In this case it is 1 for an upper sum and 0 for a lower sum, because in any \(\displaystyle [p,q]\) there are both rational and irrational numbers.

 

[The Student]

No it does not.If \(\displaystyle [p,q]\) is an interval in the partition then whether \(\displaystyle p \text{ or }q\) is rational or irrational has nothing to do with the sum. Only the evaluation point matters. In upper and lower sums we take the greatest and least value of the function any where in the interval. In this case it is 1 for an upper sum and 0 for a lower sum, because in any \(\displaystyle [p,q]\) there are both rational and irrational numbers.

Oh, I was assuming that each subinterval could have [p,q] where p = q for each rational number and each irrational number between a and b. Then it would seem that the lower sums would always equal the upper sums. But then of course x(i) - x(i-1) wouldn't exist because of the midpoint lemma - I think.

 

[pka]

Oh, I was assuming that each subinterval could have [p,q] where p = q for each rational number and each irrational number between a and b. Then it would seem that the lower sums would always equal the upper sums. But then of course x(i) - x(i-1) wouldn't exist because of the midpoint lemma - I think.

You must realize that there several different ways to define the so-called Riemann Integral. They give the same results even if the definitions look quite different.

 

[The Student]

You must realize that there several different ways to define the so-called Riemann Integral. They give the same results even if the definitions look quite different.

Thank-you, the fog is lifting.